# IRTPF_PT-I(P1)-CH.doc

STUDY INNOVATIONSEducator en Study Innovations

CHEMISTRY MULTIPLE CHOICE QUESTIONS (SINGLE OPTION CORRECT) 31. When these species are arranged in order of increasing bond energy, what is the correct sequence? (A) N2, O2, F2 (B) F2, O2, N2 (C) O2, F2, N2 (D) O2, N2, F2 31. B 32. A sample of peanut oil weighing 1.576 g is added to 25 ml of 0.421 M KOH. After saponification is complete, 8.46 ml of 0.273 M H2SO4 is needed to neutralize excess of KOH. The saponification number of peanut oil is (saponification number is defined as the milligrams of KOH consumed by 1 g of oil) (A) 209.6 (B) 108.9 (C) 98.9 (D) 218 .9 32. A 32. Equivalents of KOH used by oil = [25 ´ 0.421 – ]´10–3 Þ Moles of KOH used = 5.90 ´ 10–3 Þ Mass of KOH used in milligrams = 5.90 ´ 10–3 ´56 ´ 1000 = 330.40 \ Saponification number = = 209.64 33. The activity of a radioactive sample reduced from 20 Ci to 1.25 Ci in 2000 years. The half- life of the sample and its decay constant are, respectively, (A) 4000 years and 1.1 ´ 10–3 y–1 (B) 500 years and 1.386 ´ 10–3 y–1 (C) 1000 years and 1.386 ´ 10–3 y–1 (D) can’t be calculated 33. B 33. From 20 Ci to 1.25 Ci, the number of halves involved is 4 \ t1/2 = = = 500 years l = = 1.386 ´ 10–3 year–1. \ (B) 34. At which of the following four conditions, the density of nitrogen will be the largest? (A) STP (B) 273 K and 2 atm (C) 546 K and 1 atm (D) 546 K and 2 atm 34. B 34. Density of a gas is given r = . Obviously, the choice that has greater would have greater density. \ (B) 35. van der Waal’s constant ‘b’ of a gas is 125.57 centilitre/mol. How near can the centres of the two molecules approach each other? (A) 8 ´ 10–8 cm (B) 4.65 ´ 10–7 cm (C) 8 ´ 10–8 m (D) 1.6 ´ 10–7 m 35. B 36. Cl2 gas has been introduced in an 8 litre cubical container at pressure 5 atm and 300 K temperature. The following equilibrium has been achieved. Al(s) + 3/2Cl2(g) AlCl3(s) If the thickness of inner side of walls has been decreased by 0.04 mm at equilibrium, then find out Kp for the reaction ( = 1.13 g/cc) (A) (2.7)+3/2 (B) (2.7)–3/2 (C) (5.4)+3/2 (D) (5.4)–3/2 36. B 36. = 1.63 Mass of Al(s) = 0.05´10–1´(20´20´6)´1.13 = 13.56 gm So, mass Al(s) dissociated = 0.5 moles So, equilibrium established Al(s) + 3/2 Cl2(g) AlCl3(s) 0.5 So, Kp = = (2.7)–3/2 37. A sample of hard water contains 0.005 mole of calcium chloride per litre. What is the minimum concentration of sodium sulphate which must be added for removing the Ca2+ ions from this water sample? (Ksp(CaSO4) = 2.4´10–5) (A) 4.8´10–2 (B) 4.8´10–3 (C) 2.4×10–2 (D) 2.4´10–3 37. B 37. [Ca2+] = [CaCl2] = 0.005 (N) 4.8´10–3 38. Mass of borax (Na2B4O7.10H2O) required to prepare 100 ml 0.1 N solution is (A) 0.72 gm (B) 0.95 gm (C) 3.8 gm (D) 1.92 gm 38. D 39. During a K-electron capture mainly, (A) rays are emitted (B) particles are emitted (C) positrons are emitted (D) X-rays are emitted 39. D 40. The activation energy for the forward as well as backward re

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### IRTPF_PT-I(P1)-CH.doc

• 1. . 1 CHEMISTRY 31. When these species are arranged in order of increasing bond energy, what is the correct sequence? (A) N2, O2, F2 (B) F2, O2, N2 (C) O2, F2, N2 (D) O2, N2, F2 31. B 32. A sample of peanut oil weighing 1.576 g is added to 25 ml of 0.421 M KOH. After saponification is complete, 8.46 ml of 0.273 M H2SO4 is needed to neutralize excess of KOH. The saponification number of peanut oil is (saponification number is defined as the milligrams of KOH consumed by 1 g of oil) (A) 209.6 (B) 108.9 (C) 98.9 (D) 218 .9 32. A 32. Equivalents of KOH used by oil = [25  0.421 – 8.46 0.273 2   ]10–3  Moles of KOH used = 5.90  10–3  Mass of KOH used in milligrams = 5.90  10–3 56  1000 = 330.40  Saponification number = 330.40 1.576 = 209.64 33. The activity of a radioactive sample reduced from 20 Ci to 1.25 Ci in 2000 years. The half- life of the sample and its decay constant are, respectively, (A) 4000 years and 1.1  10–3 y–1 (B) 500 years and 1.386  10–3 y–1 (C) 1000 years and 1.386  10–3 y–1 (D) can’t be calculated 33. B 33. From 20 Ci to 1.25 Ci, the number of halves involved is 4  t1/2 = T n = 2000 4 = 500 years  = 0.693 500 = 1.386  10–3 year–1.  (B) 34. At which of the following four conditions, the density of nitrogen will be the largest? (A) STP (B) 273 K and 2 atm (C) 546 K and 1 atm (D) 546 K and 2 atm 34. B 34. Density of a gas is given  = PM RT . Obviously, the choice that has greater P T would have greater density.  (B) 35. van der Waal’s constant ‘b’ of a gas is 125.57 centilitre/mol. How near can the centres of the two molecules approach each other? (A) 8  10–8 cm (B) 4.65  10–7 cm (C) 8  10–8 m (D) 1.6  10–7 m 35. B 36. Cl2 gas has been introduced in an 8 litre cubical container at pressure 5 atm and 300 K temperature. The following equilibrium has been achieved. MULTIPLE CHOICE QUESTIONS (SINGLE OPTION CORRECT)
• 2. 2 Al(s) + 3/2Cl2(g) AlCl3(s) If the thickness of inner side of walls has been decreased by 0.04 mm at equilibrium, then find out Kp for the reaction ( Al  = 1.13 g/cc) (A) (2.7)+3/2 (B) (2.7)–3/2 (C) (5.4)+3/2 (D) (5.4)–3/2 36. B 36. 2 Cl 5 8 n 0.082 300    = 1.63 Mass of Al(s) = 0.0510–1(20206)1.13 = 13.56 gm So, mass Al(s) dissociated = 0.5 moles So, equilibrium established Al(s) + 3/2 Cl2(g) AlCl3(s) 3 1.63 0.5 2   0.5 So, Kp =   2 3/2 3/2 Cl 0.88 0.082 300 p 8            = (2.7)–3/2 37. A sample of hard water contains 0.005 mole of calcium chloride per litre. What is the minimum concentration of sodium sulphate which must be added for removing the Ca2+ ions from this water sample? (Ksp(CaSO4) = 2.410–5) (A) 4.810–2 (B) 4.810–3 (C) 2.4×10–2 (D) 2.410–3 37. B 37. [Ca2+] = [CaCl2] = 0.005 (N)   4 2 2 sp 4 CaSO K Ca SO            5 2 4 2.40 10 0.005 SO            5 2 4 3 2.4 10 SO M 5 10            4.810–3 38. 2 4 7 2 3 3 B O H O H BO OH       Mass of borax (Na2B4O7.10H2O) required to prepare 100 ml 0.1 N solution is (A) 0.72 gm (B) 0.95 gm (C) 3.8 gm (D) 1.92 gm 38. D 39. During a K-electron capture mainly, (A)  rays are emitted (B)  particles are emitted (C) positrons are emitted (D) X-rays are emitted 39. D 40. 2 2 3 N 3H 2NH heat.   The activation energy for the forward as well as backward reaction is decreased by 100 J, then the equilibrium amount of NH3 will (A) increase (B) decrease (C) remain constant (D) cannot be predicted 40. C 41. The magnetic moment of an iron compound is 5.918 BM, then the oxidation state of iron in this compound will be
• 3. . 3 (A) 0 (B) 1 (C) 2 (D) 3 41. D 42. The value of rate constant for a reaction is 10–5 at 25oC and 10–5% molecules can cross energy barrier at this temperature, then the maximum value of rate constant possible for this reaction will be (A) 10–5 (B) 10–2 (C) 102 (D) 1 42. C 43. The coefficient of S in the balanced reaction is 3 2 5 2 2 7 3 4 As S H Cr O H AsO S Cr       (A) 5 (B) 10 (C) 15 (D) 1 43. C 44. 2 3 4 Al(SCN) SO HCN     The number of electrons lost or gained in the reaction are (A) 6 (B) 10 (C) 18 (D) 36 44. C 45. A hypothetical reaction, 2 2X Y 2XY    follows the mechanism 2 X X X (fast)  2 2 X Y 2XY (slow)  then the order of reaction is (A) 1 (B) 2 (C) 3 (D) undefined 45. C 46. A mixture of two gases A and B is placed in a container at pressure 1 atmosphere. Due to a pin hole in the container, gaseous mixture effuses out. At first instant % of A in effused mixture is 10%, then what was the % of B in container present initially? (A) 81.8 (B) 22.22 (C) 18.18 (D) 50.11 46. A 47. Which of the following statement is/are not true for bond   ? 1. It is formed by the overlapping of p  p or p  d orbitals. 2. It is stronger than sigma bond. 3. It is formed when bond already exists. 4. It is resulted from sideways overlapping of orbitals. (A) 1, 2, 3, 4 (B) 2, 3 and 4 (C) 2 and 4 (D) 1, 2 and 4 47. C 48. Which of the following statement is not correct? (A) Hybridization is the mixing of atomic orbitals prior to their combining into molecular orbitals. (B) sp2 hybrid orbitals are formed from two p-atomic orbitals and one s-atomic orbital. (C) dsp3 hybrid orbitals are all at 90o to one another.
• 4. 4 (D) d2sp3 hybrid orbitals are directed towards the corner of regular octahedral. 48. C 49. A freshly cut piece of wood gives 16100 counts of ray   emission per minute per kg and old wooden bowl gives 13200 counts per minute per kg. The age of wooden bowl is (half-life of 14C is 5568 years) [Given: log 1.219 = 0.0857] (A) 576 years (B) 1591 years (C) 100 years (D) 300 years 49. B 49. 16100 0.693 2.303 log t 13200 5568   2.303 5568 161 t log 0.693 132   5568 log 1.219 0.3010  = 1590.85 years 1591 years  50. The number of - and -particles   emitted in the nuclear reaction, 228 212 90 83 Th Bi ,   are (A) 7 , 3   (B) 3 , 4   (C) 4 , 3   (D) 4 ,   50. D 51. 0.1 M acetic acid solution is titrated against 0.1 M NaOH solution. What would be the difference in pH between ¼ and ¾ stages of neutralization of acid? (A) 2 log 3/4 (B) 2 log 1/4 (C) log 1/3 (D) 2 log 3 51. D 52. The pH of a saturated solution of M(OH)2 is 10. The Ksp value at this temperature is (A) 5  10–13 (B) 5  10–10 (C) 5  10–8 (D) None of the above 52. A 52. 2 2 M(OH) M 2OH      pH = 14 – pOH 10 = 14 – pOH pOH = 14 – 10 = 4 log OH 4;   [OH–] = 10–4 M 2 sp K M OH          4 4 2 10 [10 ] 2           = 5 × 10–13 53. The equilibrium constants for the following reactions at 1400 K are given 2H2O(g) 2H2(g) + O2(g); K1 = 2.1  10–13, 2CO2(g) 2CO(g) + O2(g); K2 = 1.4  10–12. Then the equilibrium constant K for the reaction at temperature 1400 K is H2(g) + CO2(g) CO(g) + H2O(g) (A) 2.04 (B) 20.5 (C) 2.6 (D) 8.4 53. C
• 5. . 5 54. Liquid 3 NH ionizes to a slight extent. At 60°C, its ionic product, 3 36 NH 4 2 K [NH ][NH ] 10 .      The number of 2 NH ions present per ml of pure liquid 3 NH are (assuming that NA = 6  1023) (A) 300 ions (B) 400 ions (C) 600 ions (D) 500 ions 54. C 55. One litre of gaseous mixture is effused in 5 min 11 sec, while 1 litre of oxygen takes 10 mins for effusion through the same pin hole. If the gaseous mixture contains methane and hydrogen, calculate the density of the gaseous mixture at 1 atm and 298 K. (A) 5.269 gm/litre (B) 4.298 gm/litre (C) 9.567 gm/litre (D) 0.351 gm/litre 55. D 55. 2 2 O mix O mix. 1litre M r (5 60 11)sec 1litre r M 60 10sec      mixture 32 600 M 311   mix 32 311 311 M 8.5974222 600 600      PM 1 8.5974222 0.3518343 RT 0.082 298       56. A sample of water from newly dug well read pH = 7 in the field study on the night of 31st Dec. 2005 in Shimla. The sample was (A) basic (B) acidic (C) neutral (D) cannot be predicted 56. B 56. The value of K decreases with decreases in temperature. 14 w K 10  (at temperature > 25oC) H+ < 10–7 For neutral water, pH = –log H+ pH > 7 So, pH 7 indicated must be acidic. 57. 20 gm of CaCO3 was taken into a vessel of 10 litre volume and heated at 800oC. What per cent of CaCO3 will decompose if Kp for the reaction, 3 2 CaCO (s) CaO(s) CO (g),  is 1.16 atm? (A) 20% (B) 40% (C) 60% (D) 66% 57. D 58. CO2 and He are kept in a container at a partial pressure of 2 CO He P and P at temperature T. A small orifice is made in the wall of container, it is observed that both the gases effuse at the same rate. The ratio 2 CO He P :P is (A) 3.33 : 1 (B) 1 : 3.33 (C) 2 : 1 (D) 1 : 2 58. A
• 6. 6 58. 2 2 2 CO CO He He He CO r P M r P M   2 CO He P 1 4 1 P 44    2 CO He P 3.33 P 1  59. The correct order of increasing N – O bond length in 2 2 3 NO , NO and NO    is (A) 3 2 2 NO NO NO      (B) 2 2 3 NO NO NO      (C) 2 3 2 NO NO NO      (D) 2 2 3 NO NO NO      59. D 60. The circumference of the 4th orbit of hydrogen is 5.32 nm. The wavelength of electron revolving in the first Bohr orbit will be (A) 0.133 nm (B) 1.33 nm (C) 4.32 nm (D) None of the above 60. D 60. 2 r n    Circumference 4 2 r 5.32 4      4 5.32 1.33 4     Now, 2 1 1 4 1 2 r , also r r 4      Hence, (D).
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