Classes at: - Topic: Redox & Volumetric Titration

Transcription

1 PHYSICAL CHEMISTRY by: SHAILENDRA KR. Classes at: - SCIENCE TUTRIALS; pp. Khuda Baksh Library, Ashok Rajpath, Patna PIN PINT STUDY CIRCLE; House No. 5A/65, pp. Mahual Kothi, Alpana Market, Patna Topic: Redox & Volumetric Titration Meq. Approach (1) How many mol of Fe +2 ions are formed, when exess of iron treated with 50 ml of 4.0 M HCl under inert atmosphere? (Assume no change in volume) (a) 0.4 (b) 0.1 (c) 0.2 (d) 0.8 (2) The molality of 15% (wt/volume) solution of of density 1.1 gm/cm 3 approximately (a) 1.2 (b) 1.4 (c) 1.8 (d) 1.6 (3) 100 ml of 0.3 N HCl solution were mixed with 200 ml of 0.6 N solution. The final acidic normality : (a) 0.9 N (b) 0.6 N (c) 0.5 N (d) 0.4 N (4) 0.7 gm of C. x were dsolved in water and volume was made to 100 ml, 20 ml of th solution required 19.8 ml of N/10 HCl for complete neutralization. The value of x (a) 7 (b) 3 (c) 2 (d) 5 (5) The eq.wt of S 4 6 in the reaction + I 2 NaI + S 4 6 (a) (c) 2 3 (b) (d) 1 8 (6) How many gram of KMn 4 are contained in 4.0 liter of 0.05 N solution. The KMn 4 to be used as an oxidant in acid medium. (a) 1.58 gm (b) 15.8 gm (c) 6.32 gm (d) 31.6 gm (7) How many liter of Cl 2 at STP will be liberated by the oxidation of NaCl with10 gm KMn 4 : (a) 3.54 liter (b) 7.08 liter (c) 1.77 liter (d) None of these (8) What volume of 0.40 M would be required to react with the I 2 liberated by adding exess of KI to 50 ml of 0.20 M Cu solution : (a) 12.5 ml (b) 25 ml (c) 50 ml (d) 2.5 ml (9) In the reaction, V + F 2 Fe + V 2 5. The eq.wt of V 2 5 equal to its : (a) Mol.wt (c) 6 (b) 8 (d) None (10) 0.2 gm of a sample of required 10 ml of 1 N KMn 4 in a titration in the presence of. Purity of (a) 25 % (b) 85 % (c) 65 % (d) 95 % (11) The amount of dihydrated oxallic acid required to prepare 500 ml of its 0.01 N Page No.: 1

2 solution (a) gm (b) 6.3 gm (c) 63 gm (d) 3.15 gm (12) 10 ml of a solution of labelled '10 volume' just decolore 100 ml of potassium permagnate solution acidified with dilute. Calculate the amount of potassium permagnate in the given solution : (a) gm (b) gm (c) 5.63 gm (d) gm (13) Equal volume of 1 M each of KMn 4 and K 2 are used to oxides Fe(II) solution in acidic medium. The amount of Fe oxided will be (a) More with KMn 4 (b) Equal with both oxiding agent (c) More with K 2 (d) Can't be determined (14) In an experiment, 50 ml of 0.1 M solution of a salt reacted with 25 ml of 0.1 M solution of sodium sulphite. The half equation for the oxidation of sulphite ion -2-2 S (aq) + + 2H + + 2e - If the oxidation number of metal in the salt was 3, what would be the new oxidation number of the metal? (a) 0 (b) 1 (c) 2 (d) 4 (15) 100 ml of 0.1 N KMn 4 reacts with 0.45 gm of oxalic. The mot.wt of oxalic acid (a) 45 (b) 90 (c) 180 (d) 22.5 (16) Polythene can be produced from calcium carbide according to the following sequence of reaction :- CaC 2 + Ca + CH CH n(ch CH) + n (-C -C- )n The mass of polythene which can be produced from 20 kg of CaC 2 (c) 8.75 kg (d) 9.75 kg (17) A solution contains C and NaHC. 10 ml of th solution required 2.5 ml of 0.1M for neutralation using phenolphthalein as indicator. Methyl arange then added when a further 2.5 ml of 0.2M was required. The amount of C and NaHC in 1.0 liter of the solution ( of NaHC =84, of C =106) (a) 5.3 gm and 4.2 gm (b) 3.3 gm and 6.2 gm (c) 4.2 gm and 5.3 gm (d) 6.2 gm and 3.3 gm (18) The number of mol of oxalate ions oxidized - by one mol of Mn 4 ions (a) 1/5 (b) 2/5 (c) 5/2 (d) 5 (19) What volume of 0.1 M KMn 4 needed to oxided 100 mg of FeC 2 4 in acid solution : ( Mol.wt of FeC 2 4 =144) (a) 4.1 ml (b) 8.2 ml (c) 10.2 ml (d) 4.6 ml (20) When one mol of KMn 4 reacts with HCl, the volume of Cl 2 liberated at NTP will be : (a) 11.2 liter (b) 22.4 liter (c) 44.8 liter (d) 56.0 liter (21) An element A in a compound ABD has oxidation number A -n.it oxided by 2 in acid medium. In the expriment 1.68 x 10-3 mol of K 2 were used for 3.26 x 10-3 mol of ABD. The new oxidation number of A after oxidation (a) 3 (b) 3-n (c) n-3 (d) +n (22) Number of K + ions and number of mol of K + ions present in 1 liter of N/5 KMn 4 solution respectively in acidic medium are: (a) 0.04 and 2.4 x (b) 2.4 x and 0.04 (c) 200 and x (a) 6.75 kg (b) 7.75 kg (d) x and 200 Page No.: 2

3 (23) The eq.wt of Fe 3 4 in, Fe KMn 4 Fe 2 + Mn (a) M/6 (b) M (c) 2 M (d) M/3 (24) Eq.wt of NH 3 in, NH 3 + N + (a) 3.4 (b) 17 (c) 8.5 (d) None (25) How many gram of KMn 4 are contained in 4 liter of 0.05 N solution. The KMn 4 to be used as an oxidant in acid medium. (a) 1.58 gm (b) 15.8 gm (c) 31.6 gm (d) 6.32 gm (26) - When Br ion reacts with Br - ion in acid solution Br 2 liberated. The eq.wt of KBr in th solution ( M molar mass) (a) M/8 (b) M/3 (c) M/5 (d) M/6 (27) 8 gm of sulphur are burnt to form S which oxided by Cl 2 water. The solution treated with BaCl 2 solution. The amount of Ba precipitated (a) 1.0 mol (b) 0.5 mol (c) 0.24 mol (d) 0.25 mol (28) The number of equivalent per mol of S used in its oxidation to S (a) 3 (b) 6 (c) 4 (d) 2 (29) 100 ml of 1 M CH 3 CH mixed with100 ml 1 M-NaH. The resulting solution (a) Acidic (b) Slightly Basic (c) Neutral (d) Basic (30) When exess of Zn reacts with 10 ml of, 224 ml liberated at S.T.P. The strength of (a) 10 N (b) 1 N (c) 2 N (d) 4 N (31) 1 gm marble when reacted with 1 N HCl acid required 19 ml for complete reaction leaving behind some sandy materials. The percentage purity of CaC in the marble : (a) 0.95 (b) 95 (c) 5 (d) 19 (32) 5.3 gm of pure and anhydrous sodium carbonate treated with 100 ml of N/10 HCl. After the reaction over (a) Exess of HCl left (b) Unreacted sodium carbonate left (c) Neither of HCl nor of sodium carbonate left (d) None of these (33) Which of the following should be done in order to prepare 0.40 M NaCl starting with 100 ml of 0.30 M NaCl (Mol.Wt. of NaCl =58.5) (a) Add 5.85 gm NaCl (b) Add 20.0 ml water (c) Add 0.01 mol NaCl (d) Evaborate 10 ml water (34) The volume of N/5 HCl solution required to neutrale 2 gm of CaC (a) 100 ml (b) 200 ml (c) 50 ml (d) 90 ml (35) ne molar K 2 solution in acid medium equivalent to (a) 1 N K 2 (b) 2 N K 2 (c) 6 N K 2 (d) 4 N K 2 (36) In the following reaction : 2Cu + 4 KI 2 CuI + I K 2 the amount of liberated I 2 completely reacted with 100 ml of M/10 solution of. The no of milliequivalent of Cu present in solution (a) 10 (b) 5 (c) 15 (d) 20 (37) 100 ml of a dibasic acid solution contains 3.15 gm of the acid. 10ml of th solution neutrales 10 ml of 0.5 N NaH solution. The molecular weight of acid (a) 63 (b) 126 (c) 31.5 (d) 40 Page No.: 3

4 (38) Which two of the following solution contain opproximately equal Hydrogen ion concentration. (i) 50 ml of 0.1 M HCl + 25 ml (ii) 50 ml of 0.10 M HCl + 50 ml (iii) 50 ml of 0.10 M + 25 ml (iv) 25 ml of 0.1 M + 50 ml (a) (i) and (ii) (b) (i) and (iii) (c) (i) and (iv) (d) (ii) and (iii) (39) If 40.0 ml of 1.6 M HCl and 60 ml of 2 M NaH are mixed. Concentration of Na +, Cl and H in the resulting solution in the term of molarity... (a) 1.2 M, M, M respectively (b) M, M, 1.2 M respectively (c) M, M, 1.2 M respectively (d) M, M, 1.2 M respectively (40) No of molar of gas present in 1m 3 gas at NTP (a) 4.46 (b) 44.6 (c) 22.4 (d) 2.24 (41) Commercial "11.2 volume" volume solution has a molarity of (a) 1 (b) 0.5 (c) 11.2 (d) 1.12 (42) 10 ml of solution (volume strength=x) required 10 ml of N/0.56 Mn 4 solution in acidic medium. Hence x (a) 0.56 (b) 5.6 (c) 0.1 (d) 10 (43) 2 (l) 2 (l) ml of x molar gives 3.0 L of gas under the condition when 1 mol occupies 24.0 L ; value of x (a) 2.5 (b) 1.0 (c) 0.5 (d) 0.25 (44) How many ions are present in 2.0 L of a solution that 0.8 M K 4 Fe(CN) 6? (a) (b) (c) (d) (45) 20 ml of x M HCl neutralizes completely 10 ml of 0.1 M NaHC solution and a further 5 ml of 0.2 M NaC solution to methylorange end point. The value of x (a) M (b) M (c) 0.15 M (d) 0.2 M (46) Eq.wt of Mn 4 in acidic, basic and neutral medium are in the ratio of (a) 3 : 5 : 15 (b) 5 : 3 : 1 (c) 5 : 1 : 3 (d) 5 : 15 : 5 (47) 4I - + Hg HgI 4, 1 mol each of Hg +2 and I - -2 will form HgI 4 (a) 1 mol (b) 0.5 mol (c) 0.25 mol (d) 2 mol (48) Molarity of 18 M. Its density 1.8 gm/ml. Hence molality (a) 36 (b) 200 (c) 500 (d) 18 (49) 10 L of hard water require 0.56 gm of lime (Ca) for removing hardness. Hence temporary hardness in PPM (Parts Per Million(10 6 )) of CaC (a) 100 (b) 200 (c) 10 (d) 20 (50) Which of the following will contain same no of atom as in 20 gm of calcium. (a) 24 gm Magnesium (b) 12 gm Carbon (c) 8 gm xygen gas (d) 16 gm xygen gas Page No.: 4

5 A N S W E R S 01. b 02. d 03. c 04. c 05. a 06. c 07. a 08. b 09. c 10. b 11. d 12. b 13. c 14. b 15. b 16. c 17. a 18. c 19. a 20. d 21. b 22. b 23. b 24. a 25. d 26. c 27. d 28. b 29. b 30. c 31. b 32. b 33. c 34. b 35. c 36. b 37. b 38. c 39. a 40. b 41. a 42. d 43. a 44. b 45. c 46. d 47. c 48. c 49. b 50. d Page No.: 5