DPP-57 to 58-Physical Chemistry

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 57 to 58 Class : XIII Course : DPP No.57 DPP No.1 Total Marks : 30 Max. Time : 30 min. Single choice Objective (no negative marking) Q.1 to Q.3 (3 Marks, 3 Min.) [9, 9] Comprehension ('–1' negative marking) Q.4 to Q.6 (3 marks 3 min.) [9, 9] Subjective Questions (no negative marking) Q.7 to Q.9 (4 marks 4 min.) [12, 12] 1. A current is passed through 2 voltameters connected in series. The first voltameter contains XSO4 (aq.) and second has Y2SO4 the relative atomic masses of X and Y are in the ratio of 2 : 1. The ratio of the mass of X liberated to the mass of Y liberated is (1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) None of the above 2. For the electrolytic production of NaClO4 from NaClO3 as per the equation, NaClO3 + H O  NaClO + H2, how many faradays of electricity will be required to produce 0.5 mole of NaClO4, assuming 60% efficiency ? (1) 0.835 F (2) 1.67 F (3) 3.34 F (4) 1.6 F 3. To observe the effect of concentration on the conductivity, electrolytes of different nature are taken in two vessel ‘A’ and ‘B’. ‘A’ contains weak electrolyte e.g., NH4OH and ‘B’ contains strong electrolyte e.g., NaCl. In both container concentration of respective electrolyte is increased and conductivity observed : (1) in ‘A’ conductivity increases, in ‘B’ conductivity decreases (2) in ‘A’ conductivity decreases while in ‘B’ conductivity increases (3) in both ‘A’ and ‘B’ conductivity increases (4) in both ‘A’ and ‘B’ conductivity decreases Comprehension : (4 to 6) The thermal decomposition of N2O5 occurs as : 2N2O5  4NO2 + O2 Experimental studies suggest that rate of decomposition of N2O5, rate of formation of NO2 and rate of formation of O2 all becomes double if concentration of N2O5 is doubled. 4. If rate constants for decomposition of N2O5, formation of NO2 and formation of O2 are K1, K2 and K3 respectively, then : (1) K1 = K2 = K3 (2) 2K1 = K2 = 4K3 (3) K1 = 2K2 = K3 (4) K1 = K2 = 2K3 5. If rate of formation of O2 is 16 gm/hr, then rate of decomposition of N2O5 and rate of formation of NO2 respectively are : (1) cannot be calculated with out knowing rate constants (2) 108 gm/hr, 92 gm/hr (3) 32 gm/hr, 64 gm/hr (4) 54 gm/hr, 46 gm/hr 6. The container of 2 litre contains 4 mole of N2O5. On heating to 100°C, N2O5 undergoes complete dissociation to NO and O . Select the correct answers if rate constant for decomposition of N O is 6.3 x 10–4 sec–1. 2 2 2 5 (𝑙n2 = 0.693) 1. The mole ratio before and after dissociation (of total gaseous moles) is 4 : 2. 2. Half life of N2O5 is 1100 sec and it is independent of temperature. 3. Time required to complete 87.5% of reaction is 55 min. 4. If volume of container is doubled, the rate of decomposition becomes half of the initial rate : (1) 1, 3, 4 (2) 1, 2, 3, 4 (3) 3, 4 (4) 2, 3, 4 7. Find the thickness of the electro deposited silver if the surface area over which deposition occurred was 100 cm2 and a current of 0.2 A flowed for 1hr with the cathodic efficiency of 80%. Density of Ag = 10 g/cc (Ag = 108). 8. A 300 mL solution of NaCl was electrolysed for 6.00 min. If the pH of the final solution was 12 calculate the average current used. 9. An acidic solution of Cu2+ salt containing 0.635 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of the solution kept at 100 mL and the current at 1.93 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis. DPP No.58 DPP No.2 Total Marks : 34 Max. Time : 34 min. Single choice Objective (no negative marking) Q.1 to Q.6 (3 Marks, 3 Min.) [18, 18] Subjective Questions (no negative marking) Q.7 to Q.10 (4 marks 4 min.) [16, 16] 1. For a saturated solution of AgCl at 25°C,  = 3.4×10–6 S cm–1 and that of H O (𝑙) used is 2.02 × 10–6 S cm1  for AgCl is 138 S cm2 mol–1 then the solubility of AgCl in moles per liter will be - (1) 10–5 (2) 10–10 (3) 10–14 (4) 10–16 2. Given that (in S cm2 eq–1) at T = 298 K : eq for Ba(OH2), BaCl2 & NH4Cl are 228.8, 120.3 & 129.8 respectively. Specific conductance for 0.2 N NH OH solution is 4.766 × 10–4 S cm–1, then value of pH of the given solution of NH4OH will be nearly (1) 9.2 (2) 11.3 (3) 12.1 (4) 7.9 3. In which case van’t Hoff factor is maximum ? (1) KCl, 50% ionised (2) K2SO4, 40% ionised (3) SnCl4, 20% ionised (4) FeCl3, 30% ionised 4. Solution having osmotic pressure nearer to that of an equimolar solution of K4[Fe(CN)6] is: (1) Na2SO4 (2) BaCl2 (3) Al2(SO)3 (4)C12 H22O11 5. Osmotic pressure [atm] of a 0.1 M solution of K4[Fe(CN)6], which undergoes 50% dissociation, will be atm at 270C. (1) 7.38 (2) 3.69 (3) 405.9 (4) none of these 6. Two aqueous solutions, one of the NaCl in water (1) and the other of C8H15O2Na in water (2) are isotonic. If wA and wB are weight fractions of NaCl and C8H15O2Na in solution A and B respectively, then (assuming that both the salts dissociate completely) : (1) wA > wB (2) wA = wB (3) wA < wB (4) none of these 7. A conductivity cell has a resistance of 250  when filled with 0.02 M KCl at 250C and one of 105  when filled with 6 × 10–5 M NH4OH solution, the specific conductivity of 0.02 M KCl is 0.00277 S cm–1 and the NH + and OH– are 73.4 and 198 respectively. Calculate the cell constant and the degree of dissociation of NH4OH in the 6 × 10–5 M solution. 8. Calculate the osmotic pressure of a decimolar solution of cane sugar at 27°C. 9. An aqueous solution contains 18 g of glucose (mol wt. = 180) per 0.5 L. Assuming the solution to be ideal, calculate osmotic pressure at 27°C. 10. A solution of 1.73 g of ‘A’ in 100 cc of water is found to be isotonic with a 3.42% (weight/volume) solution sucrose (C12H22O11) Calculate molecular weight of A (C12H22O11 = 342) ANSWER KEY DPP No.-55 1. 2 2. 1 3. 1 4. 3 5. 4 6. 2 7. (a) 1 ; (b) 2 8. 4 9. 2 10. 1,2,4 11. – 0.0785 V. 12. E° = 0.782 V ANSWER KEY DPP No.-56 1. 4 2. 1 3. 3 4. 3 5. 3 6. 1 7. 1 8. 3 9. 3 10. 2 11. 2

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