DPP-11 to 12 With Answer

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 11 Class : XIII Course : DPP No.1 Max. Time : 29 Total Time : 29 min. Single choice Objective ('–1' negative marking) Q.1 to Q.7 (3 marks 3 min.) [21, 21] Subjective Questions ('–1' negative marking) Q.8 (4 marks 4 min.) [4, 4] Match the Following (no negative marking) Q.9 (4 marks 4 min.) [4, 4] 1. 84 g of Iron (Fe) is reacted with sufficient amount of steam to produced 44.8 L, H2 gas at S.T.P. according to the following reaction, a Fe + b H O  c Fe O + dH . The stoichiometric coefficients of the reaction is 2 (At. wt., Fe = 56, O = 16, H = 1) 3 4 2 (1) 4, 3, 1, 4 (2*) 3, 4, 1, 4 (3) 1, 4, 2, 3 (4) none of these Sol. Mole of Fe a = Mole of H2 d 84 / a = 56 1.5  2 44.8 / 22.4 d ⇒ a  3 a d d 4 So, Ans. B. 2. Which statement is wrong- (1*) Oxidation number of oxygen is +1 in peroxides (2) Oxidation number of oxygen is +2 in oxygen difluoride (3) Oxidation number of oxygen is  1 2 in superoxides (4) Oxidation number of oxygen is –2 in most of is compound Sol. Oxidation number of oxygen is (–1) in peroxide. 3. Which of the following is not a redox reaction? (1) Mg + N  Mg N (2) MnO – + C O 2–  Mn2+ + CO 2 3 2 4 2 4 2 (3) CuSO4 + Kl  Cu  + 2 + K2SO4 (4*) AgCl + NH3  [Ag(NH ) ] Cl Sol. Those reaction in which oxidation number of any element do not change not a redox reaction. AgCl + NH3  [Ag(NH3)2]Cl. 4. In the reaction, 2S O 2– + I  S O 2– + 2I–, the eq. wt. of S O –2 is equal to its - (1) Mol. wt. (2*) Mol. wt./2 (3) 2 × mol. wt. (4) Mol. wt./6. Sol. I + 2S O 2–  S O 2– + 2I–. molecular wt. M ES O2 – = = . 4 6 no.of electron gain or loss by per moleculeof oxidant or reductant 2 2S O 2–  S O 2– + 2e– . 2 3 4 6 5. In the following change - 3Fe + 4H2O  Fe3O4 + 4H2. If the atomic weight of iron is 56, then its equivalent weight will be - (1) 42 (2*) 21 (3) 63 (4) 84 0 Sol. 3 Fe + 4H2O  8 / 3 Fe3O4 + 4H2 3Fe + 4H O  Fe O + 8H+ + 8e– 8 V.F. of Fe = 3 . EFe = Atomic mass V.F. 56 = 8 / 3 = 21. 6. How many millilitres of a 9 N H2SO4 solution will be required to neutralize completely 20 mL of a 3.6 N NaOH solution? (1) 18.0 mL (2*) 8.0 mL (3) 16.0 mL (4) 80.0 mL Sol. NNaOH = 3.6 ; VNaOH = 20 ; 3.6  20 NH2SO4 = 9 ; VH2SO4 = ? VH2SO4 = 9 = 8 mL. 7. Calculate the normality of an NaOH solution, 21.5 mL of which is required to convert 0.240 g of NaH2PO4 in a solution to monohydrogen phosphate. (1) 1.093 N (2*) 0.093 N (3) 0.048 N (4) 0.93 N Sol. Equivalent of NaOH = Equivalent of NaH2PO4 0.24 21.5 × 10–3 × N = 120 × 1  N = 0.093 8. Identify the oxidant and the reductant in the following reactions : (a) KMnO4 + KCl + H2SO4  MnSO4 + K2SO4 + H2O + Cl2 (b) FeCl2 + H2O2 + HCl  FeCl3 + H2O (7) (–1) (2) (0) Ans. (a) KMnO4 + KCl + H2SO4  (7) MnSO4 + K2SO4 + H2O + Cl2 . (2) KMnO4 (oxidant ) (–1)  (0) MnSO4 (reduction half). KCl (reductant)  Cl2 (oxidant half). (2) (–1) (3) (–2) (b) FeCl2 + H2O2 + HCl  FeCl3 + H2 O (2) (3) (oxidation half) FeCl2 (reductant)  FeCl3 (oxidation half). (–1) H2O2 (oxidant)  H O2– (reduction half). 9. Match the following : Column (Ξ™) Column (ΙΙ) (A) 50 ml of 3M HCl + 150 ml of 1M FeCl3 (p) 1.85 m (B) mole fraction of NaCl in aqueous solution of NaCl is 0.1 then molality of the solution is (q) [Cl–] = 3 M (C) 10%(w/w) propanol (C H OH) solution has molality (r) [H+]= 0.75 M 3 7 (D) 10.95% (w/v) HCl (s) 6.1 m Ans. (A – q, r) ; (B – s) ; (C – p) ; (D – q) 50  3  150  1 3 Sol. (A) [Cl–] = 200 600 = 200 = 3 M (B) molality = 0.1 0.9 18 1000 = 6.17 m (C) Molality = 10  1000 60  90 = 1.85 m. 10.95 (D) Molarity of HCl = 36.5 × 1000 = 3 M 100 DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 12 Class : XIII Course : DPP No.2 Max. Marks : 30 Total Time : 30 min. Single choice Objective ('–1' negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18] Multiple choice objective ('–1' negative marking) Q.7 to Q.8 (4 marks 4 min.) [8, 8] Subjective Questions ('–1' negative marking) Q.9 (4 marks 4 min.) [4, 4] 1. How many millilitres of 0.150 M H2SO4 (sulphuric acid) are required to react with 1.68 g of sodium hydrogen carbonate, (NaHCO3), according to the following equation ? H2SO4(aq) + 2NaHCO3(aq)  Na2SO4(aq) + 2H2O(𝑙) + 2CO2(g) (1) 55.55 mL (2*) 66.66 mL (3) 25.55 mL (4) 40.88 mL Sol. Moles of H2SO4 = moles of NaHCO3 2 MV 1000 = wt 1 mol wt × 2 1.68 V = 84 1000 × 2 × 1 0.15 = 66.66 ml 2. A sample of H SO (density 1.787 g mL–1) is labelled as 86% by weight. What is molarity of acid ? What volume of acid has to be used to make 1 litre of 0.2 M H2SO4 ? (1*) 15.68 M, 12.75 m𝑙 (2) 10.70 M, 12.75 m𝑙 (3) 15.68 M, 10.75 m𝑙 (4) 25.22 M, 20.55 m𝑙 86 1.787 Sol. M = 98 × 10 = 15.68 M M1V1 = M2V2 15.68 × V1 = 0.2 × 1000 V1 = 12.75 ml 3. What is the normality of the H2SO4 solution, 18.6 mL of which neutralizes 30.0 mL of a 1.55 N KOH solution? (1) 5.0 N (2) 1.25 N Sol. VNaOH = 18.6 ; NNaOH = ? 1.55  30 NNaOH = 18.6 = 2.5 N. 4. In the equation , SnCl2 + 2HgCl2  Hg2Cl2 + SnCl4 The equivalent weight of stannous chloride (molecular weight = 190) will be - (1) 190 (2*) 95 (3) 47.5 (4) 154.5 Sol. SnCl2 + 2HgCl2  Hg2Cl2 + SnCl4 E = molecular wt. 190 = = 95. SnCl2 Sn2+ no.of electron gain or loss by per molecule of oxidant or reductant 2  Sn4+ + 2e–. 5. Four different identical vessels at same temperature contains one mole each of C2H6, CO2, C𝑙2 and H2S at pressures P1, P2, P3 and P4 respectively. The value of van der Waals constant ‘a‘ for C2H6, CO2, Cl2 and H S is 5.562, 3.640, 6.579 and 4.490 atm.L2.mol–2 respectively. Then (1*) P3 < P1 < P4 < P2 (2) P1 < P3 < P2 < P4 (3) P2 < P4 < P1 < P3 (4) P1 = P2 = P3 = P4 6. Under critical states of a gas for one mol of a gas, compressibility factor is - 3 (1*) 8 8 (2) 3 (3) 1 (4) 1 4 7. Critical temperature for a particular gas is – 177°C then for which of the following case value of compressibility factor of the gas may be more than unity. (1) at 0°C and 0.01 atm (2*) at 0°C and 2000 atm (3*) at 60°C and 0.01 atm (4*) at 60°C and 10 atm Pb Sol. at very high Pressure Z = 1+ RT Z > 1 for particular realgas above boyle temp Z > 1. 8. The vander waal gas constant ‘a’ is given by 1 1 RTC 27 R2 T2 (1) 3 VC (2*) 3PC 2 (3) 8 PC (4*) 64 PC 9. Balance the following redox equations and identify the valency factor (n-factor) for different compunds or molecule involved in the reactions at reactant and product side. (i) K2Cr2O7 + H2O2 + H2SO4  K2SO4 + Cr2(SO4)3 + H2O + O2 Ans. K2Cr2O7 + 3H2O2 + 4H2SO4  K2SO4 + Cr2(SO4)3 + 7H2O + 3O2 . Sol. Mass Balance and Charge Balance : Remove the spectator ion — 2K+ , SO42–. Cr2O72– + H2O2 + 2H+  2Cr3+ + H2O + O2 . Oxidation Half : H2O2  O2 + 2H+ + 2e– . Reduction Half : Cr2O72– + 14H+ + 6e–  2Cr3+ + 7H2O. Total loss electrons = total gain electrons. 3H2O2 + Cr2O72– + 8H+  2Cr3+ + 7H2O + 3O2 . Add the spectator ion — 2K+ , SO42–. 3H2O2 + K2Cr2O7 + 4H2SO4  Cr2(SO4)3 + K2SO4 + 7H2O + 3O2. (ii) Zn + NaNO3 + NaOH  Na2ZnO2 + H2O + NH3 Ans. 4Zn + NaNO3 + 7NaOH = 4Na2ZnO2 + 2H2O + NH3 . Sol. Mass Balance and Charge Balance : Remove the spectator ion — Na+ Zn + NO3– + OH–  ZnO22– + H2O + NH3 . Oxidation Half : Zn + 4OH–  ZnO22– + 2H2O + 2e– . Reduction Half : NO3– + 6H2O + 8e–  NH3 + 9OH–. Total loss electrons = total gain electrons. 4Zn + 7OH– + NO3–  4ZnO22– + 2H2O + NH3 . Add the spectator ion — Na+ . 4Zn + 7NaOH + NaNO3  4Na2ZnO2 + 2H2O + NH3 .

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