INSTITUTE FOR COMETITIVE EXAMINATIONS DELHI CENTRES: E-6, Defence Colony New Delhi - 11004. h.: 4614, 46144. B-1/6, Main Nazafgarh Road, Janakpuri, New Delhi - 110058. h.: 011-557111/1/1/14. NOIDA CENTRE: Amity Campus, Sector-44, Noida - 010. h.: 9510-4189, 4184 SYNCHRO XI STUDYING UNIT TEST-05 (CHEMISTRY) Time : 1 hr. 15 min. M.M. : 60 Date : 1/1/011 C O TOIC COVERED : (Chemical and Ionic Equilibrium) AER-1 SET - 1 CODE-A Section-A ANSWERS 1... 4. 5. 6. 7. 8. 9. 10. 11. Section-B and C Section-D 1. 1. 14. 15. 16. 17. 1. A-(s); B-(p); C-(p); D-(q) INSTITUTE FOR COMETITIVE EXAMINATIONS DELHI CENTRES: E-6, Defence Colony New Delhi - 11004. h.: 4614, 46144. B-1/6, Main Nazafgarh Road, Janakpuri, New Delhi - 110058. h.: 011-557111/1/1/14. NOIDA CENTRE: Amity Campus, Sector-44, Noida - 010. h.: 9510-4189, 4184 SYNCHRO XI STUDYING UNIT TEST-05 (CHEMISTRY) Time : 1 hr. 15 min. M.M. : 60 Date : 1/1/011 C O TOIC COVERED : (Chemical and Ionic Equilibrium) AER-1 SET - 1 CODE-B Section-A ANSWERS 1... 4. 5. 6. 7. 8. 9. 10. 11. Section-B and C 1. 1. 14. 15. 16. 17. Section-D 1. A-(s); B-(p); C-(p); D-(q) [5] Unit Test-05 (Syncyro XI Studying) (1-1-11)
UNIT TEST-05 (CHEMISTRY) SOLUTIONS Section-A Choose the correct answer : 1. To a 50 ml of 0.1 M HCl solution, 10 ml of 0.1 M NaOH is added and the resulting solution is diluted to 100 ml. What is change in ph of the HCl solution? 4.98 0.98 0.1M none of these. Sol.: Before adding HCl solution ph = 1 [ [HCl] = [H O] + = 10 1 M] n HCl (initially) = MV = 0.1 M 0.05 L = 5 10 mol n NaOH added = MV = 0.1 M 0.01 L = 1 10 mol HCl + NaOH NaCl + H O t = 0 5 10 mol 1 10 mol 0 4 10 mol 0 V final = 100 ml = 0.1 L n 410 mol [HCl] = = 4 10 V 0.1L M ph = log [H O + ] = log4 = log = 0.01= 0.60 = 1.98 Increase in ph = (1.98 1) = 0.98. What amount of solid sodium acetate be added into 1 litre of the 0.1 M CH COOH solution so that the resulting solution has ph almost equal to pk a (CH COOH) = 4.7 1gm 5 gm 16 gm 16.4 gm. Sol.: Since the resulting solution be acidic buffer, one may use Henderson equation. [CHCOO ] ph = pk a log [CH COOH] Let n mol of CH COONa be added to do so n mol vl or, ph = 4.70 + log 0.1 mol vl n or, 5 = 4.70 + log 0.1 n or, log 0.1 = 0. n or, = antilog 0.0 = 0. 0.1 n = 0. mol Amount of sodium acetate = 0. 8 gm = 16.4 gm. A weak base (BOH) with K b = 10 5 is titrated with a strong acid, HCl. At /4 th of the equivalent point, ph of the solution is: 5 + log 5 log 14 5 + log 8.5. Sol.: Let the initial equivalent of BOH be x BOH + HCl? BCl + H O Initial equivalent x /4x 0 0 At /4 th eqv. pt. x x x 0 4 4 x 4 x 4 [6] Unit Test-05 (Syncyro XI Studying) (1-1-11)
poh = [salt] x 4 pkb log 5 log [Base] 4 x ph = 14 5 log = 8.5 4. A reaction mixture containing H, N and NH has partial pressures atm, 1 atm and atm respectively at 75 K. If the value of K p for the reaction, N (g) + H (g) NH (g) is 4.8 10 5 atm at 75 K, in which direction the net reaction will go? Forward Backward No net reaction Direction of reaction cannot be predicted. Sol.: Q p = NH () 9 atm (1)() 8 N H = 1.15 atm. Since value of Q p is larger than K p (4.8 10 5 atm ), it indicates net reaction will proceed in backward direction. 5..0 mol of Cl 5 were introduced in a vessel of 5.0 L capacity at a particular temperature. At equilibrium, Cl 5 was found to be 5% dissociated into Cl and Cl. The value of K c for the reaction is 1.89 0.77 0.75 0.075. Sol.: Moles of Cl 5 dissociated = 5 = 0.7 100 Moles of Cl 5 left undissociated = 0.7 = 1. mol [Cl 5 ] = 1. 5 M, [Cl ] = 0.7 5 M, [Cl ] = 0.7 5 M 0.7 0.7 [Cl ][Cl ] 5 5 K [Cl 1. 5] 5 = 0.075 6. At temperature T, a compound AB (g) dissociates according to the reaction AB (g) AB(g) + B (g) with degree of dissociation, which is small compared with unity. The expression for K p, in terms of and the total pressure, T is Sol.: T T T For the given equilibrium, the equilibrium concentration are T. AB (g) AB(g) + B (g) Equilib. conc. c (1 ) c c K ( )( ) B AB ( AB ) c (c ) T [c(1 )] [c(1 )] ; K T (1 ) 1 [7] Unit Test-05 (Syncyro XI Studying) (1-1-11)
Since, is small compared to unity, so 1 1 and 1 + 1. K T 7. The equilibrium constant for equilibria, SO (g) + 1 O (g) SO (g) and SO (g) SO (g) + O (g) Sol.: are K 1 and K respectively. Then K = K 1 K = K 1 K = 1/K 1 K = 1/K 1 Factual 8. If different quantities of ethanol and acetic acid are used in the following reversible reaction, CH COOH + C H 5 OH same in all cases CH COOC H 5 + H O the equilibrium constant will have values which will be: Different in all cases Higher in cases where higher concentration of ethanol is used Sol.: Higher in cases where concentration of acetic acid is used. Equilibrium constant is constant at constant temperature, do not depends on initial amount of reactant 9. If 1.0 mole of I is introduced into 1.0 litre flask at 1000 K, at equilibrium (K c = 10 6 ), which one is correct: Sol.: [I (g)] > [I (g)] [I (g)] < [I (g)] [I (g)] = [I (g)] [I (g)] = 1 [I (g)] Since K C < 10, therefore reactant predominats over product. 10. Oxidation of SO by O to SO is an exothermic reaction. The yield of SO will be maximum if: Sol.: Temperature is increased and pressure is kept constant Temperature is reduced and pressure is increased Both temperature and pressure are increased Both temperature and pressure are reduced Factual 11. The pk a of acetylsalicylic acid (aspirin) is.5. The ph of gastric juice in human stomach is about - and the ph in the small intestine is about 8. Aspirin will be Sol.: unionised in the small intestine and in the stomach completely ionised in the small intestine and the stomach ionised in the stomach and almost unionised in the small intestine. ionised in the small intestine and almost unionised in the stomach Factual Read the following questions and choose Section-B A. Both Statement-1 and Statement- are true and Statement- is correct explanation of Statement-1. B. Both Statement-1 and Statement- are true but Statement- is not a correct explanation of Statement-1. C. Statement-1 is true and Statement- is false. D. Statement-1 is false and Statement- is true. [8] Unit Test-05 (Syncyro XI Studying) (1-1-11)
1. Statement-1: An aqueous solution made by adding CH COOH and CH COO Na + to water in molar ratio Sol.: Statement-: n 1 : n where n >> n 1, the solution can take more acid than base without getting ph changed substantially. A good buffer is made when concentration of acid and base differs substantially in a buffer solution. A B C D To make a good buffer solution the concentration of ratio of weak acid and their conjugate base should be 1 : 1 1. Statement-1: Degree of ionization of a weak electrolyte decreases with increase in concentration Sol.: Statement-: In presence of common ion the degree of ionization of a weak electrolyte is suppressed. A B C D Common ion effect. 14. Statement-1: The equilibrium (given below) attained in a closed vessel remains unaltered by the addition of Statement-: CaCO (s). CaCO CaO + CO (s) (s) The active mass of a solid is a constant independent of its mass and is always taken to be unity. A B C D Comprehension A A 4...(i) Section-C K = atm 1 81 p 1 A + C A C...(ii) A C AC...(iii) A and C are taken in : 1 mole ratio in a closed container of a certain volume at a fixed temperature and above three equilibriums are established simultaneously. K p for the first reaction is /81 atm 1. At equilibrium partial 1 pressure of A 4 (g) and AC(g) are found to be 1/ atm each and the total pressure at equilibrium is found to be 7 4 atm. Then 15. The partical pressure of A C at equilibrium is 1/ 1/4 /4 1 Sol.: x y = 9 6 1 y = 9 y = 1 atm 16. The mole ratio of gases A and AC at equilibrium is 9/ 7/ 8 9 [9] Unit Test-05 (Syncyro XI Studying) (1-1-11)
Sol.: na A x y 9 / = 9 n z 1/ AC AC 17. Equilibrium constant K p for the reaction, AC A C is /4 4 1/ Sol.: K p = / 4 (1/ ) AC AC Match the following Section-D 1. For CH COOH (K a = 10 5 ) Column I Column II A. 0.1 M 00 ml CH COOH + 0.1 M 100 ml NaOH (p) ph = 5 log B. 0.1 M 00 ml CH COOH + 0.1 M 100 ml NaOH +0.1 M 50 ml HCl (q) ph = 5 + log C. 0.1 M 00 ml CH COOH + 0.1 M 50 ml NaOH (r) ph = 6 D. 0.1 M 00 ml CH COOH + 0.1 M 150 ml NaOH (s) ph = 5 A-(s), B-(p), C-(p), D-(q) [10] Unit Test-05 (Syncyro XI Studying) (1-1-11)