{"id":9404,"date":"2021-02-16T16:00:12","date_gmt":"2021-02-16T10:30:12","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=9404"},"modified":"2021-02-16T16:21:04","modified_gmt":"2021-02-16T10:51:04","slug":"samacheer-kalvi-11th-chemistry-solutions-chapter-1","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-11th-chemistry-solutions-chapter-1\/","title":{"rendered":"Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations"},"content":{"rendered":"

The Samacheer Kalvi Class 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations Questions and Answers prevailing here are designed by the academic subject expertise in accordance with the state board prescribed syllabus and books. Aspirants who require help in preparing the Tamilnadu State Board Class 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations exercise questions can rely on the Samacheer Kalvi 11th Chemistry Solutions<\/a> pdf for Chapter 1 Basic Concepts of Chemistry and Chemical Calculations given in this article.<\/p>\n

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Samacheer Kalvi 11th Chemistry Chapter 1 Basic Concepts of Chemistry and Chemical Calculations Textual Evaluation Solved<\/h3>\n

I. Choose the Best Answer<\/span><\/p>\n

Basic Concepts Of Chemistry And Chemical Calculations Question 1.<\/strong>
\n40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature is ………..
\n(a) 40 ml CO2 <\/sub>gas
\n(b) 40 ml CO2 <\/sub>gas and 80 ml H2<\/sub>O gas
\n(c) 60 ml CO2 <\/sub>gas and 60 ml H2<\/sub>O gas
\n(d) 120 ml CO2<\/sub> gas
\nAnswer:
\n(a) 40 ml CO2 <\/sub>gas
\nSolution:
\nCH4(g)<\/sub>\u00a0 + 2O2(g)<\/sub>\u00a0 CO2(g)<\/sub>\u00a0+ 2 H2<\/sub>O(l)<\/sub><\/p>\n\n\n\n\n\n\n\n
Content<\/strong><\/td>\nCH4<\/sub><\/strong><\/td>\nO2<\/sub><\/strong><\/td>\nCO2<\/sub><\/strong><\/td>\n<\/tr>\n
Stoichiometric coefficient<\/td>\n1<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
Volume of reactants allowed to react<\/td>\n40 mL<\/td>\n80 mL<\/td>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 –<\/td>\n<\/tr>\n
Volume of reactant reacted and product formed<\/td>\n40 mL<\/td>\n80 mL<\/td>\n40 mL<\/td>\n<\/tr>\n
Volume of gas after cooling to the room temperature<\/td>\n–<\/td>\n–<\/td>\n–<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct<\/p>\n

Samacheer Kalvi 11th Chemistry Chapter 1 Solutions Question 2.<\/strong>
\nAn element X has the following isotopic composition 200<\/sup>X = 90 %, 199<\/sup>X = 8 % and 202<\/sup>X = 2 %. The weighted average atomic mass of the element X is closest to …………
\n(a) 201 u
\n(b) 202 u
\n(c) 199 u
\n(d) 200 u
\nAnswer:
\n(d) 200 u
\n= \\(\\frac {(200 \u00d7 90) + (199 \u00d7 8) + (202 \u00d7 2) }{100}\\) = 199.96 = 200 u<\/p>\n

Samacheer Kalvi Guru 11th Chemistry Question 3.<\/strong>
\nAssertion:
\nTwo mole of glucose contains 12.044 \u00d7 1023<\/sup> molecules of glucose.<\/p>\n

Reason:
\nTotal number of entities present in one mole of any substance is equal to 6.02 \u00d7 1022<\/sup>
\n(a) both assertion and reason are true and the reason is the correct explanation of assertion
\n(b) both assertion and reason are true but reason is not the correct explanation of assertion
\n(c) assertion is true but reason is false
\n(d) both assertion and reason are false
\nAnswer:
\n(c) assertion is true but reason is false<\/p>\n

Correct reason:
\nTotal number of entities present in one mole of any substance is equal to 6.022 x 1023<\/sup><\/p>\n

Basic Concepts Of Chemistry And Chemical Calculations Pdf Question 4.<\/strong>
\nCarbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
\n(a) Carbon
\n(b) Oxygen
\n(c) Both carbon and oxygen
\n(d) Neither carbon nor oxygen
\nAnswer:
\n(b) Oxygen<\/p>\n

Reaction 1:
\n2 C + O2<\/sub> \u2192 2 CO2<\/sub>
\n2 \u00d7 12 g carbon combines with 32 g of oxygen.
\nHence, Equivalent mass of carbon = \\(\\frac {2 \u00d7 12}{32}\\) \u00d7 8 = 6<\/p>\n

Reaction 2:
\nC + O2<\/sub> \u2192 2 CO2<\/sub>
\n12 g carbon combines with 32 g of oxygen.
\nHence, Equivalent mass of carbon = \\(\\frac {12}{32}\\) \u00d7 8 = 3<\/p>\n

Basic Concepts Of Chemistry And Chemical Calculations Book Back Answers Question 5.<\/strong>
\nThe equivalent mass of a trivalent metal element is 9 g eq-1<\/sup> the molar mass of its anhydrous oxide is ………..
\n(a) 102 g
\n(b) 27 g
\n(c) 270 g
\n(d) 78 g
\nAnswer:
\n(a) 102 g
\nLet the trivalent metal be M3+<\/sup>
\nEquivalent mass = mass of the metal \/ valance factor
\n9g eq-1<\/sup> = mass of the metal \/ 3 eq
\nMass of the metal = 27 g
\nOxide formed M2<\/sub>O3<\/sub>
\nMass of the oxide = (2 \u00d7 27) + (3 \u00d7 16) = 102 g<\/p>\n

11th Chemistry Evaluate Yourself Answers Chapter 1 Question 6.<\/strong>
\nThe number of water molecules in a drop of water weighing 0.018 g is ………….
\n(a) 6.022 \u00d7 1026<\/sup>
\n(b) 6.022 \u00d7 1023<\/sup>
\n(c) 6.022 \u00d7 1020<\/sup>
\n(d) 99 \u00d7 1022<\/sup>
\nAnswer:
\n(c) 6.022 \u00d7 1020<\/sup>
\nWeight of the water drop = 0.0 18 g
\nNo. of moles of water in the drop = Mass of water \/ molar mass = 0.018\/18 = 10-3<\/sup> mole
\nNo of water molecules present \u00a1n I mole of water = 6.022 \u00d7 1023<\/sup>
\n\u201cNo. water molecules in one drop of water (10 mole) = 6.022 \u00d7 1023<\/sup> \u00d7 10-3<\/sup>
\n= 6.022 \u00d7 1020<\/sup><\/p>\n

11th Chemistry Evaluate Yourself Answers Question 7.<\/strong>
\n1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is ………..
\n(a) 0 %
\n(b) 4.4 %
\n(c) 16 %
\n(d) 8.4 %
\nAnswer:
\n(c) 16%
\nMg CO3<\/sub> \u2192 MgO + CO2<\/sub>\u2191
\nMg CO3<\/sub> : (1 \u00d7 24) + (1 \u00d7 12) + (3 \u00d7 16) = 84 g
\nCO2<\/sub> : (1 \u00d7 12) + (2 \u00d7 16) 44g
\n100% pure 84 g MgCO3<\/sub> on heating gives 44 g CO2<\/sub>
\nGiven that I g of MgCO3<\/sub> on heating gives 0.44 g CO2<\/sub>
\nTherefore, 84 g MgCO3<\/sub> sample on heating gives 36.96 g CO2<\/sub> = 100%
\nPercentage of purity of the sample = \\(\\frac{100 \\%}{44 \\mathrm{g} \\mathrm{CO}_{2}}\\) \u00d7 36.96 g CO2<\/sub> = 84%
\nPercentage of impurity = 16%<\/p>\n

11th Chemistry Chapter 1 In Tamil Question 8.<\/strong>
\nWhen 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is –
\n(a) 3
\n(b) 0.75
\n(c) 0.075
\n(a) 0.3
\nAnswer:
\n(c) 0.075
\n\"Basic
\nThe amount of CO2<\/sub> released, x = 3.3 g
\nNo. of moles of CO2<\/sub> released = 3.3 \/ 44 = 0.075 mol<\/p>\n

Class 11 Chemistry Solutions Samacheer Kalvi Question 9.<\/strong>
\nWhen 22.4 liters of H2<\/sub> (g) is mixed with 11.2 liters of Cl2<\/sub> (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to ………..
\n(a) 2 moles of HCl (g)
\n(b) 0.5 moles of HCl (g)
\n(c) 1.5 moles of HCl (g)
\n(d) 1 moles of HCl (g)
\nAnswer:
\n(d) 1 moles of HCl (g)
\nSolution:
\nH2<\/sub>(g) + Cl2<\/sub>(g) \u2192 2 HCl (g)<\/p>\n\n\n\n\n\n\n\n
Content<\/strong><\/td>\nH2<\/sub>(g)<\/strong><\/td>\ncl2<\/sub>(g)<\/strong><\/td>\nHCl (g)<\/strong><\/td>\n<\/tr>\n
Stoichiometric coefficient<\/td>\n1<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
No. of moles of reactants allowed to react at 273 K and 1 atm pressure<\/td>\n22.4 L (1 mol)<\/td>\n11.2 L (0.5 mol)<\/td>\n\u2014<\/td>\n<\/tr>\n
No. of moles of reactant reacted and product formed<\/td>\n0.5<\/td>\n0.5<\/td>\n1<\/td>\n<\/tr>\n
A mount of HCl formed 1 mol<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

11 Samacheer Chemistry Solutions Question 10.<\/strong>
\nHot concentrated sulfuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior?
\n(a) Cu + 2H2<\/sub>SO4<\/sub> \u2192 CuSO4<\/sub> + SO2<\/sub> + 2H2<\/sub>O
\n(b) C + 2H2<\/sub>SO4<\/sub> \u2192 4 CO2<\/sub> + 2SO2<\/sub> + 2H2<\/sub>O
\n(c) BaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub>+ 2HCl
\n(d) none of the above
\nAnswer:
\n(c) BaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub>+ 2HCl
\n\"Samacheer<\/p>\n

11th Chemistry Chapter 1 Book Back Answers Question 11.<\/strong>
\nChoose the disproportional reaction among the following redox reactions.
\n(a) 3Mg (s) + N2<\/sub>(g) \u2192 Mg2<\/sub>N2<\/sub> (s)
\n(b) P4<\/sub> (s) + 3NaOH + 3H2<\/sub>O \u2192 PH3<\/sub>(g) + 3NaH2<\/sub>PO2<\/sub> (aq)
\n(c) Cl2<\/sub> (g) + 2Kl (aq) \u2192 2KC1 (aq) + I2<\/sub>
\n(d) Cr2<\/sub>O3<\/sub> (s) + 2Al (s) \u2192 A2<\/sub>O3<\/sub> (s) + 2Cr (s)
\nAnswer:
\n(b) P4<\/sub> (s) + 3NaOH + 3H2<\/sub>O \u2192 PH3<\/sub>(g) + 3NaH2<\/sub>PO2<\/sub> (aq)
\n\"Samacheer<\/p>\n

11th Chemistry Samacheer Kalvi Question 12.<\/strong>
\nThe equivalent mass of potassium permanganate in alkaline medium is
\nMnO4<\/sub> + 2H2<\/sub>O + 3e–<\/sup> \u2192 MnO2<\/sub> + 4OH–<\/sup>
\n(a) 31.6
\n(b) 52.7
\n(c) 79
\n(d) None of these
\nAnswer:
\n(b) 52.7
\nThe reduction reaction of the oxidizing agent(MnO4<\/sub>) involves gain of 3 electrons.
\nHence the equivalent mass = (Molar mass of KMnO4<\/sub>) \/ 3 = 158.1 \/ 3 = 52.7<\/p>\n

Samacheer Kalvi 11th Chemistry Solutions Question 13.<\/strong>
\nWhich one of the following represents 180 g of water?
\n(a) 5 Moles of water
\n(b) 90 moles of water
\n(c) \\(\\frac{6.022 \\times 10^{23}}{180}\\) Molecules of water
\n(d) 6.022 \u00d7 1024<\/sup> Molecules of water
\nAnswer:
\n(d) 6.022 x 1024<\/sup> Molecules of water
\nNo. of moles of water present in 180 g
\n= Mass of water \/ Molar mass of water
\n= 180 g \/18 g mol-1<\/sup> = 10 moles
\nOne mole of water contains
\n= 6.022 \u00d7 1023<\/sup> water molecules
\n10 mole of water contains = 6.022 \u00d7 1023<\/sup> \u00d7 10
\n= 6.022 \u00d7 1024<\/sup> water molecules<\/p>\n

11th Chemistry Chapter 1 Evaluate Yourself Answers Question 14.<\/strong>
\n7.5 g of a gas occupies a volume of 5.6 liters at 0\u00b0C and 1 atm pressure. The gas is …………
\n(a) NO
\n(b) N2<\/sub>O
\n(c) CO
\n(d) CO2<\/sub>
\nAnswer:
\n(a) NO
\n7.5 g of gas occupies a volume of 5.6 liters at 273 K and 1 atm pressure Therefore, the mass of gas that occupies a volume of 22.4 liters –
\n\\(\\frac {7.5 g}{5.6 L}\\) \u00d7 22. 4 L = 30g
\nMolar mass of NO (14 + 16) = 30g<\/p>\n

11th Chemistry Solutions Samacheer Kalvi Question 15.<\/strong>
\nTotal number of electrons present in 1.7 g of ammonia is ………..
\n(a) 6.022 \u00d7 1023<\/sup>
\n(b) \\(\\frac{6.022 \\times 10^{22}}{1.7}\\)
\n(c) \\(\\frac{6.022 \\times 10^{24}}{1.7}\\)
\n(d) \\(\\frac{6.022 \\times 10^{23}}{1.7}\\)
\nAnswer:
\n(a) 6.022 \u00d7 1023<\/sup>
\nNo. of electrons present in one ammonia (NH3<\/sub>) molecule (7 + 3) = 10
\nNo. of moles of ammonia = \\(\\frac {Mass}{Molar mass}\\)
\n= \\(\\frac{1.7 \\mathrm{g}}{17 \\mathrm{g} \\mathrm{mol}^{-1}}\\) = 0.1 mol
\nNo. of molecules present in 0ne ammonia
\n= 0.1 \u00d7 6.022 \u00d7 1023<\/sup> = 6.O22 \u00d7 1022<\/sup>
\nNo. of electrons present in 0.1 mol of ammonia
\n10\u00d7 6.022 \u00d7 1022<\/sup> = 6.022 \u00d7 1023<\/sup><\/p>\n

Samacheer Kalvi Class 11 Chemistry Solutions Question 16.<\/strong>
\nThe correct increasing order of the oxidation state of sulphur in the anions SO4<\/sub>2-<\/sup>, SO3<\/sub>2-<\/sup>, S2<\/sub>O4<\/sub>2-<\/sup>,S2<\/sub>O6<\/sub>2-<\/sup> is ………..
\n(a) SO3<\/sub>2-<\/sup> < SO3<\/sub>2-<\/sup> < S2<\/sub>O4<\/sub>2-<\/sup> < S2<\/sub>O6<\/sub>2-
\n<\/sup>(b) SO4<\/sub>2-<\/sup> < S2<\/sub>O4<\/sub>2-<\/sup> < S2<\/sub>O6<\/sub>2-<\/sup><SO3<\/sub>2-
\n<\/sup>(c) S2<\/sub>O4<\/sub>2-<\/sup> < SO3<\/sub>2-<\/sup> < S2<\/sub>O6<\/sub>2-<\/sup> < SO4<\/sub>2-
\n<\/sup>(d) S2<\/sub>O6<\/sub>2-<\/sup> < S2<\/sub>O4<\/sub>2-<\/sup> < SO4<\/sub>2-<\/sup> < SO3<\/sub>2-<\/sup>
\nAnswer:
\n(c) S2<\/sub>O4<\/sub>2-<\/sup> < SO3<\/sub>2-<\/sup> < S2<\/sub>O6<\/sub>2-<\/sup> < SO4<\/sub>2-<\/sup>
\n\"Basic<\/p>\n

Class 11 Chemistry Samacheer Solutions Question 17.<\/strong>
\nThe equivalent mass of ferrous oxalate is ……….
\n\"Basic
\nAnswer:
\n\"11th
\n\"11th<\/p>\n

Chemistry Class 11 Solutions Samacheer Kalvi Question 18.<\/strong>
\nIf Avogadro number were changed from 6.022 \u00d7 1023<\/sup> to 6.022 \u00d7 1020<\/sup>, this would change ………..
\n(a) the ratio of chemical species to each other in a balanced equation
\n(b) the ratio of elements to each other in a compound
\n(c) the definition of mass in units of grams
\n(d) the mass of one mole of carbon
\nAnswer:
\n(d) the mass of one mole of carbon<\/p>\n

Samacheer Kalvi 11 Chemistry Solutions Question 19.<\/strong>
\nTwo 22.4 liter containers A and B contains 8 g of O2<\/sub> and 8 g of SO2<\/sub> respectively, at 273 K and 1 atm pressure, then ……….
\n(a) number of molecules in A and B are same
\n(b) number of molecules in B is more than that in A
\n(c) the ratio between the number of molecules in A to the number of molecules in B is 2 : 1
\n(d) number of molecules in B is three times greater than the number of molecules in A
\nAnswer:
\n(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1<\/p>\n

Samacheer Kalvi 11th Chemistry Book Solutions Question 20.<\/strong>
\nWhat is the mass of precipitate formed when 50 ml of 8.5% solution of Ag NO3<\/sub> is mixed with 100 ml of 1.865% potassium chloride solution?
\n(a) 3.59 g
\n(b) 7 g
\n(c) 14 g
\n(d) 28 g
\nAnswer:
\n(a) 3.59 g
\nAgNO3<\/sub> + KCl \u2192 KNO3<\/sub> + AgCl
\nSolution:
\n50 mL of 8.5% solution contains 4.25 g of AgNO3<\/sub>
\nNo. of moles of AgNO3<\/sub> present in 50 mL of 8.5% AgNO3<\/sub> solution
\n= Mass \/ Molar mass = 4.25 \/ 170 = 0.025 moles
\nSimilarly, No of moles of KCl present in loo mL of 1.865% KCl solution
\n= 1.865 \/ 74.5 = 0.025 moles
\nSo total amount of AgCl formed is 0.025 moles (based on the stoichiometry calculator<\/a>)
\nAmount of AgCl present in 0.025 moles of AgCl
\n= no. of moles \u00d7 molar mass
\n= 0.025 \u00d7 143.5 = 3.59 g<\/p>\n

Question 21.
\nThe mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25\u00b0C and 1 atm pressure) is 1.1g. The molar mass of the gas is ………..
\n(a) 66.25 g mol-1<\/sup>
\n(b) 44 g mol-1<\/sup>
\n(c) 24.5 g mol -1<\/sup>
\n(d) 662.5 g mol-1<\/sup>
\nAnswer:
\n(b) 44 g mol-1
\n<\/sup>Solution:
\nNo. of moles of a gas that occupies a volume of 6 12.5 ml at room temperature and pressure
\n(25\u00b0 C and 1 atm pressure)
\n= 612.5 \u00d7 10-3<\/sup> L\/24.5 L mol-1<\/sup>
\n= 0.02 5 moles
\nWe know that,
\nMolar mass = Mass \/ no. of moles
\n= 1.1 g\/0.025 mol = 44 g mol-1<\/sup><\/p>\n

Question 22.
\nWhich of the following contain same number of carbon atoms as in 6 g of carbon -12?
\n(a) 7.5 g ethane
\n(b) 8 g methane
\n(c) both (a) and (b)
\n(d) none of these
\nAnswer:
\n(c) both (a) and (b)
\nSolution:
\nNo. of moles of carbon present in 6 g of C – 12 = Mass \/ Molar mass
\n= 6\/12 = 0.5 moles = 0.5 \u00d7 6.022 \u00d7 1023<\/sup> carbon atoms.
\nNo. of moles in 8 g of methane = 8 116 = 0.5 moles
\n= 0.5 \u00d7 6.022 \u00d7 1023<\/sup> carbon atoms.
\nNo. of moles in 7.5 g of ethane = 7.5 \/ 16 = 0.25 moles
\n= 2 \u00d7 0.25 \u00d7 6.022 \u00d7 1023<\/sup> carbon atoms.<\/p>\n

Question 23.
\nWhich of the following compound( s) has\/have percentage of carbon same as that in ethylene (C2<\/sub>H4<\/sub>)?
\n(a) propene
\n(b) ethyne
\n(c) benzene
\n(d) ethane
\nAnswer:
\n(a) propene
\nSolution:
\nMolar mass of carbon
\nPercentage of carbon in ethylene(C2<\/sub>H6<\/sub>) = \"11th
\n= \\(\\frac {24}{28}\\) \u00d7 100 = 85.71%
\nPercentage of carbon in propene (C3<\/sub>H6<\/sub>) = \\(\\frac {24}{28}\\) \u00d7 100 = 85.7 1%<\/p>\n

Question 24.
\nWhich of the following is\/are true with respect to carbon – 12?
\n(a) relative atomic mass is 12 u
\n(b) oxidation number of carbon is +4 in all its compounds.
\n(c) I mole of carbon -12 contain 6.022 \u00d7 1022<\/sup> carbon atoms.
\n(d) all of these
\nAnswer:
\n(a) relative atomic mass is 12 u<\/p>\n

Question 25.
\nWhich one of the following is used as a standard for atomic mass?
\n(a) 6<\/sub>C12<\/sup>
\n(b) 7<\/sub>C12<\/sup>
\n(c) 6<\/sub>C13<\/sup>
\n(d) 6<\/sub>C14<\/sup>
\nAnswer:
\n(a) 6<\/sub>C12<\/sup><\/p>\n

II. Write brief answer to the following questions<\/span><\/p>\n

Question 26.
\nDefine relative atomic mass.<\/p>\n

On the basis of carbon, the relative atomic mass of element is defined as the ratio of mass of one atom of the element to the mass of l\/12th mass of one atom of Carbon – 12.
\n\"Class<\/p>\n

Question 27.
\nWhat do you understand by the term mole?
\nAnswer:
\nThe mole is defined as the amount of a substance which contains 6.023 x 1023<\/sup> particles such as atoms, molecules or ions. It is represented by the symbol<\/p>\n

Question 28.
\nDefine equivalent mass.
\nAnswer:
\nThe equivalent mass of an element is the number of parts of the mass of an element which combines with or displaces 1.008 parts of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine.<\/p>\n

Question 29.
\nWhat do you understand by the term oxidation number?
\nAnswer:
\nOxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely. The oxidation numbers reflect the number of electrons \u201ctransferred\u201d.<\/p>\n

Question 30.
\nDistinguish between oxidation and reduction.
\nAnswer:
\nOxidation:
\nAccording to the classical concept, oxidation is a process of addition of oxygen or removal of hydrogen.
\nRemoval of hydrogen
\n2H2<\/sub>S + O2<\/sub> \u2192 H2<\/sub>O + 2S
\nAddition of oxygen
\nC + O2<\/sub> \u2192 CO2<\/sub>
\nAccording to the electronic concept, loss of electrons is called oxidation reaction.
\nCa \u2192 Ca2+<\/sup> + 2e–<\/sup>
\nDuring oxidation, oxidation number increases.
\nDining oxidation, reducing agent gets oxidised.<\/p>\n

Reduction:
\nReduction is a process of removal of oxygen or addition of hydrogen.
\nAddition of hydrogen
\nCa + H2<\/sub> \u2192 CaH2<\/sub>
\nRemoval of oxygen
\nZn O + C \u2192 Zn + CO
\nAccording to the electronic concept, gain of electrons is called reduction reaction.
\nZn2+<\/sup> + 2e–<\/sup> \u2192 Zn
\nDuring reduction, oxidation number decreases.
\nDuring reduction, oxidising agent gets reduced.<\/p>\n

Question 31.
\nCalculate the molar mass of the following compounds.<\/p>\n

    \n
  1. urea [CO(NH<\/span>2<\/sub>)<\/span>2<\/sub>]
    \n<\/span><\/li>\n
  2. acetone [CH<\/span>3<\/sub>COCH<\/span>3<\/sub>]
    \n<\/span><\/li>\n
  3. boric acid [H3<\/sub>BO3<\/sub>]<\/li>\n
  4. sulphuric acid [H2<\/sub>SO4<\/sub>]<\/li>\n<\/ol>\n

    Answer:
    \n1. urea [CO(NH2<\/sub>)2<\/sub>]
    \nAtomic mass of C =12
    \nAtomic mass of O =16
    \nAtomic mass of 2(N) = 28
    \nAtomic mass of 4(H) = 4
    \n\u2234 Molar mass of Urea = 60<\/p>\n

    2. Acetone [CH3<\/sub>COCH3<\/sub>]
    \nAtomic mass of 3(C) = 36
    \nAtomic mass of 1(0) = 16
    \nAtomic mass of 6(H) = 6
    \n\u2234 Molar mass of Acetone = 58<\/p>\n

    3. Boric acid [H3<\/sub>BO3<\/sub>]
    \nAtomic mass of B = 10
    \nAtomic mass of 3(H) = 3
    \nAtomic mass of 3(O) = 48
    \n\u2234 Molar mass of Boric acid = 61<\/p>\n

    4. Sulphuric acid 2<\/sub>[H2<\/sub>SO4<\/sub>]
    \nAtomic mass of 2(H) = 2
    \nAtomic mass of 1(S) = 32
    \nAtomic mass of 4(O) = 64
    \n\u2234 Molar mass of Sulphuric acid = 98<\/p>\n

    Question 32.
    \nThe density of carbon dioxide is equal to 1.977 kg m-3<\/sup> at 273 K and 1 atm pressure. Calculate the molar mass of CO2
    \n<\/sub>Answer:
    \nMolecular mass = Density x Molar volume
    \nMolar volume of CO2<\/sub> = 2.24 x 10-2<\/sup> m3<\/sup>
    \nDensity of CO2<\/sub> = 1.977 kg m-3<\/sup>
    \nMolecular mass of CO2<\/sub> = 1.977 x 103<\/sup> gm-3 <\/sup>\u00a0x 2.24 x 10-2<\/sup> m3<\/sup>
    \n= 1.977 \u00d7 10-1<\/sup> \u00d7 2.24 = 44 g<\/p>\n

    Question 33.
    \nWhich contains the greatest number of moles of oxygen atoms?<\/p>\n

      \n
    1. 1 mol of ethanol<\/li>\n
    2. 1 mol of formic acid<\/li>\n
    3. 1 mol of H2<\/sub>O<\/li>\n<\/ol>\n

      Answer:
      \n1. 1 mol of ethanol
      \nC2<\/sub>H5<\/sub>OH (ethanol) – Molar mass = 24 + 6 + 16 = 46
      \n46 g of ethanol contains 1 \u00d7 6.023 \u00d7 1023<\/sup> number of oxygen atoms.<\/p>\n

      2. 1 mol of formic acid.
      \nHCOOH (formic acid) – Molar mass = 2+12 + 32 = 46
      \n46 g of HCOOH contains 2 \u00d7 6.023 \u00d7 1023<\/sup> number of oxygen atoms.<\/p>\n

      3. 1 mol of H2<\/sub>O
      \nH2<\/sub>O (water) – Molar mass = 2 + 16 = 18
      \n18 g of water contains 1 \u00d7 6.023 \u00d7 1023<\/sup> number of oxygen atoms.
      \n\u2234 1 mole of formic acid contains the greatest number of oxygen atoms.<\/p>\n

      Question 34.
      \nCalculate the average atomic mass of naturally occurring magnesium using the following data<\/p>\n\n\n\n\n\n\n
      Isotope<\/strong><\/td>\nIsotopic atomic mass<\/strong><\/td>\nAbundance (%)<\/strong><\/td>\n<\/tr>\n
      Mg24<\/sup><\/td>\n23.99<\/td>\n78.99<\/td>\n<\/tr>\n
      Mg26<\/sup><\/td>\n24.99<\/td>\n10<\/td>\n<\/tr>\n
      Mg25<\/sup><\/td>\n25.98<\/td>\n11.01<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      Answer:
      \nIsotopes of Mg.
      \nAtomic mass = Mg24<\/sup> = 23.99 x \\(\\frac {783. 99}{100}\\) = 18.95
      \nAtomic mass = Mg26<\/sup> = 24.99 x \\(\\frac {10}{100}\\) = 2.499
      \nAtomic mass = Mg25<\/sup> = 25.98 x \\(\\frac {11.01}{100}\\) = 2.860
      \nAverage Atomic mass = 24.309
      \nAverage atomic mass of Mg = 24.309<\/p>\n

      Question 35.
      \nIn a reaction x + y + z2<\/sub> \u2192 xyz2<\/sub>, identify the limiting reagent if any, in the following reaction mixtures.
      \n(a) 200 atoms of x + 200 atoms of y + 50 molecules of z2<\/sub>
      \n(b) 1 mol of x + 1 mol of y + 3 mol of z2<\/sub>
      \n(c) 50 atoms of x + 25 atoms of y + 50 molecules of z2<\/sub>
      \n(d) 2.5 mol of x + 5 mol of y + 5 mol of z2<\/sub>
      \nAnswer:
      \nx + y + z2<\/sub>
      \n(a) 200 atoms of x + 200 atoms of y + 50 molecules of z2<\/sub> According to the reaction, 1 atom of x reacts with one atom of y and one molecule of z to give product. In the case (a) 200 atoms of x, 200 atoms of y react with 50 molecules of z2<\/sub> (4 part) i.e. 50 molecules of z2<\/sub> react with 50 atoms of x and 50 atoms of y. Hence z is the limiting reagent.<\/p>\n

      (b) 1 mol of x + 1 mol of y + 3 mol of z2<\/sub>
      \nAccording to the equation 1 mole of z2<\/sub> only react with one mole of x and one mole of y. If 3 moles of z2<\/sub> are there, z is limiting reagent.<\/p>\n

      (c) 50 atoms of x + 25 atoms of y + 50 molecules of z2<\/sub>
      \n25 atoms of y react with 25 atoms of x and 25 molecules of z2<\/sub>. So y is the limiting reagent.<\/p>\n

      (d) 2.5 mol of x + 5 mol of y + 5 mol of z2<\/sub>
      \n2.5 mol of x react with 2.5 mole of y and 2.5 mole of z2<\/sub>. So x is the limiting reagent.<\/p>\n

      Question 36.
      \nMass of one atom of an element is 6.645 \u00d7 10-23<\/sup> g. How many moles of element are there in 0.320 kg?
      \nAnswer:
      \nMass of one atom of an element = 6.645 \u00d7 10-23<\/sup> g = Atomic mass.
      \nMass of given element = 0.320 kg
      \nNumber of moles =
      \nAtomic mass
      \n\"Samacheer
      \n= 48.156 x 10-23<\/sup>
      \n= 4.8156 x 10-24<\/sup> moles.<\/p>\n

      Question 37.
      \nWhat is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
      \nAnswer:
      \nMolecular mass:<\/p>\n

        \n
      • Relative molecular mass is defined as the ratio of the mass of the molecule to the unified atomic mass unit.<\/li>\n
      • \u00a0It can be calculated by adding the relative atomic masses of its constituent atoms.<\/li>\n
      • For carbon monoxide (CO) Molecular mass = Atomic mass of carbon + Atomic mass of oxygen 12 + 16 = 28 u.<\/li>\n<\/ul>\n

        Molar mass:<\/p>\n

          \n
        • It is defined as the mass of one mole of a substance.<\/li>\n
        • The molar mass of a compound is equal to the sum of the relative atomic masses of its constituent expressed in g mol-1<\/sup>.<\/li>\n
        • \u00a0For carbon monoxide (CO) 12 + 16 = 28 g mol-1 <\/sup>Both molecular mass and molar mass are numerically same but the units are different.<\/li>\n<\/ul>\n

          Question 38.
          \nWhat is the empirical formula of the following?<\/p>\n

            \n
          1. Fructose (C6<\/sub>H12<\/sub>O6<\/sub>) found in honey<\/li>\n
          2. Caffeine (C8<\/sub>H10<\/sub>N4<\/sub>O2<\/sub>) a substance found in tea and coffee.<\/li>\n<\/ol>\n

            Answer:
            \n1. Fructose (C6<\/sub>H12<\/sub>O6<\/sub>)
            \nEmpirical formula is the simplest formula. So it is divided by 6 and so empirical formula is CH2<\/sub>O.<\/p>\n

            2. Caffeine (C8<\/sub>H10<\/sub>N4<\/sub>O2<\/sub>)
            \nSimplified formula = \\(\\frac {molecular formula}{2}\\)
            \nEmpirical formula = C4<\/sub>H5<\/sub>N2<\/sub>O.<\/p>\n

            Question 39.
            \nThe reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of AC = 21 u Atomic mass of 0 = 16 u) 2Al + Fe2<\/sub>O2<\/sub> \u2192 Al2<\/sub>O3<\/sub> + 2Fe; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide.<\/p>\n

              \n
            1. Calculate the mass of Al2<\/sub>O3<\/sub> formed.<\/li>\n
            2. How much of the excess reagent is left at the end of the reaction?<\/li>\n<\/ol>\n

              Answer:
              \n\"Samacheer
              \n1. As per balanced equation 54 g A1 is required for 112 g of iron and 102 g of Al2<\/sub>O3<\/sub>.
              \n54 g of Al gives 102 g of Al2<\/sub>O3<\/sub>.
              \n\u2234 324 g of Al will give \\(\\frac{102}{54}\\) x 324 = 612 g of Al2<\/sub>O3<\/sub>.<\/p>\n

              2. 54 g of Al requires 160 g of Fe2<\/sub>O3<\/sub> for welding reaction.
              \n\u2234 324 g of Al will require \\(\\frac {160}{54}\\) x 324 = 960 g of Fe2<\/sub>O3<\/sub>.
              \n\u2234 Excess Fe2<\/sub>O3<\/sub> – Un reacted Fe2<\/sub>O3<\/sub> = 1120 – 960 = 160 g
              \n160 g of excess reagent is left at the end of the reaction.<\/p>\n

              Question 40.
              \nHow many moles of ethane is required to produce 44 g of CO2<\/sub> (g) after combustion.
              \nAnswer:
              \n\"Samacheer
              \n\u2234 44g of CO2<\/sub> = I mole of CO2<\/sub>
              \n2 moles of CO2<\/sub> is produced by 1 mole of ethane.
              \n\u2234 1 mole of CO2<\/sub> will be produced by = ?
              \n\u2234 To produce 1 mole of CO2<\/sub>, the required mole of ethane is = \\(\\frac {1}{2}\\) x 1 = 0.5 mole of ethane.<\/p>\n

              Question 41.
              \nHydrogen peroxide is an oxidizing agent. It dioxides ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
              \nAnswer:
              \nH2<\/sub>O2<\/sub> – Oxidizing agent
              \nFe2+<\/sup> + H2<\/sub>O2<\/sub> \u2192 Fe3+<\/sup> + H2<\/sub>O (Acetic medium)
              \nFerrous ion is oxidized by H2<\/sub>O2<\/sub> to Ferric ion.
              \nThe balanced equation is Fe2+<\/sup> \u2192 Fe3+<\/sup> + e–<\/sup> x 2
              \n\"Samacheer<\/p>\n

              Question 42.
              \nCalculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
              \nAnswer:
              \n\"Samacheer
              \nEmpirical formula = C6<\/sub>H6<\/sub>O
              \nVa-pour density 47
              \n\u2234 Molecular mass = 2 x vapor density = 2 x 47 = 94
              \nMolecular formula Empirical formula x n
              \nMolecular mass x n
              \nn = \\(\\frac { Molecular mass }{ Empirical formula mass }\\) = \\(\\frac {94}{94}\\) = 1
              \n\u2234 Molecular formula = C6<\/sub>H6<\/sub>O<\/p>\n

              Question 43.
              \nA Compound on analysis gave Na = 14.31% S = 9.97% H = 6.22% and O = 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization, (molecular mass of the compound is 322).
              \nAnswer:
              \n\"Samacheer
              \nAll H combines with 10 oxygen atoms to form as 10H2<\/sub>O.
              \nSo the empirical formula is Na2<\/sub>SO4<\/sub>\u00a0.10H2<\/sub>\u00a0O
              \nEmpirical formula mass = (23 x 2) + (32 x 1) + (16 x 4) + (10 x 18)
              \n= 46 + 32 + 64 + 180 = 322
              \nn = \\(\\frac { Molecular mass }{ Empirical formula mass }\\) = \\(\\frac {322}{322}\\) = 1
              \nMolecular formula = Na2<\/sub>SO4<\/sub>. 10H2<\/sub>O<\/p>\n

              Question 44.
              \nBalance the following equations by oxidation number method<\/p>\n

                \n
              1. K2<\/sub>Cr2<\/sub> O7<\/sub> + KI + H2<\/sub>SO2<\/sub> \u2192 K2<\/sub>SO4<\/sub> + Cr2<\/sub>(SO4<\/sub>)3<\/sub> + I2<\/sub> + H2<\/sub>O<\/li>\n
              2. KMnO4<\/sub> + Na2<\/sub>SO3<\/sub> \u2192 Mn02<\/sub> + Na2<\/sub>SO4<\/sub> + KOH<\/li>\n
              3. Cu+ HNO3<\/sub> \u2192 Cu(N03<\/sub>)2<\/sub> + NO2<\/sub> + H2<\/sub>O<\/li>\n
              4. H2<\/sub>C2<\/sub>O4<\/sub> + KMnO4<\/sub> + H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + MnSO4<\/sub> + CO2<\/sub> + H2<\/sub>O<\/li>\n<\/ol>\n

                Answer:
                \n1. K2<\/sub>Cr2<\/sub> O7<\/sub> + KI + H2<\/sub>SO2<\/sub> \u2192 K2<\/sub>SO2<\/sub> + Cr2<\/sub>(SO4<\/sub>)3<\/sub> + I2<\/sub> + H2<\/sub>O
                \nStep – 1.
                \n\"Samacheer
                \nStep – 2
                \nK2<\/sub>Cr2<\/sub> O7<\/sub> + 6KI + H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + Cr2<\/sub>(SO4<\/sub>)3<\/sub> + 3I2<\/sub> + H2<\/sub>O
                \nStep – 3
                \nTo balance other atoms
                \nK2<\/sub>Cr2<\/sub> O7<\/sub> + 6KI + H2<\/sub>SO4<\/sub> \u2192 4K2<\/sub>SO4<\/sub> + Cr2<\/sub>(SO4<\/sub>)3<\/sub> + 3I2<\/sub> + H2<\/sub>O
                \nStep – 4
                \nK2<\/sub>Cr2<\/sub> O7<\/sub> + 6KI + 7H2<\/sub>SO4<\/sub> \u2192 4K2<\/sub>SO4<\/sub> + Cr2<\/sub>(SO4<\/sub>)3<\/sub> + 3I2<\/sub> + 7H2<\/sub>O<\/p>\n

                2. KMnO4<\/sub> + Na2<\/sub>SO3<\/sub> \u2192 MnO2<\/sub> + Na2<\/sub>SO4<\/sub> + KOH (Alkaline medium)
                \nStep – 1
                \n\"Samacheer
                \nStep – 2
                \n2KMnO4<\/sub> + 3Na2<\/sub>SO3<\/sub> \u2192 2MnO2<\/sub> + 3Na2<\/sub>SO4<\/sub> + KOH
                \nStep – 3
                \nbalancing potassium, KOH is multiplied by 2
                \n2KMnO4<\/sub> + 3Na2<\/sub>SO3<\/sub> \u2192 2MnO2<\/sub> + 3Na2<\/sub>SO4<\/sub> + KOH
                \nStep – 4
                \nTo balance H atom, H2<\/sub>0 is added on reactant side.
                \n2KMnO4<\/sub> + 3Na2<\/sub>SO3<\/sub> + H2<\/sub>O \u2192 2MnO2<\/sub> + 3Na2<\/sub>SO4<\/sub> + KOH<\/p>\n

                3. Cu + HNO3<\/sub> \u2192 Cu(NO3<\/sub>)2<\/sub> + NO2<\/sub> + H2<\/sub>O
                \nStep – 1
                \n\"Samacheer
                \nStep – 2
                \nCu + HNO3<\/sub> \u2192 Cu(NO3<\/sub>)2<\/sub> + NO2<\/sub> + H2<\/sub>O
                \nStep – 3
                \nTo balance Nitrogen, 2HNO3<\/sub> is multiplied by 2 and NO2<\/sub> is multiplied by 2
                \nCu + 4HNO3<\/sub> \u2192 Cu(NO3<\/sub>)2<\/sub> + 2NO2<\/sub> + H2<\/sub>O
                \nStep 4.
                \nTo balance oxygen, H2<\/sub>O is multiplied by 2
                \nCu + 4HNO3<\/sub> \u2192 Cu(NO3<\/sub>)2<\/sub> + 2NO2<\/sub> + 2H2<\/sub>O<\/p>\n

                4. H2<\/sub>C2<\/sub>O4<\/sub> + KMnO4<\/sub> + H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + MnSO4<\/sub> + CO2<\/sub> + H2<\/sub>O
                \nStep – 1
                \n\"Samacheer
                \nStep – 2
                \n5 H2<\/sub>C2<\/sub>O4<\/sub> + KMnO4<\/sub> + H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + MnSO4<\/sub> + 10 CO2<\/sub> + H2<\/sub>O
                \nStep – 3
                \nTo balance K, KMnO4<\/sub> and MnSO4<\/sub> are multiplied by 2
                \n5 H2<\/sub>C2<\/sub>O4<\/sub> + 2KMnO4<\/sub> + H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2MnSO4<\/sub> + 10 CO2<\/sub> + H2<\/sub>O
                \nStep – 4
                \nTo balance O and H, H2<\/sub>O and H2<\/sub>SO4<\/sub> are multiplied by 3 and 6.
                \n5 H2<\/sub>C2<\/sub>O4<\/sub> + 2KMnO4<\/sub> + 3H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2MnSO4<\/sub> + 10 CO2<\/sub> + 8H2<\/sub>O<\/p>\n

                Question 45.
                \nBalance the following equations by ion electron method.<\/p>\n

                  \n
                1. KMnO4<\/sub>\u00a0+ SnCl2<\/sub>\u00a0 + HCl \u2192 MnCl2<\/sub>\u00a0 + SnCl4<\/sub>\u00a0 + H2<\/sub>O + KCl<\/li>\n
                2. C2<\/sub>O4<\/sub>2-<\/sup> + Cr2<\/sub>\u00a0O7<\/sub>\u00a02-<\/sub> \u2192 Cr 3+<\/sup>\u00a0+ CO2<\/sub>\u00a0 (in acid medium)<\/li>\n
                3. Na2<\/sub>S2<\/sub>O3<\/sub> + I2<\/sub> \u2192 Na2<\/sub>S4<\/sub>O6<\/sub> + NaI (in acid medium)<\/li>\n
                4. Zn + NO3<\/sub>–<\/sup> \u2192 Zn2+<\/sup> + NO<\/li>\n<\/ol>\n

                  Answer:
                  \n1. KMnO4<\/sub> + SnCl2<\/sub> + HCl \u2192 MnCl3<\/sub> + SnCl3<\/sub> +H2<\/sub>O + KCl
                  \nOxidation half reaction: (loss of electrons)
                  \n\"Samacheer<\/p>\n

                  Reduction half reaction: (gain of electrons)
                  \n\"Samacheer
                  \nAdd H2<\/sub>O to balance oxygen atoms.
                  \n\"Samacheer
                  \nAdd HCl to balance hydrogen atoms
                  \nKMnO4<\/sub> + 5e–<\/sub> + 8HCl \u2192 MnCl2<\/sub> + 4H2<\/sub>O ………(4)
                  \nTo equalize the number of electrons equation (1) x 5 and equation (2) x 2
                  \n5SnCl2<\/sub> \u2192 5SnCl4<\/sub> + 10e–
                  \n<\/sup>2KMnO4<\/sub> + 16HCl + 10e–<\/sup> \u2192 2MnCl2<\/sub> + 4H2<\/sub>O + 2KCl
                  \n2KMnO4<\/sub> + 5SnCl2<\/sub> + 16HCl \u2192 5SnCl4<\/sub> + 2MnCl2<\/sub> + 4H2<\/sub>O + 2KCl<\/p>\n

                  2. C2<\/sub>O4<\/sub>2-<\/sup> + Cr2<\/sub>\u00a0O7<\/sub>2-<\/sup> \u2192 Cr 3+<\/sup>\u00a0+ CO2<\/sub>\u00a0 (in acid medium)
                  \nOxidation half reaction:
                  \n\"Samacheer
                  \nReduction half reaction:
                  \n\"Samacheer
                  \nTo balance oxygen atoms, H2<\/sub>O is added on RHS of equation (2)
                  \nCr2<\/sub>O7<\/sub>2-<\/sup> + 6e–<\/sup> \u2192 2Cr3+<\/sup> + 7 H2<\/sub>O ……….(3)
                  \nTo balance Hydrogen atoms, H+<\/sup> is added on LHS of equation (1)
                  \nC2<\/sub>O4<\/sub>2-<\/sup> + 14H+<\/sup> \u2192 2CO2<\/sub> + 2e–<\/sup> ……..(4)
                  \nTo equalize the number of electrons gained and lost, multiply the equation (4) x 3.
                  \n\"Samacheer<\/p>\n

                  3. Na2<\/sub>S2<\/sub>O3<\/sub> + I2<\/sub> \u2192 Na2<\/sub>S4<\/sub>O6<\/sub> + NaI (in acid medium)
                  \nOxidation half reaction: (Loss of electron)
                  \nNa2<\/sub>S2<\/sub>O3<\/sub> \u2192 Na2<\/sub>S4<\/sub>O6<\/sub> + 2e2-<\/sup> ……..(1)
                  \nReduction half reaction: (Gain of electron)
                  \nI2<\/sub> + 2e2-<\/sup>\u2192 2NaI …………(2)
                  \nAdding (1) and (2)
                  \n\"Samacheer
                  \nTo balance oxygen,
                  \n2Na2<\/sub>S2<\/sub>O3<\/sub> + I2 <\/sub>\u2192 Na2<\/sub>S2<\/sub>O2<\/sub> + 2NaI<\/p>\n

                  In acidic medium
                  \n4. Zn + NO3<\/sub>–<\/sup> \u2192 Zn2+<\/sup> + NO
                  \nHalf reactions are –
                  \n\"Samacheer<\/p>\n

                  Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations In Text Questions – Evaluate Yourself<\/span><\/strong><\/p>\n

                  Question 1.
                  \nBy applying the knowledge of chemical classification, classify each of the following Into elements, compounds or mixtures.<\/p>\n

                    \n
                  1. Sugar<\/li>\n
                  2. Sea water<\/li>\n
                  3. Distilled water<\/li>\n
                  4. Carbon dioxide<\/li>\n
                  5. Copper wire<\/li>\n
                  6. Table salt<\/li>\n
                  7. Silver plate<\/li>\n
                  8. Naphthalene balls<\/li>\n<\/ol>\n

                    Answer:<\/p>\n\n\n\n\n
                    Elements<\/strong><\/td>\nCompounds<\/strong><\/td>\nMixtures<\/strong><\/td>\n<\/tr>\n
                    Copper wire (Cu)
                    \nSilver plate (Ag)<\/td>\n                    		Sugar
                    \nDistilled water Carbon dioxide Naphthalene balls<\/td>\n                    		Sea water Table salt<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                    Question 2.
                    \nCalculate the molar mass of the following.<\/p>\n

                      \n
                    1. Ethanol (C2<\/sub>H5<\/sub>OH)<\/li>\n
                    2. Potassium permanganate (KMnO4<\/sub>)<\/li>\n
                    3. Potassium dichromate (K2<\/sub>Cr2<\/sub>O7<\/sub>)<\/li>\n
                    4. Sucrose (C12<\/sub>H22<\/sub>O11<\/sub>)<\/li>\n<\/ol>\n

                      Answer:
                      \n(1) Ethanol (C2<\/sub>H5<\/sub>OH)
                      \nMolar mass = (2 \u00d7 12) + (6 \u00d7 1) + (1 \u00d7 16)
                      \n= 24 + 6+16 = 46 g mol-1<\/sup><\/p>\n

                      (2) Potassium permanganate (KMnO4<\/sub>)
                      \nMolar mass = 39 + 55 + (4 \u00d7 16)
                      \n= 39 + 55 + 64 = 158 g mol-1<\/sup><\/p>\n

                      (3) Potassium dichromate (K2<\/sub>Cr2<\/sub>O7<\/sub>)
                      \nMolar mass = (39 \u00d7 2) + (2 \u00d7 52) + (7 \u00d7 16)
                      \n= 78 + 104 + 112 = 294 g mol-1<\/sup><\/p>\n

                      (4) Sucrose (C12<\/sub>H22<\/sub>O11<\/sub>)
                      \nMolar mass = (12 \u00d7 12) + (22 \u00d7 1) + (11 \u00d7 16) = 144 + 22 + 176 = 342 g mol-1<\/sup><\/p>\n

                      Question 3.
                      \n(a) Calculate the number of moles present in 9 g of ethane.
                      \nAnswer:
                      \nMass of ethane = 9 g
                      \nMolar mass of ethane C2<\/sub>H6<\/sub> = 30 g mol-1<\/sup>
                      \nNo. of moles = \\(\\frac {Mass}{Molar mass}\\) = \\(\\frac {9}{30}\\) = 0.3 mol.<\/p>\n

                      (b) Calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 K and 3 atm pressure.
                      \nAnswer:
                      \nMolar volume of oxygen = 22400 ml.
                      \n22400 ml of oxygen contains 6.023 x 1023<\/sup> molecules.
                      \n224 ml of oxygen contain \\(\\frac{6.023 \\times 10^{23}}{22400}\\) x 224
                      \n\\(\\frac{6.023 \\times 10^{23}}{100}\\) = 6.023 \u00d7 1021<\/sup><\/p>\n

                      Question 4.
                      \n(a) 0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal.
                      \nAnswer:
                      \nMass of the metal = W1<\/sub> = 0.456 g.
                      \nMass of the metal chloride = W2<\/sub> = 0.606 g.
                      \n\u2234Mass of chlorine = W2<\/sub> – W1<\/sub> = 0.606 – 0.456 = 0.15 g
                      \n0.15 g of chlorine combine with 0.456 g of metal.
                      \n\u2234 35.46 g of chlorine will combine with \\(\\frac {0.456}{0.15}\\) x 35.46 = 107.79 g eq-1<\/sup><\/p>\n

                      (b) Calculate the equivalent mass of potassium dichromate. The reduction half – reaction in acid medium is –
                      \nCr2<\/sub>O7<\/sub>2-<\/sup> + 14H+<\/sup> +6e–<\/sup> \u2192 2 Cr3+<\/sup> + 7H2<\/sub>O
                      \nAnswer:
                      \nCr2<\/sub>O7<\/sub>2-<\/sup> + 14H+<\/sup> +6e–<\/sup> \u2192 2 Cr3+<\/sup> + 7H2<\/sub>O
                      \n\"Samacheer
                      \nMolar mass of K2<\/sub>Cr2<\/sub>O7<\/sub> = 294.18
                      \n\u2234 Equivalent mass of K2<\/sub>Cr2<\/sub>O7<\/sub> = \\(\\frac { 294.18}{6}\\) = 49.03<\/p>\n

                      Question 5.
                      \nA Compound on analysis gave the following percentage composition C = 54.55%, H = 9.09%, O = 36.36%. Determine the empirical formula of the compound.
                      \nAnswer:
                      \n\"Samacheer
                      \nEmpirical formula = C2<\/sub>H4<\/sub>O<\/p>\n

                      Question 6.
                      \nExperimental analysis of a compound containing the elements x,y,z on analysis gave the following data, x = 32 %, y = 24 %, z = 44 %. The relative number of atoms of x, y and z are 2,1 and 0.5, respectively. (Molecular mass of the compound is 400 g) Find out.<\/p>\n

                        \n
                      1. The atomic masses of the element x,y,z.<\/li>\n
                      2. Empirical formula of the compound and<\/li>\n
                      3. Molecular formula of the compound.<\/li>\n<\/ol>\n

                        Answer:
                        \nElement x = 32% ; y = 24% ; z = 44%
                        \nRelative number of atoms x = 2 ; y = 1 ; z = 0.5
                        \nMolar mass of the compound = 400 g.<\/p>\n

                        1. Atomic mass of the element.
                        \nRelative number of atoms \"Samacheer
                        \n\u2234\u00a0 Atomic mass \"Samacheer
                        \nAtomic mass x = \\(\\frac {32}{2}\\) = 16
                        \nAtomic mass y = \\(\\frac {24}{1}\\) = 24
                        \nAtomic mass z = \\(\\frac {44}{0.5}\\) = 88
                        \n\"Samacheer<\/p>\n

                        2. Empirical formula of the compound x4<\/sub> y2<\/sub> Z1<\/sub>
                        \nMolecular mass of the compound = 400
                        \n\"Samacheer = \\(\\frac {400}{200}\\) = 2<\/p>\n

                        3. Molecular formula of the compound = x8<\/sub> y4<\/sub> z2<\/sub><\/p>\n

                        Question 7.
                        \nThe balanced equation for a reaction is given below 2x + 3y \u2192 41 + m When 8 moles of x react with 15 moles of y, then –<\/p>\n

                          \n
                        1. Which is the limiting reagent?<\/li>\n
                        2. Calculate the amount of products formed.<\/li>\n
                        3. Calculate the amount of excess reactant left at the end of the reaction.<\/li>\n<\/ol>\n

                          Answer:
                          \n2 x + 3 y \u2192 41 + m<\/p>\n

                          1. 2x reacts with 3y to give products.
                          \n5x reacts with l5y means, y is the excess because 8 moles of x should react withn 4 x 3y = 12y moles of y to give products. In this reaction 15y moles are used. Therefore, 3 moles of y is excess and it is the limiting agent.<\/p>\n

                          2. When 8 moles of x react with 12 moles of y, the product formed will be 4 x 41 i.e. 161 and 4m as product.
                          \n8x + 12y \u2192 161 + 4m<\/p>\n

                          3.\u00a0 At the end of the reaction, the excess reactant left in 3 moles of y.<\/p>\n

                          Question 8.
                          \nBalance the following equation using oxidation number method
                          \nAS2<\/sub>\u00a0S3<\/sub>\u00a0 + HNO3<\/sub>\u00a0+ H2<\/sub>O \u2192 H3<\/sub>ASO4<\/sub> + H2<\/sub>SO4<\/sub> + NO
                          \nAnswer:
                          \n\"Samacheer
                          \nTo balance oxygen and sulphur, H2<\/sub>O and H2<\/sub>SO4<\/sub> are added.
                          \n3AS2<\/sub>S3<\/sub> + 2HNO3<\/sub> + H2<\/sub>O \u2192 6H3<\/sub>ASO4<\/sub> + 2NO + H2<\/sub>SO2<\/sub>
                          \n3AS2<\/sub>S3<\/sub> + 28HNO3<\/sub> + 4H2<\/sub>O \u2192 6H3<\/sub>ASO4<\/sub> + 28NO + 9H2<\/sub>SO4<\/sub><\/p>\n

                          Samacheer Kalvi 11th Chemistry Solutions Basic Concepts of Chemistry and Chemical Calculations Textual Calculations based on Stolchlometry Solved<\/span><\/strong><\/p>\n

                          Question 1.
                          \nHow many moles of hydrogen is required to produce 10 moles of ammonia?
                          \nAnswer:
                          \nThe balanced stoichiometric equation f\u00f2r the formation of ammonia is
                          \nN2<\/sub>(g) + 3H2<\/sub>(g) \u2192 2NH2<\/sub>(g)
                          \n4s per the stoichiometric equation, to produce 2 moles of ammonia, 3 moles of hydrogen are required.
                          \n\u2234 to produce 10 moles of ammonia,
                          \n\"Samacheer
                          \n= 15 moles of hydrogen are required.<\/p>\n

                          Question 2.
                          \nCalculate the amount of water produced by the combustion of 32 g of methane.
                          \nAnswer:
                          \nCH4(g)<\/sub> + 2O2(g)<\/sub> \u2192 CO2(g)<\/sub> + 2H2<\/sub>O(g)<\/sub>
                          \nAs per the stoichiometric equation,
                          \nCombustion of 1 mole (16 g) CH4<\/sub> produces 2 moles (2 x 18 = 36 g) of water.
                          \n\"Samacheer
                          \nCombustion of 32 g CH4<\/sub> produces
                          \n\"Samacheer<\/p>\n

                          Question 3.
                          \nHow much volume of carbon dioxide is produced when 50 g of calcium carbonate is heated completely under standard conditions?
                          \nAnswer:
                          \nThe balanced chemical equation is,
                          \n\"Samacheer
                          \nAs per the stoichiometric equation,
                          \n1 mole (100g) CaCO3<\/sub> on heating produces 1 mole CO2<\/sub>
                          \n\"Samacheer
                          \nAt STP, 1 mole of CO2<\/sub> occupies a volume of 22.7 litres
                          \nAt STP, 50 g of CaCO3<\/sub> on heating produces,
                          \n\"Samacheer
                          \n= 11.35 litres of CO2<\/sub><\/p>\n

                          Question 4.
                          \nHow much volume of chlorine is required to form 11.2 L of HCl at 273 K and 1 atm pressure ?
                          \nAnswer:
                          \nThe balanced equation for the formation of HCl is,
                          \nH2<\/sub> (g) + Cl2<\/sub> (g) \u2192 2 HCl (g)
                          \nAs per the stoichiometric equation, under given conditions,
                          \nTo produce 2 moles of HCl, 1 mole of chlorine gas is required.
                          \nTo produce 44.8 liters of HCl, 22.4 liters of chlorine gas are required
                          \n\u2234 To produce 11.2 liters of HCl,
                          \n\"Samacheer
                          \n= 5.6 litres of chlorine are required.<\/p>\n

                          Question 5.
                          \nCalculate the percentage composition of the elements present in magnesium carbonate. How many kilogram of CO2<\/sub> can be obtained by heating 1 kg of 90 % pure magnesium carbonate.
                          \nAnswer:
                          \nThe balanced chemical equation is
                          \n\"Samacheer
                          \nMolar mass of MgCO3<\/sub>\u00a0is 84 g mol-1<\/sup>.
                          \n84 g MgCO3<\/sub> contain 24 g of Magnesium.
                          \n\u2234 100 MgCO3<\/sub> contain
                          \n\"Samacheer
                          \n= 28.57 g Mg i.e. percentage of magnesium = 28.57 %.
                          \n84 g MgCO3<\/sub> contain 12 g of carbon
                          \n\"Samacheer
                          \n\"Samacheer
                          \n= 14.29 g of carbon
                          \nPercentage of carbon = 14.29 %.
                          \n84 g MgCO3<\/sub> contain 48 g of oxygen
                          \n\u2234 100 g MgCO3<\/sub> contains
                          \n\"Samacheer
                          \n= 57.14 g of oxygen.
                          \n\u2234 Percentage of oxygen = 57.14 %.
                          \nAs per the stoichiometric equation,
                          \n84 g of 100 % pure Mg CO3<\/sub> on heating gives 44 g of CO2<\/sub>.
                          \n\u22341000 g of 90 % pure Mg CO3<\/sub> gives
                          \n\\(\\frac {44 g}{84 g x 100 %}\\) x 90 % x 1000 g
                          \n= 471.43 g of CO2<\/sub>
                          \n= 0. 471 kg of CO2<\/sub><\/p>\n

                          Question 6.
                          \n\"Samacheer
                          \n(1) If the entire quantity of all the reactants is not consumed in the reaction which is the limiting reagent ?
                          \n(2) Calculate the quantity of urea formed and un reacted quantity of the excess reagent. The balanced equation is
                          \n\"Samacheer
                          \nAnswer:
                          \n(1) The entire quantity of ammonia is consumed in the reaction. So ammonia is the limiting reagent. Some quantity of CO2<\/sub> remains unreacted, so CO2<\/sub> is the excess reagent.<\/p>\n

                          (2) Quantity of urea formed = number of moles of urea formed x molar mass of urea
                          \n= 19 moles x 60 g mol-1<\/sup>
                          \n= 1140 g = 1.14 kg
                          \nExcess reagent leftover at the end of the reaction is carbon dioxide. Amount of carbon dioxide leftover = number of moles of CO2<\/sub> left over x molar mass of CO2<\/sub>
                          \n= 7 moles x 44 g mol-1<\/sup> = 308 g.
                          \n\"Samacheer<\/p>\n

                          Samacheer Kalvi 11th Chemistry Solutions Basic Concepts of Chemistry and Chemical Calculations Additional Questions Solved<\/span><\/h3>\n

                          I. Choose the correct answer from the following<\/span><\/p>\n

                          Question 1.
                          \nWhich one of the following is the standard for atomic mass?
                          \n(a) 1<\/sub>H1<\/sup>
                          \n(b) 6<\/sub>6C12<\/sup>
                          \n(c) 6<\/sub>C14<\/sup>
                          \n(d) 8<\/sub>O16<\/sup>
                          \nAnswer:
                          \n(b) 6<\/sub>6C12<\/sup>
                          \nHint:
                          \nStandard element used to determine atomic mass is 6 Cl2<\/sub><\/p>\n

                          Question 2.
                          \nOne mole of CO2<\/sub> contains ………….
                          \n(a) 6.023 x 1023<\/sup> atoms of C
                          \n(b) 6.023 x 1023 <\/sup>atoms of O
                          \n(c) 18.1 x 1023<\/sup> molecules of CO2<\/sub>
                          \n(d) 3g atoms of CO2<\/sub>
                          \nAnswer:
                          \n(a) 6.023 x 1023<\/sup> atoms of C
                          \nHint:
                          \nOne mole of any substance contains Avogadro number of atoms. In this carbon one mole is present and oxygen two atoms are present. So, 6.023 x 1023<\/sup> atoms of C is correct.<\/p>\n

                          Question 3.
                          \nThe largest number of molecules is in
                          \n(a) 54 g of nitrogen pent oxide
                          \n(b) 28 g of carbon dioxide
                          \n(c) 36 g of water
                          \n(d) 46 g of ethyl alcohol
                          \nAnswer:
                          \n(c) 36 g of water
                          \nHint:(a) 54 g of N2<\/sub>O5<\/sub>
                          \nN2<\/sub>O5<\/sub> = Molecular mass = 28 + 80 = 108
                          \n108 g of N2<\/sub>O5<\/sub> contains 6.023 x 1023<\/sup> molecules.
                          \n\u2234 54 g of N2<\/sub>O5<\/sub> will contain \\(\\frac{6.023 \\times 10^{23}}{108}\\) x 54 = 3.0115 x 1023<\/sup> molecules.<\/p>\n

                          (b) 28 g of CO2<\/sub>
                          \nCO2<\/sub> = Molecular mass = 12 + 32 = 44
                          \n44 g of CO2<\/sub> contains 6.023 x 1023<\/sup> molecules.
                          \n\u2234 28 g of CO2<\/sub> will contain \\(\\frac{6.023 \\times 10^{23}}{44}\\) x 28 = 3.832 x 1023<\/sup> molecules.<\/p>\n

                          (c) 36 g of H2<\/sub>O
                          \nH2<\/sub>O = Molecular mass = 2 + 16 = 18
                          \n18 g of H2<\/sub>O contains 6.023 x 1023<\/sup> molecules.
                          \n\u2234 36 g of H2<\/sub>O will contain \\(\\frac{6.023 \\times 10^{23}}{18}\\) x 36 = 12.046 X 1023<\/sup>molecules.<\/p>\n

                          (d) 46 g of C2<\/sub>H5<\/sub>OH
                          \nC2<\/sub>H5<\/sub>OH = Molecular mass = 24 + 6 + 16 = 46
                          \n46 g of C2<\/sub>H5<\/sub>OH contains 6.023 x 1023<\/sup> molecules.
                          \nSo, among the 4, 36 g of water contain the largest number of molecules as 12.046 x 1023<\/sup>.<\/p>\n

                          Question 4.
                          \nThe number of moles of H2<\/sub> in 0.224 liter of hydrogen gas at STP is …………..
                          \n(a) 1
                          \n(b) 0.1
                          \n(c) 0.01
                          \n(d) 0.001
                          \nAnswer:
                          \n(c) 0.01
                          \n22.4 liter of hydrogen gas at STP contains 1 mole.
                          \n\u2234 0.224 liter of hydrogen gas at STP will contain \\(\\frac{1}{22.4}\\) x 0.224 = 0.01<\/p>\n

                          Question 5.
                          \n10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be ………..
                          \n(a) 3 mol
                          \n(b) 4 mol
                          \n(c) 1 mol
                          \n(d) 2 mol
                          \nAnswer:
                          \n(b) 4 mol
                          \n10 g of H2<\/sub> + 64 g of O2<\/sub> \u2192 water
                          \n2H2<\/sub> + O2<\/sub> \u2192 2H2<\/sub>O
                          \n4 g of hydrogen react with 32 g of oxygen. So 10 g of hydrogen will react with 80 g of oxygen. But we have given amount of oxygen only 64 g. It means, here oxygen is the limiting agent. Now all the oxygen react with 8 g of hydrogen and form 4 moles of water.<\/p>\n

                          Question 6.
                          \n6.023 x 1020<\/sup> molecules of urea are present in 100 ml of its solution. The concentration of the solution is –
                          \n(a) 0.02 M
                          \n(b) 0.1 M
                          \n(c) 0.01 M
                          \n(d) 0.001 M
                          \nAnswer:
                          \n(c) 0.01 M
                          \n100 ml of solution contain 6.023 x 1020<\/sup> molecules.
                          \n6.023 x 1023<\/sup> molecules in 1000 ml = 1 M.
                          \n\u2234 6.023 x 1020<\/sup> molecules in 1000 ml =\u00a0 \"Samacheer
                          \n= 1020<\/sup> x 10-23<\/sup> = 10-3<\/sup>
                          \n10-3 <\/sup>moles present in 100 ml.
                          \n\u2234 In 1000 ml the moles present is = \\(\\frac{10^{-3}}{100}\\) x 1000 = 10-2<\/sup> moles
                          \nConcentration = 0.01 M<\/p>\n

                          Question 7.
                          \nTwo containers A and 13 of equal volume contain 6 g of O2<\/sub> and SO2<\/sub> at 300 K and 1 atm. Then
                          \n(a) No. of molecules in A is less than that in B
                          \n(b) No. of molecules in A is more than that in B
                          \n(c) No. of molecules in A and B are same
                          \n(d) none of these
                          \nAnswer:
                          \n(b) No, of molecules in A is more than that in B
                          \nO2<\/sub> = Mass = 6 g
                          \nMolar mass = 32 g
                          \n32 g of O2<\/sub> contains 6.023 x 1023 molecules.
                          \n\u2234 6 g of O2<\/sub> will contain = 6.023 X 1023 x 6 = 1.129 x 1023<\/sup> molecules.
                          \nSO2<\/sub> = Mass = 6 g
                          \nMolar mass = 64 g
                          \n64 g of SO2<\/sub> contains 6.023 x 1023<\/sup> molecules.
                          \n\u2234 6 g of SO2<\/sub> will contain 6.023 x 1023 x 6 = 0.5646 x 1023 molecules.
                          \n\u2234 Number of molecules in A is more than that in B.<\/p>\n

                          Question 8.
                          \nThe number of molecules in 16 g of methane is ……….
                          \n(a) 3.023 x 1023<\/sup>
                          \n(b) 6.023 x 1023<\/sup>
                          \n(c) 16 \/ 6.023 x 1023<\/sup>
                          \n(d) 6.023 \/ 3 x 1023<\/sup>
                          \nAnswer:
                          \n(b) 6.023 x 1023<\/sup>
                          \nHint:
                          \nMethane: CH4<\/sub>
                          \nMolecular mass 12 + 4 = 16
                          \n16 g of methane contains Avogadro number of molecules 6.023 x 1023<\/sup> molecules.<\/p>\n

                          Question 9.
                          \nNumber of atoms in 4.25 g of ammonia is …………..
                          \n(a) 1 x 1023<\/sup>
                          \n(b) 2 x 1023<\/sup>
                          \n(e) 4 x 1023<\/sup>
                          \n(d) 6 x 1023<\/sup>
                          \nAnswer:
                          \n(d) 6 x 1023<\/sup>
                          \nHint:
                          \nAmmonia = NH3<\/sub> (4 atoms)
                          \nMolecular mass = 14 + 3 = 17
                          \n17 g of Ammonia contains 6.023 X 1023<\/sup> atoms.
                          \n\u2234 4.25 g of Ammonia will contain 6.0231; 1023 x 4.25 = 1.5055 x 1023<\/sup> molecules.
                          \n\u2234 1.5055 x 1023<\/sup> molecules will contain 4 x 1.5 x 1023<\/sup> = 6 x 1023<\/sup> molecules.<\/p>\n

                          Question 10.
                          \nThe number of molecules in a drop of water (0.0018 ml) at room temperature is ……….
                          \n(a) 6.02 x 1023<\/sup>
                          \n(b) 1.084 x 1023<\/sup>
                          \n(c) 4.84 x 1023<\/sup>
                          \n(d) 6.02 x 1023<\/sup>
                          \nAnswer:
                          \n(a) 6.02 x 1023<\/sup>
                          \nHint:
                          \n0.0018 ml – drop of water = 0.0018 g
                          \nH2<\/sub>O = molecular mass = 18 g.
                          \nNumber of molecules in 18 g = 6.023 x 1023<\/sup>
                          \n\u2234 Number of molecules in 0.0018 g 6.023 x 1023 x 0.0018 = 6.023 x 1023<\/sup> x 105<\/sup>
                          \n= 6.023 x l019<\/sup> molecules.
                          \nor
                          \nDensity of water at 25\u00b0C = 997.0479 g \/ L
                          \nMass of 0.0018 ml (or) 0.00 18 x 10-3<\/sup> L
                          \n= D x V
                          \n= 997.05 x 0.0018 x 10-3<\/sup>
                          \n= 1.795 x 10-3<\/sup> g
                          \nMolar mass of water = 18 g
                          \nMole = \\(\\frac{Mass}{Molecular mass}\\) = \\(\\frac{1.795 \\times 10^{-3}}{18}\\)
                          \n= 9.971 x 10-5<\/sup>
                          \n\u2234 Number of molecules in 0.0018 ml = moles x Avogadro number
                          \n= 9.971 x 10-5<\/sup> x 6.023 x 1023<\/sup>
                          \n= 6 x 1019<\/sup> molecules.<\/p>\n

                          Question 11.
                          \n7.5 g of a gas occupy 5.6 liters of volume at STP. The gas ……….
                          \n(a) NO
                          \n(b) N2<\/sub>O
                          \n(c) CO
                          \n(d) CO2<\/sub>
                          \nAnswer:
                          \n(a) NO
                          \nHint:
                          \n22.4 liters = 1 mole
                          \n5.6 liters = \\(\\frac {1}{22.4}\\) x 5.6 = 0.25 mole.
                          \nNO = molar mass = 14 + 16 = 30 g = 1 mole
                          \nN2<\/sub>O = molar mass = 28 + 16 = 44g = 1 mole
                          \nCO = molar mass = 12 + 16 = 28 g = 1 mole
                          \nCO2<\/sub>=molar mass = 12 + 32 = 44g = 1 mole
                          \nAmong the four gases, 0.25 mole = 7.5 g is equal to NO gas.<\/p>\n

                          Question 12.
                          \nThe mass in grams of 0.45 mole of CO2<\/sub> ions ……….
                          \n(a) 1.8
                          \n(b) 40
                          \n(c) 36
                          \n(d) 18
                          \nAnswer:
                          \n(d) 18
                          \nHint:
                          \nCa = Atomic mass = 40
                          \nCa \u2192 Ca2+<\/sup> + 2e–<\/sup>
                          \n41 g of Ca = 1 mole (for Ca2+<\/sup> Atomic mass remains same)
                          \n1 mole of Ca2+<\/sup> = 40 g
                          \n\u2234 0.45 mole of Ca2+<\/sup> = \\(\\frac {40}{1}\\) x 0.45 = 18g<\/p>\n

                          Question 13.
                          \nThe mass of one molecule of HI in grams is ……….
                          \n(a) 2.125 x 10-22<\/sup>
                          \n(b) 128
                          \n(c) 127
                          \n(d) 6.02 x 10-23<\/sup>
                          \nAnswer:
                          \n(b) 128
                          \nHint:
                          \nHI = 1 mole = 1 + 127 = 128 g<\/p>\n

                          Question 14.
                          \nAvogadro\u2019s number is the number of molecules present in ……….
                          \n(a) 1 g of molecule
                          \n(b) 1 g atom of molecule
                          \n(c) gram molecular mass
                          \n(d) I lit of molecule
                          \nAnswer:
                          \n(c) gram molecular mass
                          \nby definition (c) is correct<\/p>\n

                          Question 15.
                          \nWhich of the following contains same number of carbon atoms as are in 6.0 g of carbon (C-12)?
                          \n(a) 6.0 g ethane
                          \n(b) 8.0 g methane
                          \n(c) 21.0 g Propane
                          \n(d) 28.0 g CO
                          \nAnswer:
                          \n(b) 8.0 g methane<\/p>\n

                          Hint:
                          \n(a) 6.0 g of ethane (C2<\/sub>H6<\/sub>)
                          \nC2<\/sub>H6<\/sub> = molar mass = 24 + 6 = 30 g
                          \n30 g of ethane contains 2 x 6.023 x 1023<\/sup> Carbon atoms.<\/p>\n

                          (b) 8.0 g of methane (CH4<\/sub>)
                          \nCH4<\/sub> molar mass = 12 + 4 = 16g
                          \n16 g of methane contains 6.023 x 1023<\/sup> Carbon atoms.<\/p>\n

                          (c) 21.0 g of propane (C3<\/sub>H8<\/sub>)
                          \nC3<\/sub>H8<\/sub> = molar mass = 36 + 8 = 44 g
                          \n44 g of propane contains 3 x 6.023 x 1023<\/sup> Carbon atoms.<\/p>\n

                          (d) 28.0 g of Carbon monoxide (CO)
                          \nCO = molar mass = 12+ 16 = 28 g
                          \n28 g of Carbon monoxide contains 6.023 x 1023<\/sup> Carbon atoms.
                          \n6.0 g of Carbon contains = 6.023 x 10 x 6 = 3.0115 x 1023<\/sup> Carbon atoms.
                          \nAmong the (a), (b), (c), (d) – 8 g of CH4<\/sub> contains x 8 = 3.0115 x 1023<\/sup>\u00a0Carbon atoms.<\/p>\n

                          Question 16.
                          \nEquivalent mass of KMnO4<\/sub> when it is converted to MnSO4<\/sub> is equal to molar mass divided by ………..
                          \n(a) 6
                          \n(b) 4
                          \n(c) 5
                          \n(d) 2
                          \nAnswer:
                          \n(c) 5
                          \nHint:
                          \n\"Samacheer<\/p>\n

                          Question 17.
                          \nHow many equivalents of Sodium sulphate is formed when Sulphuric acid is completelyn neutralized with a base NaOH?
                          \n(a) 0.2
                          \n(b) 2
                          \n(e) 0.1
                          \n(d) 1
                          \nAnswer:
                          \n(d) 1
                          \nHint:
                          \n\"Samacheer<\/p>\n

                          Question 18.
                          \nCl2<\/sub> changes to Cl–<\/sup> and ClO–<\/sup> in cold NaOH. Equivalent mass of Cl2<\/sub> will be ………..
                          \n(a) Molar mass \/ 2
                          \n(b) Molar mass \/ 1
                          \n(c) Molar mass \/ 3
                          \n(d) 2 x Molar mass \/ 2
                          \nAnswer:
                          \n(a) Molar mass \/ 2
                          \n\"Samacheer<\/p>\n

                          Question 19.
                          \nEquivalent mass of KMnO4<\/sub> in acidic medium, concentrated alkaline medium and dilute basic medium respectively are M, M, M. Reduced products can be ……………
                          \n(a) MnO2<\/sub>, MnO2<\/sub>2-<\/sup>, Mn2+<\/sup>
                          \n(b) MnO2<\/sub>, Mn2+<\/sup>, MnO4<\/sub>2-<\/sup>
                          \n(c) Mn2+<\/sup>, MnO2<\/sub>, MnO4<\/sub>2-<\/sup>
                          \n(d) Mn2+<\/sup>, MnO4<\/sub>2-<\/sup>, MnO2<\/sub>
                          \nAnswer:
                          \n(c) Mn2+<\/sup>, MnO2<\/sub>, MnO4<\/sub>2-<\/sup>
                          \nHint:
                          \nMnO4<\/sub>–<\/sup> + 8H+<\/sup> + 5e–<\/sup> \u2192 Mn2+<\/sup> + 4H2<\/sub>O (acidic medium)
                          \nMnO4<\/sub> + 4H+<\/sup> + 3e–<\/sup> \u2192 MnO2<\/sub> + 2H2<\/sub>O (concentrated basic medium)
                          \nMnO4<\/sub> + e–<\/sup> \u2192 MnO4<\/sub>2-<\/sup> (dilute basic medium)<\/p>\n

                          Question 20.
                          \nThe empirical formula of hydrogen peroxide is
                          \n(a) HO
                          \n(b) H2<\/sub>O
                          \n(e) H3<\/sub>O
                          \n(d) H2<\/sub>O2<\/sub>
                          \nAnswer:
                          \n(a) HO
                          \nHint:
                          \nMolecular formula of hydrogen peroxide = H2<\/sub>O2<\/sub>
                          \nH2<\/sub>O2<\/sub> \u00f7 2 = HO = Empirical formula.<\/p>\n

                          Question 21.
                          \nMolecular mass =
                          \n(a) Vapour Density \u00d7 2
                          \n(b) Vapour Density \u00f7 2
                          \n(c) Vapour Density \u00d7 3
                          \n(d) Vapour Density
                          \nAnswer:
                          \n(a) Vapour Density \u00d7 2<\/p>\n

                          Question 22.
                          \n20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage of purity of magnesium carbonate in the sample?
                          \n(a) 60
                          \n(b) 84
                          \n(e) 75
                          \n(d) 96
                          \nAnswer:
                          \n(b) 84
                          \nHint:
                          \n\"Samacheer
                          \n84 g of MgCO3<\/sub> gives 40 g of MgO. (100% purity)
                          \n20 g of MgCO3<\/sub> will give = \\(\\frac{40 \\times 20}{84}\\) = 9.52 g
                          \n9.52 g of MgO is given by 100% pure MgCO3<\/sub>.
                          \n8.0 g of MgO will be given by = \\(\\frac{100 \\times 8}{9.52}\\) = 84.03%<\/p>\n

                          Question 23.
                          \nWhat is the mass of the precipitate formed when the preparation of alkyl halides 50 ml of 16.9 % solution of AgNO3<\/sub> is mixed with 50 ml of 5.8 % NaCl solution?
                          \n(a) 7 g
                          \n(b) 14 g
                          \n(c) 28 g
                          \n(d) 35 g
                          \nAnswer:
                          \n(a) 7 g
                          \nHint:
                          \n\"Samacheer<\/p>\n

                          Question 24.
                          \nWhen 22 L of hydrogen gas is mixed with 11.2 L of chlorine gas, each at STP, the moles of HCl gas formed is equal to ……….
                          \n(a) 2
                          \n(b) 0.5
                          \n(c) 1.5
                          \n(d) 1
                          \nAnswer:
                          \n(d) 1
                          \nHint:
                          \nH2<\/sub> + Cl2<\/sub> \u2192 2HCl
                          \n22 liters + 11.2 litres
                          \n1 mole + \\(\\frac {1}{2}\\) mole = 1 mole of HCl and \\(\\frac {1}{2}\\) mole of H2<\/sub> is remained.<\/p>\n

                          Question 25.
                          \n5.6 L of a gas at STP are found to have mass of 11 g. The molecular mass of the gas is
                          \n(a) 36
                          \n(b) 48
                          \n(c) 40
                          \n(d) 44
                          \nAnswer:
                          \n(d)44
                          \nHint:
                          \n5.6 L of gas has the mass of 11 g.
                          \n\u2234 22.4 L of gas will have the mass \\(\\frac {11}{5.6}\\) x 22.4 = 44<\/p>\n

                          Question 26.
                          \nOxidation number f Fluorine in all compounds is ……….
                          \n(a) + 1
                          \n(b) -1
                          \n(c) 0
                          \n(d) – 2
                          \nAnswer:
                          \n(b) – 1<\/p>\n

                          Question 27.
                          \nIn redox reaction which of the following is true?
                          \n(a) Number of electrons lost is more than number of electrons gained
                          \n(b) Number of electrons lost is less than number of electrons gained
                          \n(c) Number of electrons lost s equal number of electrons gained
                          \n(d) No transfer and gain of electrons during the reaction.
                          \nAnswer:
                          \n(c) Number of electrons lost is equal number of electrons gained<\/p>\n

                          Question 28.
                          \nWhich of the following is a mono-atomic molecule?
                          \n(a) Hydrogen
                          \n(b) Oxygen
                          \n(c) Sodium
                          \n(d) Ozone
                          \nAnswer:
                          \n(c) Sodium<\/p>\n

                          Question 29.
                          \nWhich one of the following is a diatomic molecule?
                          \n(a) Ozone
                          \n(b) Copper
                          \n(c) Hydrogen
                          \n(d) Gold
                          \nAnswer:
                          \n(c) Hydrogen<\/p>\n

                          Question 30.
                          \nThe value of Avogadro Number N is equal to ……….
                          \n(a) 2.24 x 10-2<\/sup>L
                          \n(b) 22400 cm3<\/sup>
                          \n(c) 6.023 x 10-23<\/sup>
                          \n(d) 6.023 x 1023<\/sup>
                          \nAnswer:
                          \n(d) 6.023 x 1023<\/sup><\/p>\n

                          Question 31.
                          \n46 g of ethanol contains ……….
                          \n(a) 2 x 6.023 x 1023<\/sup> C atoms
                          \n(b) 3 x 6.023 x 1023<\/sup> atoms
                          \n(c) 9 x 6.023 x 1023<\/sup> H atoms
                          \n(d) 6.023 x 1023<\/sup> Carbon atoms
                          \nAnswer:
                          \n(a) 2 x 6.023 x 1023<\/sup> C atoms
                          \nHint:
                          \nC2<\/sub>H5<\/sub>OH = Molecular mass = (12 x 2) + (1 x 6) + (1 x 16)
                          \n=24 + 6 + 16 = 46
                          \n2 Carbon atoms are present.
                          \n\u2234 2 x 6.023 x 1023<\/sup> C atoms is correct.<\/p>\n

                          Question 32.
                          \nThe mass of one mole of CaCl2<\/sub> is ……….
                          \n(a) 55.5 g mol-1<\/sup>
                          \n(b) 111 g mol-1<\/sup>
                          \n(c) 222 g mol -1<\/sup>
                          \n(d) 77.5 g mol-1<\/sup>
                          \nAnswer:
                          \n(b) 111 g mol-1
                          \n<\/sup>Hint:
                          \nCaCl2<\/sub> = Mass = 40 + 71 = 111 g mol-1<\/sup><\/p>\n

                          Question 33.
                          \n22 g of CO2<\/sub> contains molecules of CO2 <\/sub>
                          \n(a) 6.023 x 1023<\/sup>
                          \n(b) 6.023 x 1023<\/sup>
                          \n(c) 3.0115 x 1023<\/sup>
                          \n(d) 3.0115 x 1023<\/sup>
                          \nAnswer:
                          \n(c) 3.0115 x 1023<\/sup>
                          \nHint:
                          \n44 g of CO2 <\/sub>contains 6.023 x 1023<\/sup> molecules.
                          \n\u2234 22 g of CO2<\/sub> will contain = \\(\\frac{6.023 \\times 10^{23}}{44}\\) x 22 = 3.0115 x 1023<\/sup><\/p>\n

                          Question 34.
                          \nThe formula weight of ethanol (C2<\/sub>H5<\/sub>OH) is ……….
                          \n(a) 56.5 amu
                          \n(b) 16 amu
                          \n(c) 60 amu
                          \n(d) 46 amu
                          \nAnswer:
                          \n(d) 46 amu
                          \nHint:
                          \nFormula weight of C2<\/sub>H5<\/sub>OH = (12 x 2) + (6 x 1) + (1 x 16)
                          \n= 24 + 6 + 16 = 46 amu<\/p>\n

                          Question 35.
                          \nThe number of moles of ethane in 60 g is ……….
                          \n(a) 2
                          \n(b) 4
                          \n(c) 0.5
                          \n(d) 1
                          \nAnswer:
                          \n(a) 2
                          \nHint:
                          \nC2<\/sub>H6<\/sub> – Ethane – molar mass = 24 + 6 = 30 g
                          \n30 g of C2<\/sub>H6<\/sub> contains = 1 mole.
                          \n\u2234 60 g of C2<\/sub>H6<\/sub> will contains = x 1 = 2 moles.<\/p>\n

                          Question 36.
                          \nWhich of the following method is used to prevent rusting of iron?
                          \n(a) Galvanization
                          \n(b) Painting
                          \n(c) Chrome plating
                          \n(d) all the above
                          \nAnswer:
                          \n(d) all the above<\/p>\n

                          Question 37.
                          \nWhich of the following is not a redox reaction?
                          \n(a) H2<\/sub> + F2<\/sub> \u2192 2HF
                          \n(b) Cu + 4HNO3<\/sub> \u2192 Cu(NO3<\/sub>)2<\/sub> + 2NO2<\/sub> + 2H2<\/sub>O
                          \n(c) 2H2<\/sub> + O2<\/sub> \u2192 2H2<\/sub>O
                          \n(d) AgCl + NH3<\/sub> \u2192 [Ag(NH3<\/sub>)2<\/sub>]Cl
                          \nAnswer:
                          \n(d) AgCl + NH3<\/sub> \u2192 [Ag(NH3<\/sub>)2<\/sub>]Cl<\/p>\n

                          Question 38.
                          \nHow many H2<\/sub>O molecules are there in a snowflake weighing 1 mg?
                          \n(a) 3.35 x 1019<\/sup>
                          \n(b) 6.023 x 1023<\/sup>
                          \n(c) 335 x 10-19<\/sup>
                          \n(d) 100
                          \nAnswer:
                          \n(a) 3.35 x 1019<\/sup>
                          \nHint:
                          \n1 mg of H2<\/sub>O
                          \nMolar mass of H2<\/sub>O = 2 + 16 = 18 g
                          \n1 Mole contains 6.023 x 1023<\/sup> water molecules
                          \n18 g contains 6.023 x 1023<\/sup> water molecules
                          \nl mg contains \\(\\frac{6.023 \\times 10^{23}}{18}\\) x 1g \/1000 mg = 3.35 x 10-19<\/sup> H2<\/sub>O molecule.<\/p>\n

                          Question 39.
                          \nThe volume of HCl gas weighing 73 g at STP is ……….
                          \n(a) 2.24 x 10-2<\/sup> m3<\/sup>
                          \n(b) 4.48 x 10-2<\/sup> m3<\/sup>
                          \n(c) 4.48 x 102<\/sup>m3<\/sup>
                          \n(d) 2.24 x 102<\/sup>m3<\/sup>
                          \nAnswer:
                          \n(b) 4.48 x 10-2<\/sup> m3<\/sup>
                          \nHint:
                          \nHCl = Molar mass = 1 + 35.5 = 36.5 g
                          \n36.5 g of HCl occupies 2.24 x 10-2<\/sup>m3<\/sup>
                          \n\u2234 73 g of HCl at STP will occupy \\(\\frac{2.24 \\times 10^{-2}}{36.5}\\) x 73 = 4.48 x 10-2<\/sup>m3<\/sup><\/p>\n

                          Question 40.
                          \nThe molar volume of 22 g of CO2<\/sub> is ……….
                          \n(a)2.24 x 10-2<\/sup>m3<\/sup>
                          \n(b)4.48 x 10-2<\/sup>m3<\/sup>
                          \n(c) 1.12 x 10-2<\/sup>m3<\/sup>
                          \n(d)2.24 x 10-2<\/sup>m3<\/sup>
                          \nAnswer:
                          \n(c) 1.12 x 10-2<\/sup>m3<\/sup>
                          \nHint:
                          \nCO2<\/sub> = Molar\u2019mass of 12 + 32 = 44 g
                          \n44 g of CO2<\/sub> occupies molar volume = 2.24 x 10-2<\/sup>m3<\/sup>
                          \n\u2234 2g of CO2<\/sub> will occupy = \\(\\frac{2.24 \\times 10^{-2}}{44}\\) x 2= 1.12 x 10-2<\/sup> L<\/p>\n

                          Question 41.
                          \nThe equivalent mass of Aluminium is ……….
                          \n(a) 27
                          \n(b) 13.5
                          \n(c) 54
                          \n(d) 9
                          \nAnswer:
                          \n(d) 9<\/p>\n

                          Question 42.
                          \nThe equivalent mass of HSO4<\/sub> is ……….
                          \n(a) 98
                          \n(b) 97
                          \n(c) 48
                          \n(d) 96
                          \nAnswer:
                          \n(a) 98
                          \nHint:
                          \nHSO4<\/sub> = Molar mass= 1 + 32 + 64 + 1 = 98
                          \nEquivalent mass = Molar mass \/ 1 = 98<\/p>\n

                          Question 43.
                          \nThe equivalent mass of NaCl is ……….
                          \n(a) 40
                          \n(b) 58.5
                          \n(c) 35.5
                          \n(d) 23
                          \nAnswer:
                          \n(b) 58.5
                          \nHint:
                          \nNaCl = Salt Molar mass 23 + 35.5 = 58.5
                          \nEquivalent mass of Salt = Molar mass of Salt.<\/p>\n

                          Question 44.
                          \nMatch the List-I and List-lI using the correct code given below the list.
                          \n\"Samacheer
                          \nAnswer:
                          \n\"Samacheer<\/p>\n

                          Question 45.
                          \nMatch the List-I and List-Il using the correct code given below the list.
                          \n\"Samacheer
                          \n\"Samacheer
                          \nAnswer:
                          \n\"Samacheer<\/p>\n

                          Question 46.
                          \nConsider the following statements ……….
                          \n(i) Empirical formula shows the actual number of atoms of different elements in one molecule of the compound.
                          \n(ii) Ozone is a diatomic molecule.
                          \n(iii) Gases are easily compressible.
                          \nWhich of the above statement is\/are not correct?
                          \n(a) (i), (ii), (iii)
                          \n(b) (i) & (ii)
                          \n(c) (ii) & (iii)
                          \n(d) (iii) only
                          \nAnswer:
                          \n(b) (i) & (ii)<\/p>\n

                          Question 47.
                          \nHow many molecules of hydrogen is required to produce 4 moles of ammonia?
                          \n(a) 15 moles
                          \n(b) 20 moles
                          \n(c) 6 moles
                          \n(d) 4 moles
                          \nAnswer:
                          \n(c) 6 moles
                          \nHint:
                          \n3H2<\/sub> + N2<\/sub> \u2192 2NH3<\/sub>
                          \nTo get 2 moles of ammonia, 3 mole of H2<\/sub> is required.
                          \nTo get 4 moles of ammonia = \\(\\frac {3}{2}\\), x 4
                          \n= 6 moles of H2<\/sub> is required.<\/p>\n

                          Question 48.
                          \nThe number of moles of oxygen required to prepare 1 mole of water is …………..
                          \n(a) I mole
                          \n(b) 0.5 mole
                          \n(c) 2 moles
                          \n(d) 0.4 mole
                          \nAnswer:
                          \n(b) 0.5 mole
                          \nHint:
                          \n\"Samacheer
                          \n0.5 mole of oxygen is required to prepare 1 mole of H2<\/sub>O.<\/p>\n

                          Question 49.
                          \nHow much volume of CO2<\/sub> is produced when 50 g of CaCO3<\/sub> is heated strongly?
                          \n(a) 2.24 x 10-2<\/sup> m3<\/sup>
                          \n(b) 22.4
                          \n(c) 11.2 L
                          \n(d) 22400 cm3<\/sup>
                          \nAnswer:
                          \n(c) 11.2 L
                          \n\"Samacheer
                          \n100 g of CaCO3<\/sub> produces 22.4 litres of CO2<\/sub>.
                          \n50 g of CaCO3<\/sub> will produce = \\(\\frac{22.4}{100} \\times 50\\) = 11.2 litres<\/p>\n

                          Question 50.
                          \nWhich one of the following is not a redox reaction?
                          \n(a) Rusting of iron
                          \n(b) Extraction of metal Na
                          \n(c) Electroplating
                          \n(d) Aluminothermic process
                          \nAnswer:
                          \n(a) Rusting of Iron<\/p>\n

                          Question 51.
                          \nIn the reaction 2 AuCl3<\/sub> + 3 SnCl2<\/sub> \u2192 2 Au + 3 SnCl4<\/sub> which is an oxidising agent?
                          \n(a) AuCl3<\/sub>
                          \n(b) Au
                          \n(c) SnCl2<\/sub>
                          \n(d) Both AuCl3<\/sub> and SnCl2<\/sub>
                          \nAnswer:
                          \n(a) AuCl3<\/sub>
                          \nHint:
                          \nAuCl3 <\/sub>undergoes reduction. So, it is an oxidising agent.<\/p>\n

                          Question 52.
                          \nIdentify the compound formed during the rusting of iron.
                          \n(a) Fe2<\/sub>O3<\/sub>
                          \n(b) Fe2<\/sub>O3<\/sub>. x H2<\/sub>O
                          \n(c) FeO. x H2<\/sub>O
                          \n(d) FeO
                          \nAnswer:
                          \n(b) Fe2<\/sub>O3<\/sub>. x H2<\/sub>O – Hydrated iron oxide is rust.<\/p>\n

                          Question 53.
                          \nThe oxidation state of a substance in its elementary state is equal to ………..
                          \n(a) -1
                          \n(b) -2
                          \n(c) zero
                          \n(d) charge of the ion
                          \nAnswer:
                          \n(c) zero<\/p>\n

                          Question 54.
                          \nThe oxidation number of fluorine in all its compounds is equal to ………….
                          \n(a) -1
                          \n(b) +1
                          \n(c) – 2
                          \n(d) +2
                          \nAnswer:
                          \n(a) -1<\/p>\n

                          Question 55.
                          \nConsider the following statements.
                          \n(i) The sum of the oxidation number of all the atoms in neutral molecule is equal to zero.
                          \n(ii) Fluorine has an oxidation number +1 in all its compounds. .
                          \n(iii) The oxidation number of a substance in its elementary state is equal to zero.
                          \nWhich of the above statement is\/are not correct?
                          \n(a) (i), (ii) & (iii)
                          \n(b) (ii) & (iii)
                          \n(c) (i) only
                          \n(d) (ii) only
                          \nAnswer:
                          \n(d) (ii) only<\/p>\n

                          Question 56.
                          \nThe oxidation number of Cr in K2<\/sub>Cr2<\/sub>O7<\/sub> is ………….
                          \n(a) +4
                          \n(b) + 6
                          \n(c)\u00a0 O
                          \n(d) + 7
                          \nAnswer:
                          \n(b) + 6
                          \nK2<\/sub>Cr2<\/sub>O2<\/sub>
                          \n2 + 2x – 140
                          \n2x – 12 = 0
                          \n2x = + 12
                          \nx = + 6<\/p>\n

                          Question 57.
                          \nThe oxidation number of N in NH4<\/sub> ion is ……….
                          \n(a) +4
                          \n(b) + 3
                          \n(c) – 3
                          \n(d) – 4
                          \nAnswer:
                          \n(c) – 3
                          \nNH2<\/sub>+<\/sup>
                          \nx + 4 = + 1
                          \nx = + 1 – 4
                          \nx = – 3<\/p>\n

                          Question 58.
                          \nZn(s)<\/sub> + Cu2+<\/sup>(aq)<\/sub> \u2192 Zn2+<\/sup>(aq)<\/sub> + Cu(s)<\/sub>. In this reaction, which gets oxidised?
                          \n(a) Cu2+<\/sup>
                          \n(b) Zn2+<\/sup>
                          \n(c) Zn
                          \n(d) Zn, Cu2+<\/sup>
                          \nAnswer:
                          \n(c) Zn
                          \nHint:
                          \nZn \u2192 Zn2+<\/sup> + 2e–<\/sup> (loss of electron = oxidation)
                          \nZn gets oxidised<\/p>\n

                          Question 59.
                          \nWhich one of the following is an example for disproportionation reaction?
                          \n(a) CuSO4<\/sub> + Zn \u2192 ZnSO4<\/sub>\u00a0+ Cu
                          \n(b) 2KClO3<\/sub> \u2192 2KCI + 3O2<\/sub>
                          \n(c) PCl5<\/sub> \u2192 PCl3<\/sub> + Cl2<\/sub>
                          \n(d) 4H3<\/sub>PO3<\/sub> \u2192 3H3<\/sub>PO4<\/sub> + PH3<\/sub>
                          \nAnswer:
                          \n(d) 4H3<\/sub>PO3<\/sub> \u2192 3H3<\/sub>PO4<\/sub> + PH3 <\/sub>(Auto oxidation and reduction reaction)<\/p>\n

                          Question 60.
                          \nThe number of molecules in 40 g of sodium hydroxide is ……….
                          \n(a) 6.023 x 1023<\/sup>
                          \n(b) 3.0115 x 1023<\/sup>
                          \n(c) 6.023 x 1023<\/sup>
                          \n(d) 2 x 6.023 x 1023<\/sup>
                          \nAnswer:
                          \n(c) 6.023 x 1023<\/sup>
                          \nSodium hydroxide = NaOH = 23 + 16+ 1 =40 g
                          \n40g = 1 mole = 6.023 1023<\/sup><\/p>\n

                          Question 61.
                          \nThe mass of one molecule of AgCl in grams is ……….
                          \n(a) 108 g
                          \n(b) 143.5 g
                          \n(c) 35.5 g
                          \n(d) 243.5 g
                          \nAnswer:
                          \n(b) 143.5 g
                          \nHint:
                          \nMass of AgCl = 108 + 35.5 = 143.5 g.<\/p>\n

                          Question 62.
                          \nThe empirical formula of Alkene is ……….
                          \n(a) CH
                          \n(b) CH2<\/sub>
                          \n(c) CH3<\/sub>
                          \n(d) CH3<\/sub>O
                          \nAnswer:
                          \n(b) CH2<\/sub>
                          \nHint:
                          \nAlkene Cn<\/sub>H2n<\/sub> Molecular formula
                          \nE.F. = M.F.\/2
                          \n\u2234 Empirical formula = CH2<\/sub><\/p>\n

                          Question 63.
                          \n22 g of a gas occupies 11.2 litres of volume at STP. The gas is ……….
                          \n(a) CH4<\/sub>
                          \n(b) NO
                          \n(c) CO
                          \n(d) CO2<\/sub>
                          \nAnswer:
                          \n(d) CO2<\/sub>
                          \nHint:
                          \n22 g of a gas occupies 11.2 litres.
                          \n11.2 liters is occupied by 22 g of a gas.
                          \n\u2234 Molar volume 22.4 liter will be occupied by \\(\\frac {22}{11.2}\\) x 22.4 = 44 g
                          \n\u2234 The gas is CO2<\/sub>.<\/p>\n

                          Question 64.
                          \nThe number of moles of H2<\/sub> in 2.24 liter of hydrogen gas at STP is ……….
                          \n(a) 1
                          \n(b) 0.1
                          \n(c) 0.01
                          \n(d) 0.001
                          \nAnswer:
                          \n(b) 0.1
                          \nHint:
                          \n22.4 liter = 1 molar volume = 1 mole.
                          \n\u2234 2.24 liter = \\(\\frac {1}{22.4}\\) x 2.24 = 01 mole<\/p>\n

                          Question 65.
                          \nHow many molecules are present in 32 g of methane?
                          \n(a) 2 x 6.023 x 1023<\/sup>
                          \n(b) 6.023 x 1023<\/sup>
                          \n(c) 6.023 x 1023<\/sup>
                          \n(d) 3.011 x 1023<\/sup>
                          \nAnswer:
                          \n(a) 2 x 6.023 x 1023
                          \n<\/sup>Hint:
                          \nMethane (CH4<\/sub>) – Molar mass = 12 + 4 = 16g.
                          \n16 g contains 6.023 x 1023<\/sup> molecules.
                          \n\u2234 32 g of methane will contain = \\(\\frac{6.023 \\times 10^{23}}{16} \\times 32^{2}\\) = 2 x 6.023 x 1023<\/sup><\/p>\n

                          Question 66.
                          \nThe empirical formula of glucose is ……….
                          \n(a) CH
                          \n(b) CH2<\/sub>O
                          \n(c) CH2<\/sub>O2<\/sub>
                          \n(d) CHO
                          \nAnswer:
                          \n(b) CH2<\/sub>O
                          \nHint:
                          \nGlucose = Molecular formula = C6<\/sub>H12<\/sub>O6<\/sub>
                          \nEmpirical formula = \\(\\frac { Molecular formula}{6}\\) = \\(\\\\frac{\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}}{6}\\) = CH2<\/sub>O<\/p>\n

                          Question 67.
                          \nHow many moles of water is present in I L of water?
                          \n(a) 1
                          \n(b) 18
                          \n(c) 55.55
                          \n(d) 5.555
                          \nAnswer:
                          \n(c) 55.55
                          \nHint:
                          \n1 Liter of water = 1000 g.
                          \nMolar mass of water 18 g
                          \nNumber of moles = \\(\\frac {Mass}{Molar mass}\\) = \\(\\frac {1000}{18}\\) = 55.55 moles<\/p>\n

                          Question 68.
                          \nHow many moles of Hydrogen atoms are present in 1 mole of C2<\/sub> H6<\/sub> ?
                          \n(a) 18 moles
                          \n(b) 6 moles
                          \n(c) 3 moles
                          \n(d) 1 mole
                          \nAnswer:
                          \n(b) 6 moles
                          \nHint:
                          \nC2<\/sub>H6<\/sub> contains 6H atoms. 6 moles.<\/p>\n

                          Question 69.
                          \nThe molar mass of Na2<\/sub>SO4 <\/span>is ……….
                          \n(a).129
                          \n(b) 142
                          \n(c) 110
                          \n(d) 70
                          \nAnswer:
                          \n(b) 142
                          \nHint:
                          \nNa2<\/sub> SO4<\/sub> = Molar mass
                          \n= (23 x 2) + (32 x 1) + (16 x 4)
                          \n= 46 + 32 + 64 = 142<\/p>\n

                          Question 70.
                          \nMatch the List-I with List-Il using the correct code given below the list.
                          \n\"Samacheer
                          \nAnswer:
                          \n\"Samacheer<\/p>\n

                          Question 71.
                          \nOre mole of CO2<\/sub> contains ………….
                          \n(a) 6.02 x 1023<\/sup> atoms of C
                          \n(b) 3 g of CO2<\/sub>
                          \n(c) 6.02 1023<\/sup> atoms of O
                          \n(d) 18.1 x 1023<\/sup> molecules of CO2<\/sub>
                          \nAnswer:
                          \n(a) 6.02 x 1023<\/sup> atoms of C<\/p>\n

                          Question 72.
                          \n5.6 liters of oxygen at STP is equivalent to ……….
                          \n(a) 1 mole
                          \n(b) 1\/4 mole
                          \n(c) 1\/8 mole
                          \n(d) 1\/2 mole
                          \nAnswer:
                          \n(b) 1\/4 mole
                          \nHint:
                          \n22.4 litres of O2<\/sub> = 1 mole
                          \n\u2234 5.6 litres of O2<\/sub> = \\(\\frac {1}{22.4}\\) x 5.6 = 0.25 mole = 1\/4 mole.<\/p>\n

                          Question 73.
                          \nHow many grams are contained in 1 gram atom of Na?
                          \n(a) 13 g
                          \n(b) 1 g
                          \n(c) 23 g
                          \n(d) 1\/23 g
                          \nAnswer:
                          \n(c) 23 g
                          \nHint:
                          \n1 gram atom of Na
                          \nNa = Atomic mass 23 g (or) 23 amu
                          \n1 gram atom of Na = 1 mole = 23 g.<\/p>\n

                          Question 74.
                          \n12 g of Mg will react completely with an acid to give ……….
                          \n(a) 1 mole of O2<\/sub>
                          \n(b) 1\/2 mole of H2<\/sub>
                          \n(c) I mole of H2<\/sub>
                          \n(d) 2 mole of H2<\/sub>
                          \nAnswer:
                          \n(b) 1\/2 mole of H2<\/sub>
                          \nHint:
                          \n\"Samacheer
                          \n\u2234 12 g of Mg \\(\\frac {1}{2}\\) mole of H2<\/sub>O<\/p>\n

                          Question 75.
                          \nwhich of the following has the highest mass?
                          \n(a) I g atom of C
                          \n(b) 1\/2 mole of CH4<\/sub>
                          \n(c) 10 ml of water
                          \n(d) 3.011 x 1023<\/sup> atoms of oxygen
                          \nAnswer:
                          \n(a) I g atom of C
                          \n(a) 1 g atom of C = 12 g
                          \n(b) 1\/2 mole of CH4<\/sub> = \\(\\frac {12 + 4}{2}\\) = 8 g.
                          \n(c) 10 ml of water (H2<\/sub>O) = 1 x 10 = 10 g
                          \n(d) 3.011 x 1023<\/sup> atoms of oxygen = 0.5 mole of oxygen = 8 g<\/p>\n

                          Question 76.
                          \nThe empirical formula of sucrose is ……….
                          \n(a) CH2<\/sub>O
                          \n(b) CHO
                          \n(c) C12<\/sub>H22<\/sub>O11<\/sub>
                          \n(d) C(H2<\/sub>O)2<\/sub>
                          \nAnswer:
                          \n(a) CH2<\/sub>O
                          \nHint:
                          \nSucrose Molecular formula = C12<\/sub>H22<\/sub>O11<\/sub>
                          \nE.F.= \\(\\frac{\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}}{12}\\) = CH2<\/sub>O<\/p>\n

                          Question 77.
                          \nThe number of grams of oxygen in 0.10 mol of Na2<\/sub>CO3<\/sub>. 10H2<\/sub>O is ………..
                          \n(a) 20.8 g
                          \n(b) 18 g
                          \n(c) 108 g
                          \n(d) 13 g
                          \nAnswer:
                          \n(a) 20.8 g
                          \nHint:
                          \nNa2<\/sub>CO2<\/sub>.10H2<\/sub>O = 1 mole
                          \n1 mole of Na2<\/sub>CO3<\/sub>. 10H2<\/sub>O contains 13 oxygen atoms.
                          \nMass of 13 oxygen atoms = 13 x 16 = 208
                          \n1 mole of Na2<\/sub>CO3<\/sub>.10H2<\/sub>O contains 208 g of oxygen.
                          \n\u2234 0.10 mole of Na2<\/sub>CO3<\/sub>.10H2<\/sub>O contains \\(\\frac {208}{1}\\) x 0.12 = 08 g.<\/p>\n

                          Question 78.
                          \nThe mass of an atom of nitrogen is ……….
                          \n\"Samacheer
                          \nAnswer:
                          \n\"Samacheer<\/p>\n

                          \"Samacheer<\/p>\n

                          Question 79.
                          \nWhich of the following halogens do not exhibit positive oxidation number in its compounds?
                          \n(a) Fluorine
                          \n(b) Chlorine
                          \n(c) Iodine
                          \n(d) Bromine
                          \nAnswer:
                          \n(a) Fluorine<\/p>\n

                          Question 80.
                          \nWhich of the following is the most powerful oxidising agent?
                          \n(a) KMnO4<\/sub>
                          \n(b) K2<\/sub>Cr2<\/sub>O7<\/sub>
                          \n(c) O3<\/sub>
                          \n(d) H2<\/sub>O2<\/sub>
                          \nAnswer:
                          \n(a) KMnO4<\/sub><\/p>\n

                          Question 81.
                          \nOn the reaction 2Ag + H2<\/sub>SO4<\/sub> \u2192 Ag2<\/sub>SO4<\/sub> + 2H2<\/sub>O + SO2<\/sub>. Sulphuric acid acts as ……………
                          \n(a) oxidising agent
                          \n(b) reducing agent
                          \n(c) a catalyst
                          \n(d) an acid as well as an oxidant
                          \nAnswer:
                          \n(d) an acid as well as an oxidant<\/p>\n

                          Question 82.
                          \nThe oxidation number of carboxylic carbon atom in CH3<\/sub>COOH is ……….
                          \n(a) + 2
                          \n(b) + 4
                          \n(c) + 1
                          \n(d) + 3
                          \nAnswer:
                          \n(d) + 3
                          \nCH3<\/sub>COOH
                          \n-3 + 3 + x – 4 + 1
                          \nx – 3 = 0
                          \nx = + 3
                          \nCarboxylic carbon oxidation number = + 3<\/p>\n

                          Question 83.
                          \nWhen methane is burnt in oxygen to produce CO2<\/sub> and H2<\/sub>O, the oxidation number of carbon changes by ……….
                          \n(a) – 8
                          \n(b) + 4
                          \n(c) zero
                          \n(d) + 8
                          \nAnswer:
                          \n(b) + 4
                          \n\"Samacheer<\/p>\n

                          Question 84.
                          \nThe oxidation number of carbon is zero in ……….
                          \n(a) HCHO
                          \n(b) C12<\/sub>H22<\/sub>O11<\/sub>
                          \n(c) C6<\/sub>H12<\/sub>O6<\/sub>
                          \n(d) all the above
                          \nAnswer:
                          \n(d) all the above<\/p>\n

                          Question 85.
                          \nThe oxidation number of Fe in Fe2<\/sub>(SO4<\/sub>)3<\/sub> is ……….
                          \n(a) + 2
                          \n(b) + 3
                          \n(c) + 2, + 3
                          \n(d) O
                          \nAnswer:
                          \n(b) + 3
                          \nFe2<\/sub>(SO4<\/sub>)3<\/sub>
                          \n2x – 6 = 0
                          \nx = + 3<\/p>\n

                          Question 86.
                          \nAmong the following molecules in which Chlorine shows maximum oxidation state?
                          \n(a) Cl21<\/sub>
                          \n(b) KCl
                          \n(c) KClO3<\/sub>
                          \n(d) Cl2<\/sub>O7<\/sub>
                          \nAnswer:
                          \n(d) Cl2<\/sub>O7<\/sub>
                          \nCl2<\/sub>O7<\/sub>
                          \n2x – 14 = 0
                          \n2x = + 14
                          \nx = + 7<\/p>\n

                          Question 87.
                          \nThe oxidation number of carbon in CH3 <\/sub>\u2192 CH2<\/sub>OH is ………….
                          \n(a) + 2
                          \n(b) – 2
                          \n(e) O
                          \n(d) + 4
                          \nAnswer:
                          \n(b) – 2
                          \nC2<\/sub>H5<\/sub>OH
                          \n2x +6 – 2 = O
                          \n2x + 4 = 0
                          \n2x = -4
                          \nx = -2<\/p>\n

                          Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 2 – Mark Questions<\/span><\/strong><\/p>\n

                          I. Write brief answer to the following questions:<\/span><\/p>\n

                          Question 1.
                          \nState Avogadro\u2019s Hypothesis.
                          \nAnswer:
                          \nIt states that \u2018Equal volume of all gases under the same conditions of temperature and pressure contain the same number of molecules\u2019.<\/p>\n

                          Question 2.
                          \nWhat is molar volume?
                          \nAnswer:
                          \nMolar volume is the volume occupied by one mole of a substance in the gaseous state at STP. It is equal to 2.24 x 10-2<\/sup>m3<\/sup> (22.4 L).<\/p>\n

                          Question 3.
                          \nThe approximate production of Na2<\/sub>CO3<\/sub> per month is 424 x 106<\/sup> g while that of methyl alcohol is 320 x 106<\/sup> g. Which is produced more in terms of moles?
                          \nAnswer:
                          \nNa2<\/sub>CO3<\/sub> mass = 424 x 106<\/sup>g
                          \nMolecular mass of Na2<\/sub>CO3<\/sub> = (23 x 2) + 12 + (16 x 3)
                          \n= 46+ 12 +48
                          \n= 106 g
                          \nNo. of moles of Na2<\/sub>CO3<\/sub> \"Samacheer
                          \n= 4 x 106<\/sup> moles
                          \nMethyl alcohol mass = 320 x 106<\/sup> g
                          \nMolecular mass of CH3<\/sub>OH = 12 + (1 x 4)+ 16 = 32 g
                          \n= 12 + 4 + 16 = 32 g
                          \nNo. of moles of Methyl alcohol \"Samacheer
                          \n= 10 x 106<\/sup> moles.
                          \n\u2234 Methyl alcohol is more produced in terms of moles.<\/p>\n

                          Question 4.
                          \nCalculate number of moles of carbon atoms \u00a1n three moles of ethane.
                          \nAnswer:
                          \nEthane – Molecular formula = C2<\/sub>H6<\/sub>
                          \n1 mole of ethane contains 2 atoms of carbon (6.023 x 1023<\/sup> C)
                          \n\u2234 3 moles of ethane contains 6 atoms of Carbon.
                          \n\u2234 No. of moles of Carbon atoms = 3 x 6.023 x 1023<\/sup> Carbon atoms.
                          \n= 18.069 x 1023<\/sup> Carbon atoms.<\/p>\n

                          Question 5.
                          \nFind the molecular mass of FeSO4<\/sub>.7H2<\/sub>O
                          \nAnswer:
                          \nSum of Atomic mass of all elements = Molecular mass
                          \nAtomic mass of Fe = 56.0
                          \nAtomic mass of S = 32.0
                          \nAtomic mass of 4[O] = 64.0
                          \nAtomic mass of 14[H] = 14.0
                          \nAtomic mass of 7[O] = 112.0 = 278.0
                          \nMolecular mass of FeSO4<\/sub>.7H2<\/sub>O = 278 g.<\/p>\n

                          Question 6.
                          \nMass of one atom of an element \u00a1s 6.66 x 1023<\/sup> g. How many moles of element are there in 0.320 kg?
                          \nAnswer:
                          \nMass of one atom of an element = 6.66 x 1023<\/sup>g
                          \nNo. of moles = \\(\\frac {Mass}{Molecular mass}\\) 3
                          \nMolecular mass = Mass of 1 atom x Avogadro number
                          \n6.66 x 1023<\/sup> x 6.023 x 1023<\/sup>
                          \n= 6.66 x 6.023 = 40.11318
                          \nNumber of moles = \\(\\frac {Mass}{Molecular mass}\\) = \\(\\frac{0.320 \\mathrm{kg} \\times 10^{3}}{40}\\) = 8 moles.<\/p>\n

                          Question 7.
                          \nHow many moles of glucose are present in 720 g of glucose?
                          \nAnswer:
                          \nGlucose = C6<\/sub>H4<\/sub>O4<\/sub>
                          \nMolecular mass of Glucose = (12 x 6) + (1 x 12) + (16 x 6)
                          \n= 72 + 12 + 96 = 180
                          \n\\(\\frac {720}{180}\\) = 4 moles.<\/p>\n

                          Question 8.
                          \nCalculate the weight of 0.2 mole of sodium carbonate.
                          \nAnswer:
                          \nSodium carbonate = Na2<\/sub>CO3<\/sub>
                          \nMolecular mass of Na2<\/sub>CO3<\/sub> = (23 x 2)+(12 x 1)+(16 x 3)
                          \n= 46 + 12 + 48 = 106 g
                          \nMass of 1 mole of Na2<\/sub>CO3<\/sub> = \\(\\frac{106 \\times 0.2}{1}\\) = 21.2 g<\/p>\n

                          Question 9.
                          \nWhat do you understand by the terms acidity and basicity?
                          \nAnswer:
                          \nAcidity:
                          \nThe number of hydroxyl ions present in one mole of a base is known as the acidity of the base.<\/p>\n

                          Basicity:
                          \nThe number of replaceable hydrogen atoms present in a molecule of the acid is referred to as its basicity.<\/p>\n

                          Question 10.
                          \nCalculate the equivalent mass of bicarbonate ion.
                          \nAnswer:
                          \nBicarbonate ion = HCO3<\/sub>
                          \n\"Samacheer
                          \nFormula mass of HCO3<\/sub> = 1 + 12 + 48 = 61
                          \nEquivalent mass of HCO3<\/sub> = \\(\\frac {61}{1}\\) = 61<\/p>\n

                          Question 11.
                          \nCalculate the equivalent mass of barium hydroxide.
                          \nBarium hydroxide = Ba(OH)2<\/sub>
                          \nMolecular mass of Ba(OH)2<\/sub> = 137 + (16 x 2) + (1 x 2)
                          \n= 171.0 g \/ mol.
                          \nAcidity = 2
                          \nEquivalent mass of Ba(OH)2<\/sub> =
                          \n= \\(\\frac {17 1.0}{2}\\) = 85.5<\/p>\n

                          Question 12.
                          \nCalculate the equivalent mass of hydrated sodium carbonate.
                          \nAnswer:
                          \nHydrated sodium carbonate = Na2<\/sub>CO3<\/sub>. 10H2<\/sub>O
                          \nMolecular mass of Na2<\/sub>CO3<\/sub>. 10H2<\/sub>O = (23 x 2) + (2 x 1) + (16 x 13) + (1 x 20)
                          \n= 46 + 12 + 208 + 20 = 286
                          \nEquivalent mass of Na2<\/sub>CO3<\/sub>. 10H2<\/sub>O = \\(\\frac {Molecular mass}{Acidity}\\)
                          \n= \\(\\frac {286}{2}\\) = 143<\/p>\n

                          Question 13.
                          \nWhat do you understand by the terms empirical formula and molecular formula?
                          \nAnswer:
                          \nEmpirical Formula:<\/p>\n

                            \n
                          • It is the simplest formula.<\/li>\n
                          • It shows the ratio of number of atoms of different elements in one molecule of the compound.<\/li>\n<\/ul>\n

                            Molecular Formula:<\/p>\n

                              \n
                            • It is the actual formula.<\/li>\n
                            • It shows the actual number of different types of atoms present in one molecule of the compound.<\/li>\n<\/ul>\n

                              Question 14.
                              \nBoric acid, H3<\/sub>BO3<\/sub> is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H3<\/sub>BO3<\/sub>. What is the mass of boric acid in the sample?
                              \nAnswer:
                              \nMolecular mass of H3<\/sub>BO3<\/sub> = (1 x 3) + (11 x 1) + (16 x 3) = 62
                              \nBoric acid sample contains 0.543 mole.
                              \nMass of 0.543 mole of Boric acid = Molecular mass x mole
                              \n= 62 x 0.543
                              \n= 33.66 g<\/p>\n

                              Question 15.
                              \nA compound contains 50% of X (atomic mass 10) and 50% Y (atomic mass 20). Give its molecular formula
                              \nAnswer:
                              \n\"Samacheer
                              \n\u2234 The Empirical Formula is X3<\/sub>Y
                              \nEmpirical Formula mass = 20 + 20 = 40
                              \nMolecular mass = Sum of atomic mass = 40
                              \nn = 1, Molecular formula = (Empirical Formula )n<\/sub> = (X2<\/sub>Y)1<\/sub> = X2<\/sub>Y.<\/p>\n

                              Question 16.
                              \nCalculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
                              \nAnswer:
                              \nSodium Nitrate = NaNO3<\/sub>
                              \nMolecular mass of Sodium Nitrate = 23 + 14 + 48 = 85
                              \n100% pure 85 g of NaNO3<\/sub> contains 23 g of Sodium.
                              \n100% pure 95 x 103 g of NaNO3<\/sub> will contains \\(\\frac {23}{85}\\) x 95 x 103<\/sup>
                              \n= 25.70 x 103<\/sup> g of Sodium.
                              \n100% pure NaNO3<\/sub> \u00a0contains 25.70 x 103<\/sup> g of Sodium.
                              \n\u2234 70% pure NaNO3<\/sub> \u00a0will contains = img
                              \n= 17990 g (or) 17.99 Kg of Na.<\/p>\n

                              Question 17.
                              \nDefine matter. What are the types of matter?
                              \nAnswer:<\/p>\n

                                \n
                              • A matter is anything which has mass and occupies space.<\/li>\n
                              • Matters exist in all three states such as solid, liquid and gas.<\/li>\n<\/ul>\n

                                Question 18.
                                \nProve that states of matter are inter convertible.
                                \nAnswer:
                                \nStates of matter are inter convertible by changing temperature and pressure.
                                \n\"Samacheer<\/p>\n

                                Question 19.
                                \nWhat is meant by Plasma state? Give an example.
                                \nAnswer:
                                \nGaseous state of matter at very high temperature containing gaseous ions and free electron is referred to as the Plasma state. e.g. Lightning.<\/p>\n

                                Question 20.
                                \nDifferentiate an element and an atom.
                                \nAnswer:<\/p>\n

                                  \n
                                • An atom is the ultimate smallest electrically neutral, being made up of fundamental particles such as proton, neutron and electron.<\/li>\n
                                • An element consists of only one type of atoms. Elements are further divided into metals, non-metals, and metaloids.<\/li>\n<\/ul>\n

                                  Question 21.
                                  \nDistinguish between a molecule and a compound
                                  \nAnswer:
                                  \nMolecule:<\/p>\n

                                    \n
                                  • A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence.<\/li>\n
                                  • e.g. Na – Mono atomic molecule O2<\/sub> – Diatomic molecule P4<\/sub> – Poly atomic molecule<\/li>\n<\/ul>\n

                                    Compound:<\/p>\n

                                      \n
                                    • A molecule which contains two or more atoms of different elements are called a compound molecule.<\/li>\n
                                    • e.g. CO2<\/sub> – Carbon dioxide CH4<\/sub> – Methane H2<\/sub>O – Water<\/li>\n<\/ul>\n

                                      Question 22.
                                      \nChlorine has fractional average atomic mass. Justify this statement.
                                      \nAnswer:
                                      \nChlorine molecule has two isotopes as in 17<\/sub>Cl35<\/sup> , 17<\/sub> Cl37<\/sup> in the ratio of 77 : 23, so when we are calculating the average atomic mass, it becomes fractional.
                                      \nThe average relative atomic mass of Chlorine = \\(\\frac {(35 x 77) + (37 x 23)}{100}\\) = 35.46 amu<\/p>\n

                                      Question 23.
                                      \nDefine molecular mass of a substance.
                                      \nAnswer:
                                      \nMolecular mass of a substance (element or compound) represents the number of times the molecule of that substance is heavier than 1 \/ 12th<\/sup> of the mass of an atom of C-12 isotope. Molecular mass = 2 x Vapour density<\/p>\n

                                      Question 24.
                                      \nCalculate the molecular mass of Sulphuric acid (H2<\/sub>SO4<\/sub>). Element
                                      \nAnswer:
                                      \n\"Samacheer<\/p>\n

                                      Question 25.
                                      \nDefine Avogadro Number.
                                      \nAnswer:
                                      \nAvogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 x 1023<\/sup><\/p>\n

                                      Question 26.
                                      \nCalculate the number of moles present in 60 g of ethane.
                                      \nAnswer:
                                      \nNo. of moles =\"Samacheer\u00a0= \\(\\frac {W}{M}\\)
                                      \nMolar Mass of ethane (C2<\/sub>H6<\/sub>) = 24 + 6 = 30
                                      \nNumber of moles in 60 g of ethane = \\(\\frac {60}{30}\\) = 2 moles.<\/p>\n

                                      Question 27.
                                      \nCalculate the equivalent mass of Copper. (Atomic mass of copper = 63.5)
                                      \nAnswer:
                                      \nEquivalent mass = \\(\\frac {Atomic mass}{Valency}\\)
                                      \nEquivalent mass of Copper = \\(\\frac {63.5}{2}\\) = 31.75 g eq-1<\/sup>.<\/p>\n

                                      Question 28.
                                      \nCalculate the equivalent mass of (i) Sulphate ion (ii) Phosphate ion.
                                      \nAnswer:
                                      \n(i) Sulphate ion (SO4<\/sub>2-<\/sup>).
                                      \nEquivalent mass of Sulphate = \"Samacheer\u00a0= \\(\\frac {32 + 64}{2}\\) = \\(\\frac {96}{2}\\) = 48 g eq-1<\/sup><\/p>\n

                                      (ii) Phosphate ion (P04<\/sub>3-<\/sup>)
                                      \nMolar mass of Phosphate ion = \"Samacheer\u00a0= \\(\\frac {31 + 64}{3}\\) = \\(\\frac {95}{2}\\) = 31.6 = 31.6g eq-1<\/sup><\/p>\n

                                      Question 29.
                                      \nCalculate the equivalent mass of sulphuric acid.
                                      \nAnswer:
                                      \nSulphuric acid = H2<\/sub>SO4<\/sub>
                                      \nMolar mass of Sulphuric acid = 2 + 32 + 64 = 96
                                      \nBasicity of Sulphuric acid = 2
                                      \nEquivalent mass of acid = \"Samacheer\u00a0= \\(\\frac {96}{2}\\) = 49g eq-1<\/sup><\/p>\n

                                      Question 30.
                                      \nHow many moles of hydrogen is required to produce 20 moles of ammonia?
                                      \nAnswer:
                                      \n3H2<\/sub> + N2<\/sub> \u2192 2NH3<\/sub>
                                      \nA per stoichiometric equation,
                                      \nNo. of moles of hydrogen required for 2 moles of ammonia 3 moles
                                      \nNo. of moles of hydrogen required for 20 moles of ammonia = \\(\\frac {3}{2}\\) x 20 = 30 moles.<\/p>\n

                                      Question 31.
                                      \nCalculate the amount of water produced by the combustion of 32 g of methane.
                                      \nAnswer:
                                      \n\"Samacheer
                                      \nAs per stoichiometric equation,
                                      \n16 g of methane produces 36 g of H2<\/sub>O
                                      \n\u2234 32 g of methane will produce = \\(\\frac {36}{16}\\) x 32 = 72 g of water.<\/p>\n

                                      Question 32.
                                      \nHow much volume of Carbon dioxide is produced when 25 g of calcium carbonate is heated completely under standard conditions?
                                      \nAnswer:
                                      \n\"Samacheer
                                      \n100 g of CaCO3<\/sub> produces 22.4 L of CO2<\/sub>.
                                      \n\u2234 25 g of CaCO3<\/sub> will produce = \\(\\frac {22.4}{100}\\) x 25 = 5.6 L of CO2<\/sub>.<\/p>\n

                                      Question 33.
                                      \nHow much volume of chlorine is required to prepare 89.6 L of HCl gas at STP?
                                      \nAnswer:
                                      \n\"Samacheer
                                      \n2 x 22.4 L of HCl is produced by 22.4 L of Cl2<\/sub>.
                                      \n\u2234 89.6 L of HCl will be produced by = img = 89.6 L = 44.8 L of chlorine.<\/p>\n

                                      Question 34.
                                      \nWhat is meant by limiting reagent?
                                      \nAnswer:
                                      \nA large excess of one reactant is supplied to ensure the more expensive reactant is completely converted to the desired product. The reactant used up first in a reaction is called the limiting reagent.<\/p>\n

                                      Question 35.
                                      \nOn the formation of SF6<\/sub> by the direct combination of S and F2<\/sub>, which is the limiting reagent? Prove it.
                                      \nAnswer:
                                      \nSF6<\/sub> is formed by burning Sulphur in an atmosphere of Fluorine. Suppose 3 moles of S is allowed to react with 12 moles of Fluorine.
                                      \nS(l)<\/sub> +3F2(g)<\/sub><\/sub> \u2192 SF6(g)<\/sub><\/sub>
                                      \nAs per the stoichiometric reaction, one mole of S reacts with 3 moles of fluorine to complete the reaction. Similarly, 3 moles of S requires only 9 moles of fluorine.
                                      \n\u2234 It is understood that the limiting reagent is Sulphur and the excess reagent is Fluorine.<\/p>\n

                                      Question 36.
                                      \nMention any 4 redox reaction that takes place in our daily life.
                                      \nAnswer:<\/p>\n

                                        \n
                                      1. Burning of cooking gas, wood<\/li>\n
                                      2. Rusting of iron articles<\/li>\n
                                      3. Electroplating<\/li>\n
                                      4. Galvanic and electrolytic cells<\/li>\n<\/ol>\n

                                        Question 37.
                                        \nCalculate the oxidation number of underlined elements in the following.<\/p>\n

                                          \n
                                        1. KMnO4<\/sub><\/li>\n
                                        2. Cr2<\/sub>O7<\/sub>2-<\/sup><\/li>\n<\/ol>\n

                                          Answer:
                                          \n1. KMnO4<\/sub>
                                          \n1(+1) + x + 4 (-2) = 0
                                          \nx – 7 = 0
                                          \n\u2234 x = + 7
                                          \nOxidation state of Mn = +7.<\/p>\n

                                          2. Cr2<\/sub>O7<\/sub>2-<\/sup>
                                          \n2x + 7(-2) = -2
                                          \n2x – 14 = – 2
                                          \n2x = +l2
                                          \n\u2234 x = + 6
                                          \nOxidation state of Cr = +6.<\/p>\n

                                          Question 38.
                                          \nIf 10 volumes of H2<\/sub> gas react with 5 volumes of O2<\/sub> gas, how many volumes of water vapour would be produced?
                                          \nAnswer:
                                          \n\"Samacheer
                                          \nThus 2 volumes of H2<\/sub> reacts with 1 volume of O2<\/sub> to produce 2 volumes of H2<\/sub>O(g)<\/sub>.
                                          \n10 volumes of H2<\/sub> would react with 5 volumes of O2<\/sub> to produce 10 volumes of H2<\/sub>O(g)<\/sub>
                                          \nThus 10 volumes of H2<\/sub>O will be produced.<\/p>\n

                                          Question 39.
                                          \nWhich one of the following will have largest number of atoms?<\/p>\n

                                            \n
                                          1. 1 g of Au(s)<\/li>\n
                                          2. 1 g of Na(s)<\/li>\n
                                          3. 1 g of Li(s)<\/li>\n
                                          4. 1 g of Cl2<\/sub> (g)<\/li>\n<\/ol>\n

                                            Answer:
                                            \n1. Molar mass of Au = 197 g mol-1<\/sup>.
                                            \nNo: of atoms in 1 g of Au = \\(\\frac {1}{197}\\) x 6.023 x 1023<\/sup><\/p>\n

                                            2. Molar mass of Na = 23 g mol-1<\/sup>.
                                            \nNo of atoms in 1 g of Na = \\(\\frac {1}{23}\\) x 6.023 x 1023<\/sup><\/p>\n

                                            3. Molar mass of Li = 7 g mol-1<\/sup>
                                            \nNo. of atoms in 1 g of Li = \\(\\frac {1}{7}\\) x 6.023 x 1023<\/sup><\/p>\n

                                            4.\u00a0 Molar mass of Cl2<\/sub> = 35.5 g mol-1<\/sup>
                                            \nNo. of atoms in 1 g of Cl2<\/sub> = \\(\\frac {1}{35.5}\\) x 6.023 x 1023<\/sup><\/p>\n

                                            Comparing the number of atoms, the largest number of atoms will be present in 1 g of Li. Since the mass is same in each case, the element with the lowest molar mass would have the largest number of atoms.
                                            \n\u2234 Li with lowest molar mass would have the largest number of atoms.<\/p>\n

                                            Question 40.
                                            \nWhat will be the mass of one 12<\/sup>C atom in g?
                                            \nAnswer:
                                            \nMolar mass of 12<\/sup>C = 12.00 g mol-1<\/sup>.
                                            \n\u2234 Mass of 6.023 x 1023<\/sup> carbon atom = 12.0 g
                                            \n\u2234 Mass of 1 carbon atom = \\(\\frac{12}{6.023 \\times 10^{23}}\\) = 1.992 x 10 g.<\/p>\n

                                            Question 41.
                                            \nJustify the following reaction is a redox reaction.
                                            \nAnswer:
                                            \n\"Samacheer
                                            \nIn the above reaction, oxygen is removed from CuO. So CuO gets reduced. Oxygen is added to H2<\/sub> to form water. So H2<\/sub> gets oxidised. i.e. In CuO, oxidation number of Cu +2 is reduced to O whereas in H2<\/sub>, oxidation of H2<\/sub> O is increased to + 1. So the above reaction is a redox reaction.<\/p>\n

                                            Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 3 – Mark Questions<\/span><\/strong><\/p>\n

                                            Question 1.
                                            \nDistinguish among the different physical states of matter.
                                            \nAnswer:
                                            \nDifferences among three physical states of matter (solid, liquid and gas) are as follows<\/p>\n\n\n\n\n\n\n\n\n
                                            S.No.<\/strong><\/td>\nProperties<\/strong><\/td>\nSolid<\/strong><\/td>\nLiquid<\/strong><\/td>\nGas<\/strong><\/td>\n<\/tr>\n
                                            1.<\/td>\nVolume<\/td>\nDefinite<\/td>\nDefinite<\/td>\nNo definite<\/td>\n<\/tr>\n
                                            2.<\/td>\nShape<\/td>\nDefinite<\/td>\nNo definite<\/td>\nNo definite<\/td>\n<\/tr>\n
                                            3.<\/td>\nMolecular arrangement<\/td>\nVery closely packed<\/td>\nLoosely packed<\/td>\nVery loosely packed<\/td>\n<\/tr>\n
                                            4.<\/td>\nFreedom of movement<\/td>\nNot much freedom<\/td>\nMove around and better than solid<\/td>\nMove easy and fast<\/td>\n<\/tr>\n
                                            5.<\/td>\nCompressibility<\/td>\nNon compressible<\/td>\nLess compressible<\/td>\nEasily compressible<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                            Question 2.
                                            \nDefine equivalent mass of a salt.
                                            \nAnswer:
                                            \nEquivalent mass of a salt:
                                            \nIt is defined as the number of parts by mass of the salt that is produced by the neutralization of one equivalent of an acid by a base. Therefore the equivalent mass of the salt is equal to its molar mass.<\/p>\n

                                            Question 3.
                                            \nHow much copper can be obtained from 100 g of anhydrous copper sulphate?
                                            \nAnswer:
                                            \nAnhydrous copper sulphate = CuSO4<\/sub>
                                            \nMolecular mass of CuSO4<\/sub> = 63.5 + 32 + (16 x 4)
                                            \n= 63.5 + 32 + 64
                                            \n= 159.5 g
                                            \n159.5 g of CuSO4<\/sub> contains 63.5 g of copper.
                                            \n\u2234 100 g of CuSO4<\/sub> contains \\(\\frac{6.35}{159.5}\\) x 100 = 0.39811 x 100 = 39.81 g of Copper.<\/p>\n

                                            Question 4.
                                            \nCalculate the equivalent mass of hydrated ferrous sulphate.
                                            \nAnswer:
                                            \nHydrated ferrous sulphate = FeSO4<\/sub>.7H2<\/sub>O
                                            \nFerrous sulphate – Reducing agent
                                            \nFerrous sulphate reacts with an oxidising agent in acid medium according to the equation.
                                            \n\"Samacheer
                                            \n16 parts by mass of oxygen oxidised 304 g of FeSO4<\/sub>.
                                            \n8 parts by mass of oxygen will oxidise \\(\\frac{304}{16}\\) x 8 parts by mass of FeSO4<\/sub>.
                                            \n= 152
                                            \nEquivalent mass of Ferrous sulphate (Anhydrous) = 152
                                            \nEquivalent mass of crystalline Ferrous sulphate FeSO4<\/sub> .7H2<\/sub>O = 152 + 126 = 278<\/p>\n

                                            Question 5.
                                            \nGive difference between empirical and molecular formula
                                            \nAnswer:
                                            \nEmpirical Formula:<\/p>\n

                                              \n
                                            • Empirical formula is the simplest formula.<\/li>\n
                                            • It shows the ratio of number of atoms of different elements in one molecule of the compound.<\/li>\n
                                            • It is calculated from the percentage of composition of the various elements in one molecule.<\/li>\n
                                            • For example, Empirical formula of Benzene = CH.<\/li>\n<\/ul>\n

                                              Molecular Formula:<\/p>\n

                                                \n
                                              • Molecular Formula is the actual formula.<\/li>\n
                                              • It shows the actual number of different types of atoms present in one molecule of the compound.<\/li>\n
                                              • It is calculated from the Empirical formula Molecular Formula = (Empirical formula)n<\/sub><\/li>\n
                                              • Molecular formula of Benzene = C6<\/sub>H6<\/sub>.<\/li>\n<\/ul>\n

                                                Question 6.
                                                \nA sample of hydrated copper sulphate is heated to drive off the water of crystallization, cooled and reweighed 0.869 g of CuSO4<\/span>\u00a0aH2<\/sub>O gave a residue of 0.556 g. Find the molecular formula of hydrated copper sulphate.
                                                \nAnswer:
                                                \n0.869 g of CuSO4<\/sub>.aH2<\/sub>O gave a residue of 0.556 g of Anhydrous CuSO4<\/sub>.
                                                \n\u2234 Weight of a H2<\/sub>O molecule = 0.869 – 0.556 = 0.313 g
                                                \nMolecular weight of H2<\/sub>O = (1 x 2) + 16 = 2 + 16 = 18
                                                \nNo. of moles of water = \\(\\frac{Mass}{Molecular mass}\\)
                                                \nCuSO4<\/sub>.5H2<\/sub>O – Molecular mass = 63.5 + 32 + 64 + 90 = 249.5 g
                                                \n249.5 g of CuSO4<\/sub>.5H2<\/sub>O on heating gives 159.5 g of CuSO4<\/sub>.
                                                \n0.869 g of CuSO4<\/sub>.aH2<\/sub>O on heating gives = \\(\\frac{159.5}{249.5}\\) x 0.869
                                                \n= 0.556 g of anhydrous CuSO4<\/sub> \u2234 a = 5
                                                \nThe molecular formula of hydrated copper sulphate = CuSO4<\/sub>.5H2<\/sub>O<\/p>\n

                                                Question 7.
                                                \nBalance by oxidation number method: Mg + HNO3<\/sub> \u2192 Mg(NO3<\/sub>)2<\/sub> + NO2<\/sub> + H2<\/sub>O
                                                \nStep – 1
                                                \n\"Samacheer
                                                \nStep – 2
                                                \nMg + 2HNO3<\/sub> -\u00bb Mg(NO3<\/sub>)2<\/sub> + NO2<\/sub> + H2<\/sub>O
                                                \nStep – 3
                                                \nTo balance the number of oxygen atoms and hydrogen atoms 2HNO3<\/sub> is multiplied by 2.
                                                \nMg + 4HNO3 <\/sub>\u2192 Mg(NO3<\/sub>)2<\/sub> + 2NO2<\/sub> + H2<\/sub>O
                                                \nStep – 4.
                                                \nTo balance the number of hydrogen atoms, the H2<\/sub>O molecule is multiplied by 2.
                                                \nMg + 4HNO3<\/sub> \u2192 Mg(NO3<\/sub>)2<\/sub> + 2NO2<\/sub> + 2H2<\/sub>O<\/p>\n

                                                Question 8.
                                                \nExplain about the classification of matter.
                                                \nAnswer:
                                                \n\"Samacheer<\/p>\n

                                                Question 9.
                                                \nCalculate the mass of the following atoms in amu,
                                                \n(a) Helium (mass of He = 6.641 x 10-24<\/sup> g)
                                                \n(b) Silver (mass of Ag = 1.790 x 10-22<\/sup> g)
                                                \n1 amu = 1.66056 x 10-24<\/sup>
                                                \nAnswer:
                                                \n(a) The mass of Helium atom in amu = \\(\\frac{6.641 \\times 10^{-24}}{1.66056 \\times 10^{-24}}\\) = 3.9992 amu.
                                                \n(b) The mass of Silver atom in amu = \\(\\frac{1.790 \\times 10^{-22}}{1.66056 \\times 10^{-24}}\\) = 107.79 amu.<\/p>\n

                                                Question 10.
                                                \nCalculate the number of atoms present in 1 Kg of gold.
                                                \nAnswer:
                                                \nThe atomic mass of Gold = 197 g mol-1<\/sup>.
                                                \n197 g of gold contains 6.023 x 1023<\/sup> atoms of gold.
                                                \n\u2234 1000 g of gold will contain = \\(\\frac{1000 \\times 6.023 \\times 10^{23}}{197}\\)
                                                \n= 3.055 x 1024<\/sup> atoms of Gold.<\/p>\n

                                                Question 11.
                                                \nCalculate the molar volume of 146 g of HCl gas and the number of molecules present in it.
                                                \nAnswer:
                                                \nMolar mass of HCl = 36.5 g
                                                \nThe molar volume of 36.5 g (1 mole) of HCl = 2.24 x 102<\/sup> m3<\/sup>.
                                                \n\u2234 The volume of 146 g (4 moles) of HCl = \\(\\frac{2.24 \\times 10^{-2}}{36.5}\\) x 146
                                                \n= 8.96 x 10 m3<\/sup>
                                                \nNo. of molecules in 146 g of HCl = 4 N
                                                \n= 4 x Avogadro Number
                                                \n= 4 x 6.023 x 1023<\/sup>
                                                \n= 24.092 x 1023<\/sup>
                                                \n= 2.4092 x 1024<\/sup> molecules.<\/p>\n

                                                Question 12.
                                                \nCalculate the molar mass of 20 L of gas weighing 23.2 g at STP.
                                                \nMolar mass =\"Samacheer<\/p>\n

                                                Molar volume at STP = 2.24 x 10-2<\/sup> m3 <\/sup>= 22.4 L (or) 22400 cc.
                                                \nMolar mass of the gas at STP = \\(\\frac{23.2 x 22.4}{20}\\) = 25.984 g.<\/p>\n

                                                Question 13.
                                                \n0.6 g of a metal gives on oxidation 1 g of its oxide. Calculate its equivalent mass.
                                                \nAnswer:
                                                \nMass of metal = 0.6
                                                \nMass of metal oxide = 1 g
                                                \nMass of oxygen = 1 – 0.6 = 0.4 g
                                                \n0.4 g of oxygen combines with 0.6 g of metal.
                                                \n\u2234 8 g of oxygen will combine with = \\(\\frac{0.6}{0.4}\\) x 8
                                                \nEquivalent mass of the metal = 12 g eq-1<\/sup>.<\/p>\n

                                                Question 14.
                                                \nHow would you calculate the equivalent mass of anhydrous oxalic acid and hydrated oxalic acid.
                                                \nAnswer:
                                                \nIn acid medium,
                                                \n\"Samacheer
                                                \n16 g of oxygen is used for oxidation of 90 g of oxalic acid.
                                                \n\u2234 8 g of oxygen will oxidize = \\(\\frac{90}{16}\\) x 8 = 45 g eq-1<\/sup>.
                                                \nEquivalent mass of Anhydrous oxalic acid = 45 g eq-1<\/sup>
                                                \n\"Samacheer
                                                \n= \\(\\frac{126}{2}\\) = 63 g eq-1<\/sup>.<\/p>\n

                                                Question 15.
                                                \nA compound on decomposition in the laboratory produces 24.5 g of nitrogen and 70 g of oxygen. Calculate the empirical formula of the compound.
                                                \nAnswer:
                                                \n\"Samacheer
                                                \nthe empirical formula is N2<\/sub>O5<\/sub><\/p>\n

                                                Question 16.
                                                \nWhat is the steps involve in the calculation of molecular formula from empirical formula?
                                                \nAnswer:
                                                \nMolecular mass and empirical formula are used to deduce molecular formula of the compound.
                                                \nSteps to calculate molecular formula:<\/p>\n

                                                  \n
                                                • if Empirical formula is found out from the percentage composition of elements<\/li>\n
                                                • Empirical formula mass can be found from the empirical formula<\/li>\n
                                                • Molecular mass is found out from the given data<\/li>\n
                                                • Molecular formula = (Empirical formula)n<\/sub><\/li>\n
                                                • where, n = \"Samacheer<\/li>\n<\/ul>\n

                                                  Question 17.
                                                  \nWhat is combination reaction? Give example.
                                                  \nWhen two or more substances combine to form a single substance, the reactions are called combination reactions.
                                                  \nA + B \u2192 C
                                                  \nExample:
                                                  \n2 Mg + O2<\/sub> \u2192 2MgO<\/p>\n

                                                  Question 18.
                                                  \nWhat is decomposition reaction? Give two examples.
                                                  \nAnswer:
                                                  \nChemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
                                                  \nAB \u2192 A + B
                                                  \nExample:
                                                  \n2KCO3<\/sub> \u2192 2KCl + O2<\/sub>
                                                  \nPCl5<\/sub> \u2192 PCl3<\/sub> + Cl2<\/sub><\/p>\n

                                                  Question 19.
                                                  \nWhat is displacement reactions? Give its types. Explain with example.
                                                  \nAnswer:
                                                  \nThe reactions in which one ion or atom in a compound is replaced (or substituted) by an ion or atom of the other element are called displacement reactions.
                                                  \nAB + C \u2192 AC + B<\/p>\n

                                                  Example:
                                                  \nMetal displacement
                                                  \nCuSO4<\/sub> + Zn \u2192 ZnSO4<\/sub> + Cu<\/p>\n

                                                  Example:
                                                  \nNon-metal displacement
                                                  \n2KBr + Cl2<\/sub> \u2192 2KCl + Br2<\/sub><\/p>\n

                                                  Question 20.
                                                  \nWhat is disproportionation reactions? Give example.
                                                  \nAnswer:
                                                  \nThe reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
                                                  \nExample:
                                                  \nP4<\/sub> + 3NaOH + 3H2<\/sub>O \u2192 PH3<\/sub> + 3NaH2<\/sub>PO2<\/sub>
                                                  \n2HCHO + H2<\/sub>O \u2192 CH3<\/sub>OH + HCOOH<\/p>\n

                                                  Question 21.
                                                  \nWhat are competitive electron transfer reaction? Give example.
                                                  \nAnswer:
                                                  \nThese are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
                                                  \nExample:
                                                  \nZn releases electrons to Cu and Cu releases electrons to Silver and so on.
                                                  \nZn(s)<\/sub> + Cu2+<\/sup> \u2192 Zn2+<\/sup>(aq)<\/sub> + Cu(s)<\/sub> (Here Zn – oxidised; Cu2+<\/sup> – reduced)
                                                  \nCu(s)<\/sub> + 2Ag+<\/sup> \u2192 Cu2+<\/sup>(aq)<\/sub> + 2Ag(s)<\/sub> (Here Cu – oxidised; Ag+<\/sup> – reduced)<\/p>\n

                                                  Question 22.
                                                  \nBalance the following equation using oxidation number method.
                                                  \nAnswer:
                                                  \n1. S + HNO3<\/sub> \u2192 H2<\/sub>SO4<\/sub> + NO2<\/sub> + H2<\/sub>O
                                                  \n\"Samacheer
                                                  \n2. S + 6HNO3<\/sub> \u2192 H2<\/sub>SO4<\/sub> + NO2<\/sub> + H2<\/sub>O
                                                  \n3. Balance the equation (except O and H)
                                                  \nS + 6HNO3<\/sub> \u2192 H2<\/sub>SO4<\/sub> + 6NO2<\/sub> + H2<\/sub>O
                                                  \n4. Balance O atoms by adding 2H2<\/sub>O
                                                  \n5 + 6HNO3<\/sub> \u2192 H2<\/sub>SO4<\/sub> + 6NO2<\/sub> + 2H2<\/sub>O<\/p>\n

                                                  Question 23.
                                                  \nDetermine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.
                                                  \nAnswer:
                                                  \n\"Samacheer
                                                  \nthe empirical formula is Fe2<\/sub>O3<\/sub><\/p>\n

                                                  Question 24.
                                                  \nIn three moles of ethane (C2<\/sub>H6<\/sub>) calculate the following:<\/p>\n

                                                    \n
                                                  1. Number of moles of carbon atoms.<\/li>\n
                                                  2. Number of moles of hydrogen atoms.<\/li>\n
                                                  3. Number of molecules of ethane.<\/li>\n<\/ol>\n

                                                    Answer:<\/p>\n

                                                    (1) 1 mole of C2<\/sub>H6<\/sub> contains 2 moles of Carbon atoms.
                                                    \n\u2234 3 moles of C2<\/sub>H6<\/sub> will have 6 moles of Carbon atoms.
                                                    \n(2) 1 mole of C2<\/sub>H6<\/sub> contains 6 moles of Hydrogen atoms.
                                                    \n\u2234 3 moles of C2<\/sub>H6<\/sub> will have 18 moles of Hydrogen atoms.
                                                    \n(3) 1 mole of C2<\/sub>H6<\/sub> contains 6.023 x 1023<\/sup> number of molecules.
                                                    \n\u2234 3 moles of C2<\/sub>H6<\/sub> will contain 3 x 6.023 x 1023<\/sup>molecules.<\/p>\n

                                                    Question 25.
                                                    \nChlorine is prepared in the laboratory by treating manganese dioxide (MnO2<\/sub>) with aqueous hydrochloric acid according to the reaction.
                                                    \n4HCl(aq)<\/sub> + MnO2(s)<\/sub><\/sub> \u2192 2H2<\/sub>O(l)<\/sub> + MnCl2(aq)<\/sub><\/sub> + Cl2(aq)<\/sub><\/sub>
                                                    \nHow many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 g).
                                                    \nAnswer:
                                                    \n1 mole of MnO2<\/sub> = 55 + 32 = 87 g.
                                                    \n87 g of MnO2<\/sub> reacts with 4 moles of HCl. i.e. = 4 x 36.5 = 146 g of HCl.
                                                    \n\u2234 5 g of MnO2<\/sub> will react with \\(\\frac{146}{87}\\) x 5.0 = 8.40 g.<\/p>\n

                                                    Question 26.
                                                    \nThe density of water at room temperature is 1.0 g\/ml. How many molecules are there in a drop of water if its volume is 0.05 ml?
                                                    \nAnswer:
                                                    \nVolume of drop of water = 0.05 ml
                                                    \nMass of a drop of water = Volume x Density
                                                    \n= 0.05 ml x 1.0 g \/ ml.
                                                    \n= 0.05 g
                                                    \nMolar mass of water (H2<\/sub>O) = 18 g
                                                    \n18 g of water = 1 mole
                                                    \n0. 05 g of water = \\(\\frac{1}{18}\\) x 0.05 = 0.0028 mol.
                                                    \nNo. of molecules present in one mole of water = 6.023 x 1023<\/sup>
                                                    \nNo. of molecules present in 0.0028 mole of water = \\(\\frac{6.023 \\times 10^{23} \\times 0.0028}{1}\\) = 1.68 x 1021 <\/sup>water molecules<\/p>\n

                                                    Question 27.
                                                    \nBalance the following equation by oxidation number method. MnO4<\/sub>–<\/sup> + Fe2+<\/sup> \u2192 Mn 2+<\/sup> + Fe3+<\/sup> (Acidic medium)
                                                    \nAnswer:
                                                    \n\"Samacheer
                                                    \nMnO4<\/sub>–<\/sup> + 5e–<\/sup> \u2192 Mn2+<\/sup> ……….(1)
                                                    \nFe2+<\/sup> \u2192 Fe3+<\/sup> + e–<\/sup> ……….(2)
                                                    \n\"Samacheer
                                                    \nto balance O and H atoms H2<\/sub>O and H+<\/sup> are added.
                                                    \nMnO4<\/sub>–<\/sup> + 5Fe2+<\/sup> + 8H+<\/sup> \u2192 5Fe3+<\/sup> + Mn2+<\/sup> + 4H2<\/sub>O<\/p>\n

                                                    Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 5-Mark Questions<\/span><\/strong><\/p>\n

                                                    I. Answer the following questions in detail<\/span>
                                                    \nQuestion 1.
                                                    \nDefine the following (a) equivalent mass of an acid (b) equivalent mass of a base (c) equivalent mass of an oxidising agent (cl) equivalent mass of a reducing agent.
                                                    \nAnswer:
                                                    \n(a) Equivalent mass of an acid:
                                                    \nEquivalent mass of an acid is the number of parts by mass of the acid which contains 1.008 part by mass of replaceable hydrogen atom.
                                                    \nEquivalent mass of an acid = \"Samacheer<\/p>\n

                                                    (b) Equivalent mass of a base:
                                                    \nIt is defined as the number of parts by mass of the base which contains one replaceable hydroxyl ion or which completely neutralizes one gram equivalent of an acid.
                                                    \nEquivalent mass of a base = \"Samacheer<\/p>\n

                                                    (c) Equivalent mass of an oxidising agent:
                                                    \nIt is defined as the number of parts by mass of an oxidising agent which can furnish 8 parts by mass of oxygen for oxidation.<\/p>\n

                                                    (d) Equivalent mass of a reducing agent:
                                                    \nIt is defined as the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or with one equivalent of any oxidising agent.<\/p>\n

                                                    Question 2.
                                                    \nCalculate the percentage composition of the elements present in lead nitrate. How many Kg of 02 can be obtained from 50 kg of 70% pure lead nitrate?
                                                    \nAnswer:
                                                    \nLead nitrate = Pb (NO3<\/sub>)2<\/sub>
                                                    \nMolecular mass of lead nitrate = 207 + (14 x 2) + (16 x 6)
                                                    \n= 207 + 28 + 96 = 331 g \/ mol.
                                                    \n331 g of lead nitrate contains 96 g of oxygen.
                                                    \n\u2234 50 x 103<\/sup> g of lead nitrate will contain \\(\\frac {96}{331}\\) x 50 x 103<\/sup>
                                                    \n= 14501.5 g
                                                    \n= 14.501 Kg of oxygen.
                                                    \n100 % pure lead nitrate contains 14.501 Kg of oxygen.
                                                    \n70 % pure lead nitrate will contain = \\(\\frac {14.501}{100}\\) x 70 = 10.15 Kg of oxygen.
                                                    \n.’. 70 % pure lead nitrate will contain 10.15 Kg of oxygen.<\/p>\n

                                                    Question 3.
                                                    \nDetermine the empirical formula of a compound containing K = 24.15%, Mn = 34.77% and rest is oxygen.
                                                    \nAnswer:
                                                    \n\"Samacheer
                                                    \nempirical formula of a compound = KMnO4<\/sub><\/p>\n

                                                    Question 4.
                                                    \nWrite the steps to be followed for writing empirical formula.
                                                    \nAnswer:
                                                    \nEmpirical formula shows the ratio of number of atoms of different elements in one molecule of the compound.
                                                    \nSteps for finding the Empirical formula:
                                                    \nThe percentage of the elements in the compound is determined by suitable methods and from the data collected; the empirical formula is determined by the following steps.<\/p>\n

                                                      \n
                                                    1. Divide the percentage of each element by its atomic mass. This will give the relative number of atoms of various elements present in the compound.<\/li>\n
                                                    2. Divide the atom value obtained in the above step by the smallest of them so as to get a simple ratio of atoms of various elements.<\/li>\n
                                                    3. Multiply the figures so obtained, by a suitable integer if necessary in order to obtain whole number ratio.<\/li>\n
                                                    4. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.<\/li>\n
                                                    5. Percentage of Oxygen = 100 – Sum of the percentage masses of all the given elements.<\/li>\n<\/ol>\n

                                                      Question 5.
                                                      \nAn organic compound was found to contain carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour density was found to be 29.5. What is the molecular formula of the compound?
                                                      \nAnswer:
                                                      \n\"Samacheer
                                                      \nEmpirical formula = C2<\/sub>H5<\/sub>NO
                                                      \nMolecular mass = 2 x vapour density = 2 x 29.5 = 59.0
                                                      \nEmpirical formula mass of C4<\/sub>H5<\/sub>NO = 24 + 5 + 14 + 16 = 59
                                                      \nn = \"Samacheer\u00a0= \\(\\frac {59}{59}\\) = 1
                                                      \nEmpirical formula mass 59
                                                      \n.’. Molecular formula = (Empirical formula)n<\/sub>
                                                      \n= (C2<\/sub>H5<\/sub>NO)1<\/sub>
                                                      \nMolecular formula = C2<\/sub>H5<\/sub>NO<\/p>\n

                                                      Question 6.
                                                      \nCalculate the empirical and molecular formula of a compound containing 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75.
                                                      \nAnswer:
                                                      \n\"Samacheer
                                                      \nEmpirical formula = C2<\/sub>H3<\/sub>O3<\/sub>
                                                      \nEmpirical formula mass = 24 + 3 + 48 = 75
                                                      \nMolecular mass = 2 x Vapour density = 2 x 75 = 150
                                                      \nn = \"Samacheer\u00a0= \\(\\frac {150}{75}\\) = 2
                                                      \nMolecular formula = (Empirical formula)n<\/sub>
                                                      \nMolecular formula = C2<\/sub>H3<\/sub>O3<\/sub> x 2
                                                      \nMolecular formula = C4<\/sub>H6<\/sub>O6<\/sub><\/p>\n

                                                      Question 7.
                                                      \nExplain the different types of redox reactions with example.
                                                      \nAnswer:
                                                      \nRedox reactions are classified into the following types:
                                                      \n(1) Combination reactions:
                                                      \nWhen two or more substances combine to form a single substance, the reactions are called combination reactions.
                                                      \nExample:
                                                      \n2Mg + O2<\/sub> \u2192 2MgO<\/p>\n

                                                      (2) Decomposition reactions:
                                                      \nChemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
                                                      \nExample:
                                                      \n2KClO3<\/sub> \u2192 2KCl + 3O2<\/sub><\/p>\n

                                                      (3) Displacement reactions:
                                                      \nThe reactions in which one ion or atom in a compound is replaced by an ion or atom of the other element are called displacement reactions.
                                                      \nExample:
                                                      \nCuSO4<\/sub> + Zn \u2192 ZnSO4<\/sub> + Cu<\/p>\n

                                                      (4) Disproportionation reactions:
                                                      \nThe reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
                                                      \nExample:
                                                      \n2HCHO + H2<\/sub>O \u2192 CH3<\/sub>OH + HCOOH<\/p>\n

                                                      (5) Competitive Electron transfer reactions:
                                                      \nThese are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
                                                      \nExample:
                                                      \nZn(s)<\/sub> + Cu2+<\/sup> \u2192 Zn2+<\/sup>(aq)<\/sub> + Cu(s)<\/sub>\u00a0 Here Zn – oxidised; Cu2+<\/sup> reduced<\/p>\n

                                                      Question 8.
                                                      \nWrite the steps to be followed while balancing redox equation by oxidation number method.
                                                      \nAnswer:
                                                      \nOxidation number method:
                                                      \nThis method is based on the fact that
                                                      \nNumber of electrons lost by atoms = Number of electrons gained by atoms<\/p>\n

                                                      Steps to be followed while balancing Redox reactions by Oxidation Number method:<\/p>\n

                                                        \n
                                                      1. Write skeleton equation representing redox reaction<\/li>\n
                                                      2. Write the oxidation number of atoms undergoing oxidation and reduction.<\/li>\n
                                                      3. Calculate the increase or decrease in oxidation numbers per atom.<\/li>\n
                                                      4. Make increase in oxidation number equal to decrease in oxidation number by multiplying the formula of oxidant and reductant by suitable numbers.<\/li>\n
                                                      5. Balance the equation atomically on both sides except O and H atoms.<\/li>\n
                                                      6. Balance oxygen atoms by adding required number of water molecules to the side deficient
                                                        \nin oxygen atoms.<\/li>\n
                                                      7. Add required number of H+ ions to the side deficient in hydrogen atom if the reaction is in acidic medium.<\/li>\n
                                                      8. For reactions in basic medium, add H2<\/sub>O molecules to the side deficient in hydrogen atoms and simultaneously add equal number of OH ions on the other side of the equation.<\/li>\n
                                                      9. Finally balance the equation by cancelling common species present on both sides of the equation.<\/li>\n<\/ol>\n

                                                        Question 9.
                                                        \nBalance the following equation by oxidation number method:
                                                        \nAnswer:
                                                        \nK2<\/sub>Cr2<\/sub>O7<\/sub> + KCl + H2<\/sub>SO4<\/sub> \u2192 KHSO4<\/sub> + CrO2<\/sub>Cl 2<\/sub> + H2<\/sub>O
                                                        \nStep – 1
                                                        \n\"Samacheer
                                                        \nStep – 2
                                                        \nK2<\/sub>Cr2<\/sub>O7<\/sub> + KCl + H2<\/sub>SO4<\/sub> \u2192 KHSO4<\/sub> + 2CrO2<\/sub>C2<\/sub> + H2<\/sub>O<\/p>\n

                                                        Step 3.
                                                        \nTo balance Cl atom, KC1 is multiplied by 4
                                                        \nK2<\/sub>Cr2<\/sub>O7<\/sub> + 4KCl + H2<\/sub>SO4<\/sub> \u2192 2CrO2<\/sub>Cl2<\/sub> + KHSO2<\/sub> + H2<\/sub>O<\/p>\n

                                                        Step 4.
                                                        \nTo balance K atom, KHSO4<\/sub> is multiplied by 6.
                                                        \nK2Cr2O7<\/sub> + 4KCl + H2SO4<\/sub> \u2192 2CrO2<\/sub>Cl2<\/sub> + 6KHSO4<\/sub>+ H2<\/sub>O<\/p>\n

                                                        Step 5.
                                                        \nTo balance O and H atoms, HSO4<\/sub> is multiplied by 6, H20 is multiplied by 3.
                                                        \nAnswer:
                                                        \nK2<\/sub>Cr2<\/sub>O7<\/sub> + 4KCl + 6H2<\/sub>SO4<\/sub> \u2192 2CrO2<\/sub>Cl2<\/sub> + 6KHSO4<\/sub> + 3H2<\/sub>O<\/p>\n

                                                        (2) P + HNO3<\/sub> \u2192 H3<\/sub>PO4<\/sub> + NO2<\/sub> + H2<\/sub>O
                                                        \nStep – 1
                                                        \n\"Samacheer
                                                        \nStep 2.
                                                        \nP + 5HNO3<\/sub> \u2192 H3<\/sub>PO4<\/sub> + 5NO2<\/sub> + H2<\/sub>O<\/p>\n

                                                        (3) CuO + NH3<\/sub> \u2192 Cu + N2<\/sub> + H2<\/sub>O
                                                        \nStep 1.
                                                        \n\"Samacheer
                                                        \nStep 2.
                                                        \n3CuO + 2NH3<\/sub> \u2192 3Cu + N2<\/sub> + H2<\/sub>O<\/p>\n

                                                        Step 3.
                                                        \nTo balance O and N, water is multiplied by 3.
                                                        \n3CuO + 2NH3<\/sub> \u2192 3Cu + N2<\/sub> + 3H2<\/sub>O<\/p>\n

                                                        (4) Zn + HNO3<\/sub> \u2192Zn(N03<\/sub>)2<\/sub> + NH4<\/sub>NO3<\/sub> + H2<\/sub>O
                                                        \nStep – 1.
                                                        \n\"Samacheer
                                                        \nStep 2.
                                                        \n4Zn + HNO3<\/sub> \u2192 4Zn(NO3<\/sub>)3<\/sub> + NH4<\/sub>NO2<\/sub> + H2<\/sub>O<\/p>\n

                                                        Step 3.
                                                        \nTo balance N, HNO3<\/sub> is multiplied by 10
                                                        \n4Zn + 10HNO3 \u2192 4Zn(NO3<\/sub>)2<\/sub> + NH4<\/sub>NO3<\/sub> + H3<\/sub>O<\/p>\n

                                                        Step 4.
                                                        \nTo balance oxygen, H2<\/sub>O is multiplied by 3
                                                        \n4Zn + 10HNO3<\/sub> \u2192 4Zn(NO3<\/sub>)2<\/sub> + NH4<\/sub>NO3<\/sub> + 3H3<\/sub>O<\/p>\n

                                                        Question 10.
                                                        \nBalance the following equation by ion-electron method In acidic medium.
                                                        \nAnswer:
                                                        \n(i) S2<\/sub>O3<\/sub>2-<\/sup> + I2<\/sub> \u2192 S2<\/sub>O4<\/sub>2-<\/sup> + SO2<\/sub> + I–<\/sup>
                                                        \nOxidation half reaction:
                                                        \n\"Samacheer
                                                        \nReduction half reaction:
                                                        \n\"Samacheer
                                                        \nTo balance, SO2<\/sub> is added on RHS of the equation.
                                                        \nS2<\/sub>O3<\/sub>2-<\/sup> + I2<\/sub> \u2192 S2<\/sub>O4<\/sub>2-<\/sup> + 2I–<\/sup> + SO2<\/sub>
                                                        \nTo balance oxygen atom, S2<\/sub>O3<\/sub>2-<\/sup> and SO2<\/sub> is multiplied by 2.
                                                        \n2S2<\/sub>O3<\/sub>2-<\/sup> + I2<\/sub> \u2192 S2<\/sub>O4<\/sub>2-<\/sup> + 2I – + 2SO2<\/sub><\/p>\n

                                                        (2) Sb3+<\/sup> + MnO4<\/sub> \u2192 Sb5+<\/sup> + Mn2+<\/sup>
                                                        \nOxidation half reaction:
                                                        \nSb3+<\/sup> \u2192 Sb5+<\/sup> + 2e –<\/sup> ……….(1)
                                                        \nReduction half reaction:
                                                        \n\"Samacheer
                                                        \nIn equation (2), H2<\/sub>O is added on L.HS to balance oxygen atom.
                                                        \nMnO2<\/sub> + 5e–<\/sup> \u2192 Mn2+<\/sup> + 4H2<\/sub>O ………(3)
                                                        \nTo balance Hydrogen atoms, H\u2019 is added on RHS.
                                                        \nMnO4<\/sub>–<\/sup> + 5e–<\/sup> + 8H+<\/sup> \u2192 Mn2+<\/sup> + 4H2<\/sub>O ………(4)
                                                        \nEquation (4) is multiplied by 2 and equation (I) is multiplied by 5 to equalise the electrons gained and electrons lost.
                                                        \n\"Samacheer<\/p>\n

                                                        (3) MnO4<\/sub>–<\/sup> + I–<\/sup> \u2192 MnO2<\/sub> + I2<\/sub>
                                                        \nOxidation half reaction:
                                                        \n2I–<\/sup> + 2e–<\/sup> \u2192 I–<\/sup> ………..(1)
                                                        \nReduction half reaction:
                                                        \nMnO4<\/sub>–<\/sup> \u00a0\u2192 MnO2<\/sub> + e–<\/sup> ………..(2) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sup>
                                                        \nEquation (1)is multiplied by 3
                                                        \n\"Samacheer
                                                        \nEquation (2)is multiplied by 2.
                                                        \n\"Samacheer
                                                        \nTo balance oxygen and hydrogen atoms, H+<\/sup> is added on RHS and H2<\/sub>O is added on LHS.
                                                        \n2MnO4<\/sub> + 6I–<\/sup> + 4H+<\/sup> \u2192 2MnO2<\/sub> + 2I2<\/sub> + 2H2<\/sub>O<\/p>\n

                                                        In acidic medium
                                                        \n(4) MnO4<\/sub>–<\/sup> + Fe2+<\/sup> \u2192 Mn2+<\/sup> + Fe3+<\/sup>
                                                        \nOxidation half reaction:
                                                        \nFe2+<\/sup> \u2192 Fe3+<\/sup> + e–<\/sup> ……..(1)
                                                        \nReduction half reaction:
                                                        \n\"Samacheer
                                                        \nTo balance oxygen, H+<\/sup> is added on RHS and H2<\/sub>O is added on LHS.
                                                        \nMnO4<\/sub>–<\/sup> + 5Fe2+<\/sup> + 8H+<\/sup> \u2192 Mn2+<\/sup> + 5Fe3+<\/sup> + 4H2<\/sub>O<\/p>\n

                                                        (5) Cr(OH)4<\/sub>–<\/sup> + H2<\/sub>O2<\/sub> \u2192 CrO4<\/sub>–<\/sup>
                                                        \nOxidation half reaction:
                                                        \n\"Samacheer
                                                        \nReduction half reaction:
                                                        \n\"Samacheer
                                                        \nTo balance oxygen and hydrogen atoms, OH and H2<\/sub>O are added.
                                                        \n2Cr(OH)4<\/sub> + 3H2<\/sub>O2 <\/sub>+ 20H–<\/sup> \u2192 2CrO2<\/sub>–<\/sup> + 8H2<\/sub>O<\/p>\n

                                                        Question 11.
                                                        \n(a) Define equivalent mass of an oxidising agent.
                                                        \n(b) How would you calculate the equivalent mass of potassium permanganate?
                                                        \n(a) The equivalent mass of an oxidizing agent is the number of parts by mass which can furnish 8 parts by mass of oxygen for oxidation.
                                                        \n(b) Potassium permanganate is an oxidizing agent.
                                                        \n\"Samacheer
                                                        \n80 parts by mass of oxygen are given by 316 g of KMnO4<\/sub>.
                                                        \n.’. 8 parts by mass of oxygen will be furnished by \\(\\frac {316}{80}\\) x 8 = 31.6
                                                        \nEquivalent mass of KMnO4<\/sub> = 31.6 g eq-1<\/sup>.<\/p>\n

                                                        Question 12.
                                                        \n(a) Define equivalent mass of an reducing agent.
                                                        \n(b) How would you determine the equivalent mass of Ferrous sulphate?
                                                        \nAnswer:
                                                        \n(a) The equivalent mass of a reducing agent is the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or one equivalent of any oxidising agent.
                                                        \n(b) Ferrous sulphate is a reducing agent.
                                                        \n\"Samacheer
                                                        \n16 parts by mass of oxygen oxidised 304 parts by mass of FeSO4<\/sub>.
                                                        \n\u2234 8 parts by mass of oxygen will oxidise \\(\\frac {316}{80}\\) x 8 parts by mass of ferrous sulphate = 152 .
                                                        \nThe equivalent mass of ferrous sulphate (anhydrous) = 152.
                                                        \nThe equivalent mass of crystalline ferrous sulphate is (FeSO4<\/sub>.7H2<\/sub>O) = 152 + 126 = 278
                                                        \nThe equivalent mass of crystalline ferrous sulphate = 278.<\/p>\n

                                                        Question 13.
                                                        \nA compound on analysis gave the following percentage composition: C = 24.47%, H = 4.07 %, Cl = 7 1.65%. Find out its empirical formula.
                                                        \nAnswer:
                                                        \n\"Samacheer
                                                        \nthe empirical formula is CH2<\/sub>Cl.<\/p>\n

                                                        Question 14.
                                                        \nA laboratory analysis of an organic compound gives the following mass percentage composition: C = 60%, H = 4.48% and remaining oxygen.
                                                        \nAnswer:
                                                        \n\"Samacheer
                                                        \n\u2234 the empirical formula is C9<\/sub>H8<\/sub>O4<\/sub>.<\/p>\n

                                                        Question 15.
                                                        \nAn insecticide has the following percentage composition by mass: 47.5% C, 2.54% H, and 50.0% Cl. Determine its empirical formula and molecular formulae. Molar mass of the substance is 354.5 g mol-1<\/sup>
                                                        \nAnswer:
                                                        \n\"Samacheer
                                                        \nThe empirical formula is C14<\/sub>H9<\/sub>C15<\/sub>.
                                                        \nCalculation of Molecular formula:
                                                        \nThe empirical formula mass (C14<\/sub>H9<\/sub>C15<\/sub>) = (14 X 12) + (9 X 1) + (5 X 35.5)
                                                        \nn = \"Samacheer\u00a0= \\(\\frac {354.5}{354.5}\\) = 1
                                                        \nEmpirical formula mass 354.5
                                                        \nMolecular formula = (Empirical formu1a)n<\/sub>
                                                        \n= (C14<\/sub>H9<\/sub>Cl5<\/sub>)1<\/sub>
                                                        \n\u2234 Molecular formula = C14<\/sub>H9<\/sub>Cl5<\/sub><\/p>\n

                                                        Question 16.
                                                        \nAn organic fruit smelling compound on analysis has the following composition by mass: C = 54.54%, H = 9.09%, O = 36.36%. Find out the molecular formula of the compound. The vapour density of the compound was found to be 44.
                                                        \nAnswer:
                                                        \n\"Samacheer
                                                        \n\u2234 The empirical formula is C2<\/sub>H4<\/sub>O
                                                        \nTheempirical formula mass(C2<\/sub>H4<\/sub>O) = (12 x 2)+(1 x 4)+(16 x 1) = 44
                                                        \nMolecular mass = 2 x Vapour density = 2 x 44 = 88
                                                        \nn = \"Samacheer\u00a0= \\(\\frac {88}{44}\\) = 2
                                                        \nMolecular formula = (Empirical formula)n<\/sub>
                                                        \n= (C2<\/sub>H4<\/sub>O)2<\/sub>
                                                        \n\u2234 Molecular formula = C2<\/sub>H4<\/sub>O2<\/sub><\/p>\n

                                                        Question 17.
                                                        \nCalculate the percentage composition of the elements present in magnesium carbonate. How many Kg of CO2<\/sub> can be obtained from 100 Kg of is 90% pure magnesium carbonate.
                                                        \nMolar mass of MgCO3<\/sub> = 84.32 g mol-1<\/sup>
                                                        \nPercentage of Mg = \\(\\frac {24}{84.32}\\) x 100 = 28.46%
                                                        \nPercentage of C = \\(\\frac {12}{84.32}\\) x 100 = 14.23%
                                                        \nPercentage of O3<\/sub> = \\(\\frac {48}{84.32}\\) x 100 = 57.0%
                                                        \n\"Samacheer
                                                        \n84.32 g of 100% pure MgCO3<\/sub> gives 44g of CO2<\/sub>
                                                        \n\u2234 100 x 103<\/sup>g of 100% pure MgCO3<\/sub> gives = \\(\\frac {44}{84.32}\\) x 100 x 103<\/sup>
                                                        \n= 52.182 x 103<\/sup> g CO2<\/sub>
                                                        \n100% pure MgCO3<\/sub> gives 52.182 x 103<\/sup> g CO2<\/sub>
                                                        \n\u2234 90% pure MgCO3<\/sub> will give \\(\\frac{52.182 \\times 10^{3}}{100}\\) x 90 = 46963.8 g CO2<\/sub><\/p>\n

                                                        Question 18.
                                                        \nUrea is prepared by the reaction between ammonia and carbon dioxide.
                                                        \n\"Samacheer
                                                        \nIn one process, 637.2 g of NH3<\/sub> are allowed to react with 1142 g of CO2<\/sub>.
                                                        \n(a) Which of the two reactants is the limiting reagent?
                                                        \n(b) Calculate the mass of(NH4<\/sub>)2<\/sub>CO formed.
                                                        \n(c) how much of the excess reagent in grams is left at the end of the reaction?
                                                        \n\"Samacheer
                                                        \nNo. of moles of ammonia = \\(\\frac{637.2}{17}\\) = 37.45 mole
                                                        \nNo. of moles of CO2<\/sub> = \\(\\frac{1142}{44}\\) = 25.95 moles
                                                        \nAs per the balanced equation, one mole of CO2<\/sub> requires 2 moles of ammonia.
                                                        \n\u2234 No. of moles of NH3<\/sub> required to react with 25.95 moles of CO2<\/sub> is = \\(\\frac {2}{1}\\) x 25.95 = 51.90 moles.
                                                        \n\u2234 37.45 moles of NH3<\/sub> is not enough to completely react with CO2<\/sub> (25.95 moles).
                                                        \nHence, NH3<\/sub> must be the limiting reagent, and CO2<\/sub> is excess reagent.<\/p>\n

                                                        (b) 2 moles of ammonia produce 1 mole of urea.
                                                        \n\u2234 Limiting reagent 37.45 moles of NH3<\/sub> can produce \\(\\frac {1}{2}\\) x 37.45 moles of urea.
                                                        \n= 18.725 moles of urea.
                                                        \n\u2234 The mass of 18.725 moles of urea = No. of moles x Molar mass
                                                        \n= 18.725 x 60
                                                        \n= 1123.5 g of urea.<\/p>\n

                                                        (c) 2 moles of ammonia requires 1 mole of CO2<\/sub>.
                                                        \n\u2234 Limiting reagent 37.45 moles of NH3<\/sub> will require x 37.45 moles of CO2<\/sub>.
                                                        \n= 18.725 moles of CO2<\/sub>.
                                                        \n\u2234 No. of moles of the excess reagent (CO2<\/sub>) left = 25.95 – 18.725 = 7.225
                                                        \nThe mass of the excess reagent (CO2<\/sub>) left = 7.225 x 44 = 317.9 g CO2<\/sub>.<\/p>\n

                                                        Question 19.
                                                        \n(a) Define oxidation number.
                                                        \n(b) What are the rules used to assign oxidation number?
                                                        \nAnswer:
                                                        \n(a) Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely.
                                                        \n(b) Rules to assign oxidation number:<\/p>\n

                                                          \n
                                                        • Oxidation number of a substance in its elementary state is equal to zero (H2<\/sub>, Br2<\/sub>, Na)<\/li>\n
                                                        • Oxidation number of a mono-atomtic ion is equal to the charge on the Ton (Na4<\/sub> =+ 1,<\/li>\n
                                                        • Oxidation number of hydrogen in a compound is +1 (except hydrides).<\/li>\n
                                                        • Oxidation number of hydrogen in metal hydrides is -1 (NaH, CaH2<\/sub>).<\/li>\n
                                                        • Oxidation number of oxygen in a compound is -2 (except OF2<\/sub> and peroxides).<\/li>\n
                                                        • Oxidation number of oxygen in peroxides is -1 (He,, Na\u00bb,) = I.<\/li>\n
                                                        • Oxidation number of oxygen in fluorinated compounds is either +1 or +2.(OF2<\/sub> = +2, O,F2<\/sub> + 1).<\/li>\n
                                                        • Fluorine has an oxidation number -1 in all \u00ects compounds.<\/li>\n
                                                        • The sum of the oxidation number of all the atoms in neutral molecules is equal to Zero.<\/li>\n
                                                        • For all ions, the sum of the oxidation number of all atoms is equal to the charge of the ion.<\/li>\n<\/ul>\n

                                                          Question 20.
                                                          \nBalance the following equation by oxidation number method.
                                                          \nC6<\/sub>H6<\/sub> + O2<\/sub> \u2192 CO2<\/sub> + H2<\/sub>O
                                                          \n\"Samacheer
                                                          \n(ii) Balance the changes in O.N. by multiplying the oxidant and reductant by suitable numbers
                                                          \n2 C6<\/sub>H6<\/sub> + 15 O2<\/sub> \u2192 CO2<\/sub> + H2<\/sub>O<\/p>\n

                                                          (iii) Balance the equation atomically (except O and H).
                                                          \n2C6<\/sub>H6<\/sub> + 15 O2<\/sub> \u2192 12 CO2<\/sub> + H2<\/sub>O<\/p>\n

                                                          (iv) Balance O atoms by adding one HzO molecule to the RHS for making the number of molecules of H2<\/sub>O to be 6.
                                                          \n2C6<\/sub>H6<\/sub> + 15 O2<\/sub> \u2192 12 CO2<\/sub> + 6H2<\/sub>O<\/p>\n

                                                          Question 21.
                                                          \nBalance the following equation by oxidation number method.
                                                          \nKMnO4<\/sub> + HCl \u2192 KCl + MnCl2<\/sub> + H2<\/sub>O + Cl2<\/sub>
                                                          \n\"Samacheer<\/p>\n

                                                          (ii) 2KMnO4<\/sub> + 1OHCl \u2192 KCl + MnCl2<\/sub> + H2<\/sub>O + Cl2<\/sub><\/p>\n

                                                          (iii) Balance the equation atomically (except O and H).
                                                          \n2KMnO4<\/sub> + 1OHCl \u2192 2KCl + 2MnCl2<\/sub>, + H2<\/sub>O + Cl2<\/sub><\/p>\n

                                                          (iv) Balance chlorine atoms by adding HC1 and multiplying Cl2<\/sub> by 5.
                                                          \n2KMnO4<\/sub> + 16HCl \u2192 2KCl + 2MnCl2<\/sub> + H2<\/sub>O + 5Cl2<\/sub><\/p>\n

                                                          (v) To balance O and H, H2<\/sub>O is multiplied by 8.
                                                          \n2KMnO4<\/sub> \u00a0+ I6HCl \u2192 2KCl + 2MnCl4<\/sub> + 8H2<\/sub>O + 5Cl2<\/sub><\/p>\n

                                                          Question 22.
                                                          \nBalance the following equation by oxidation number method.
                                                          \nKMnO4<\/sub> + FeSO4<\/sub> + H2S04<\/sub> \u2192 K2SO4<\/sub> + MnSO4<\/sub> + Fe2(S04<\/sub>)3<\/sub> + H2<\/sub>O
                                                          \nAnswer:
                                                          \n\"Samacheer
                                                          \n(ii) 2KMnO4<\/sub> + 10FeSO4<\/sub> + H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + MnSO4<\/sub> + Fe2(SO4<\/sub>)3<\/sub>+ H2<\/sub>O<\/p>\n

                                                          (iii) Balance the equation atomically (except O and H) and sulphate ions.
                                                          \n2KMnO4<\/sub> + 1OFeSO4<\/sub> + 8H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2MnSO4<\/sub> + 5Fe4<\/sub>(SO4<\/sub>)3<\/sub> + H2<\/sub>O<\/p>\n

                                                          (iv) Balance O atoms by multiplying H2<\/sub>O by 8.
                                                          \n2KMnO4<\/sub> + 1OFeSO4<\/sub> + 8H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2MnSO4<\/sub> + 5Fe2<\/sub>(SO4<\/sub>)3<\/sub> + 8H2<\/sub>O<\/p>\n

                                                          Question 23.
                                                          \nBalancing of molecular equation in alkaline medium.
                                                          \nMnO2<\/sub> + O2<\/sub> + KOH \u2192 K2<\/sub>MnO4<\/sub> + H2<\/sub>O
                                                          \n\"Samacheer<\/p>\n

                                                          (ii) Balance the changes in O, N, by multiplying the oxidant and reductant by suitable numbers.
                                                          \n4MnO2<\/sub> + 2O2<\/sub>+ KOH \u2192 K2<\/sub>MnO4<\/sub> + H2<\/sub>O<\/p>\n

                                                          (iii) Balance the equation atomically (except O and H).
                                                          \n4MnO2<\/sub> + 2O2<\/sub> + KOH \u2192 K2<\/sub> MnO4<\/sub> + H2<\/sub>O<\/p>\n

                                                          (iv) Balance oxygen and hydrogen atoms by multiplying H20 by 4.
                                                          \n4MnO2<\/sub> + 2O2<\/sub> + 8KOH \u2192 4K2<\/sub>MnO4<\/sub> + 4H2<\/sub>O<\/p>\n

                                                          Question 24.
                                                          \nExplain the steps involved in ion-electron method for balancing redox reaction.
                                                          \nAnswer:
                                                          \nIon electron method makes use of the Half reactions. Steps involved in this method are,<\/p>\n

                                                            \n
                                                          1. Write the equation in the net ionic form without attempting to balance it.<\/li>\n
                                                          2. Write and locate the oxidation number of atoms undergoing oxidation and reduction from the knowledge of calculation of oxidation number.<\/li>\n
                                                          3. Write two half reactions showing oxidation and reduction separately.<\/li>\n
                                                          4. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.<\/li>\n
                                                          5. Add required number of H+<\/sup> ions to the side deficient in hydrogen atom if the reaction is in acidic medium<\/li>\n
                                                          6. Add electrons to whichever side is necessary to make up the difference in oxidation number.<\/li>\n
                                                          7. Add the two half reactions. The resulting equation is a net balanced equation.<\/li>\n
                                                          8. For reactions in basic medium, add H2<\/sub>O and hydrogen ion to balance H and O.<\/li>\n
                                                          9. Finally balance the equation by cancelling common species present on both sides of the equation.<\/li>\n<\/ol>\n

                                                            Question 25.
                                                            \nWrite balanced equation for the oxidation of Ferrous ions to Ferric ions by permanganate ions in acid solution. The permanganate ion forms Mn2+ ions under these conditions.
                                                            \nAnswer:
                                                            \nNet ionic reaction:
                                                            \nMnO4<\/sub>–<\/sup> + Fe2+<\/sup> + H+<\/sup> \u2192 Mn2+<\/sup> + Fe3+<\/sup>
                                                            \nOxidation half reaction:
                                                            \nFe2+<\/sup> \u2192 Fe3+<\/sup> + e–<\/sup> ……….(1)
                                                            \nReduction half reaction:
                                                            \nMnO4<\/sub>–<\/sup> + 5e–<\/sup> \u2192 Mn2+<\/sup> ……….(2)
                                                            \n\"Samacheer
                                                            \nTo balance O and H, H+<\/sup> and H2<\/sub>O are added.
                                                            \nMnO4<\/sub>–<\/sup> + 5Fe2+<\/sup> + 8H+<\/sup> \u2192 Mn2+<\/sup> + 5Fe3+<\/sup> +4H2<\/sub>O<\/p>\n

                                                            Question 26.
                                                            \nA flask A contains 0.5 mole of oxygen gas. Another flask B contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms.
                                                            \nAnswer:
                                                            \nFlask A:
                                                            \n1 mole of oxygen gas = 6.023 x 1023<\/sup> molecules
                                                            \n\u2234 0.5 mole of oxygen gas = 6.023 x 1023<\/sup> x 0.5 molecules
                                                            \nThe number of atoms in flask A = 6.023 x 1023<\/sup> x 0.5 x 2 = 6.023 x 1023<\/sup> atoms.<\/p>\n

                                                            Flask B:
                                                            \n1 mole of ozone gas = 6.023 x 1023<\/sup> molecules
                                                            \n0. 4 mole of ozone gas = 6.023 x 1023<\/sup> x 0.4 molecules
                                                            \nThe number of atoms in flask B = 6.023 x 1023<\/sup> x 0.4 x 3 = 7.227 x 1023<\/sup> atoms.
                                                            \n\u2234 Flask B contains a great number of oxygen atoms as compared to flask A.<\/p>\n

                                                            Question 27.
                                                            \n(a) Formulate possible compounds of \u2018Cl\u2019 in its oxidation state is:
                                                            \n0, – 1, + l, + 3,+ 5, + 7<\/p>\n

                                                            (b) H2<\/sub>O2<\/sub> act as an oxidising agent as well as reducing agent where as O3<\/sub> act as only oxidizing agent. Prove it.
                                                            \n(a) (1) Cl oxidation number O in Cl2<\/sub>.
                                                            \n(2) Cl oxidation number -1 in HCl.
                                                            \n(3) Cl oxidation number +3 in HCl O2<\/sub>.
                                                            \n(4) Cl oxidation number +5 in KClO3<\/sub>.
                                                            \n(5) Cl oxidation number +7 in C12<\/sub>O7<\/sub>.<\/p>\n

                                                            (b) In H2<\/sub>O2<\/sub>, oxidation number of oxygen is -1 and it can vary from O to -2 (+ 2 is possible in OF2<\/sub>). The oxidation number can decrease or increase, because of this H2<\/sub>O2<\/sub> can act both oxidising and reducing agent. Ozone (O3<\/sub> ) only acts as oxidising agent since it decomposes to give nascent oxygen.<\/p>\n

                                                            Question 28.
                                                            \nThe Mn3+<\/sup> ion is unstable in solution and undergoes disproportionation to give Mn2+<\/sup>, MnO2<\/sub> and H ion. Write a balanced ionic equation for the reaction.
                                                            \nThe skeletal equation is:
                                                            \nMn3+<\/sup>(aq)<\/sub> \u2192 Mn2+<\/sup>(aq)<\/sub> + MnO2(aq)<\/sub> + H+<\/sup>(aq)<\/sub>
                                                            \nOxidation half equation:
                                                            \n\"Samacheer
                                                            \nBalance O.N. by adding electrons,
                                                            \nMn3+<\/sup>(aq)<\/sub> \u2192 \u00a0MnO2(s)<\/sub> + e–<\/sup>
                                                            \nBalance charge by adding 4H ions,
                                                            \nMn3+<\/sup>(aq)<\/sub> \u2192 \u00a0MnO2(s)<\/sub> + 4H+<\/sup>(aq)<\/sub> +\u00a0e–<\/sup>
                                                            \nBalance O atoms by adding 2H2<\/sub>O
                                                            \nMn3+<\/sup>(aq)<\/sub> + 22H2<\/sub>O(l)<\/sub> \u2192 MnO2(s)<\/sub> + 4H+<\/sup>(aq)<\/sub> +\u00a0e–<\/sup> ……….(1)
                                                            \nReduction half equation:
                                                            \n\"Samacheer
                                                            \nBalance ON. by adding electrons:
                                                            \nMn3+<\/sup>(aq)<\/sub> + e–<\/sup> \u2192 Mn2+<\/sup>(aq)<\/sub> …………(2)
                                                            \nAdding Equation (1) and (2), the balanced equation for the disproportionation reaction is
                                                            \n2MH3+<\/sup>(aq)<\/sub> + 2H2<\/sub>O(l)<\/sub> \u2192 MnO2(s)<\/sub> + Mn2+<\/sup>(aq)<\/sub> + H+<\/sup>(aq)<\/sub><\/p>\n

                                                            Question 29.
                                                            \nChlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating with sulphur dioxide. Present a balanced equation for the reaction for this redox change taking place in water.
                                                            \nThe skeletal equation is:
                                                            \nCl2(aq)<\/sub> + SO2(aq)<\/sub> + H2<\/sub>O(l)<\/sub> \u2192 Cl–<\/sup>(aq)<\/sub> + SO4<\/sub>2-<\/sup>(aq)<\/sub>
                                                            \nReduction half equation:
                                                            \nCl2(aq)<\/sub> \u2192 Cl2-<\/sup>(aq)<\/sub>
                                                            \nBalance Cl atoms,
                                                            \n\"Samacheer
                                                            \nBalance O.N. by adding electrons
                                                            \nCl2(aq)<\/sub> + 2e–<\/sup> \u2192 2Cl + 2e–<\/sup>
                                                            \nOxidation half equation:
                                                            \n\"Samacheer
                                                            \nBalance O,N. by adding electrons
                                                            \nSO2(aq)<\/sub> \u2192 SO4<\/sub>2-<\/sup>(aq)<\/sub> + 2e–<\/sup>
                                                            \nBalance charge by adding 4H+<\/sup> ions:
                                                            \nSO2(aq)<\/sub> \u2192 SO4<\/sub>2-<\/sup>(aq)<\/sub> + 4H+<\/sup>(aq)<\/sub> + 2e–<\/sup>
                                                            \nBalance O atoms by adding 2H2<\/sub>O
                                                            \nSO2(aq)<\/sub> + 2H2<\/sub>O(l)<\/sub> \u2192 SO4<\/sub>2-<\/sup>(aq)<\/sub> + 4H+<\/sup>(aq)<\/sub> + 2e–<\/sup>
                                                            \nAdding equation (1) and (2), we have,
                                                            \nCl2(aq)<\/sub> + SO2(aq)<\/sub> + 2H2<\/sub>O(l)<\/sub> \u2192 Cl–<\/sup>2(aq)<\/sub> + SO4<\/sub>2-<\/sup>(aq)<\/sub> + 4H+<\/sup>(aq)<\/sub>
                                                            \nThis represents the balanced redox reaction.<\/p>\n

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