DPP-9 to 10 With Answer

FAC ULTY DAILY PRACTICE PROBLEMS (DPP) COPY Subject : Physical/Inorg.Chemistry Date : DPP No. 9 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 to Q.9 (4 marks 4 min.) [8, 8] 1. Oxidation number of underlined elements are N O , SO 2–, NH+ 2 5 3 4– (1) +5, +2, –3 (2) +6, –2, +3 (3) +6, +2, –3 (4*) +5, +4, –3 Sol. ( 5) N2O5 (4) > SO2– (–3) > NH 2. Which of the following reactions involves oxidation and reduction- (1) NaBr + HCl  NaCl + HBr (2) HBr + AgNO3  AgBr +HNO3 (3*) H2 + Br2  2HBr (4) Na2O + H2SO4  Na2SO4 + H2O Sol. Redox Reaction — Those reaction in which oxidation & reduction takes place. 3. The reaction 3ClO–(aq.)  ClO – + 2Cl– (aq.) is an example of (1) oxidation (2) reduction (3*) disproportionation (4) decomposition reaction Sol. (1) 3ClO (aq)  (5) ClO– (–1) (aq) + 2Cl (aq). Disproportionation reaction. In this reaction Cl atom present in intermediate oxidation state and Cl undergoes both oxidation and reduction. 4. The oxidation number of oxygen in F2O is : (1) +1 (2) –2 (3*) +2 (4) –3 5. In which of the following reactions, is there a change in the oxidation number of nitrogen atoms ? (1) 2NO2  N O (2*) 2NO + H O  HNO + HNO (3) NH3 + H2O  NH + + OH– (4) N O + H O  2HNO 6. Which of the following can not be oxidized ? (1) NO – (2) CO 2– (3) PO 3– (4*) All of these 3 3 4 7. Phosphorous has the oxidation state of +3 in – (1*) Phosphorus acid (H3PO3) (2) Ortho phosphoric acid (H3PO4) (3) Meta phosphoric acid (HPO3) (4) Pyro phosphoric acid (H4P2O7) Sol. Sub. 8. How much volume of 63% w/w aq. HNO3 solution (d = 1.5 g/ml) is diluted with sufficient amount of water to prepare 1 L of 3 M HNO3 solution Ans. 200 ml. Sol. M1V1 = M2V2  63   63   100 1000  × V = 3 × 1000 ml   1  1.5  V1 = 200 ml. 9. A bottle of 1 litre capacity is labelled as 1 molar Al2(SO4)3 (aq). If the bottle is half filled and density of solution is 1.342 g/ml, molality of Al3+ (aq) in this solution will be : (A) 1 (B*) 2 (C) 3 (D) 4 1 Sol. m = (1000 1.342  342) = 1 1000 As each mole of Al (SO ) gives two mole of Al3+ ion mAl3 = 2 × 1 = 2 m FAC ULTY DAILY PRACTICE PROBLEMS (DPP) COPY Subject : Physical/Inorg.Chemistry Date : DPP No. 10 Class : XIII Course : DPP No.2 Max. Marks : 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 to Q.9 (4 marks 4 min.) [8, 8] 1. Which of the following reactions does not involve either oxidation or reduction ? (1) VO2+  V O (2) Na  Na+ (3) Zn+2  Zn (4*) CrO –2  Cr O –2 Sol. Oxidation number of both element Cr & O does not change. 2. The oxidation number and covalency of sulphur in the sulphur molecule (S8) are respectively- (1*) 0 and 2 (2) +6 and 8 (3) 0 and 8 (4) +6 and 2 Sol. S8 Crown Sulphur Oxidation number and covalency of sulphur in the sulphur molecule (S8) are respectively O & 2 3. Oxidation number of iodine varies from - (1) –1 to +1 (2*) –1 to +7 (3) +3 to +5 (4) –1 to +5 Sol. oxidation number of iodine varies from (–1 to +7). minimum oxidation number of I = – 1. maximum oxidation number of I = + 7. 4. In the reaction x FeCl3 + yH S  FeCl + S + HCl (1*) x = 2, y = 1 (2) x = 3, y = 2 (3) x = 4, y = 3 (4) x = 2, y = 2. Sol. Mass Balance and Charge Balance : Remove the spectator ion — Cl–. Fe3+ + H S Oxidation Half :  Fe2+ + S + HCl. Fe3+ + 1e–  Reduction Half : Fe2+ H2S  S + 2H + 2e . + – Total loss electrons = total gain electrons. 2Fe3+ + H S  2Fe2+ + S + 2H+. Add the spectator ion — Cl–. 2FeCl3 + H2S  2FeCl2 + S + 2HCl. 5. Which of the following is a redox reaction ? (1) 2CrO 2– + 2H+  Cr O 2– + H O (2) CuSO + 4NH  [Cu(NH ) ]SO (3*) Na S O +   Na S O + Na (4) Cr O 2– + 2OH–  2CrO 2– + H O 6. Which of the following can show disproportionation reaction ? (1) ClO – (2) Cl– (3*) ClO – (4*) ClO – 4 2 3 7. 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. Subjective 8. Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure limestone (CaCO3). Ans. 106.4 kg Sol. CaCO (s)  CaO(s) + CO (g) K.mole of CaCO (pure) = 200  95 = K mole of CaO 3 100 100  200  95  so wt. of CaO =  100 100  × 56 = 106.4 kg. 9. A sample of hydrazine sulphate [N2H6SO4] was dissolved in 100 mL water. 10 mL of this solutions was treated with excess of FeCl3 solution and named to complete the reaction. Ferrons ions formed were esti- mated and it required 20 mL of M/50 KMnO4 solution. Estimate the amount of hydrazine sulphate in one litre of solution. Given : 4Fe3+ + N H  N + 4Fe2+ + 4H+ MnO – + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H O 1 Sol. Amount of hydrazine sulphate in 1 litre solution = 20  50 1  5  10–3  4  1000 10  130 = 6.5 g