NATIONAL STANDARD EXAMINATION IN CHEMISTRY ANSWER KEYS SET No. C C 2. C 3. D 4. A & C 5. A 6. D 7. D 8. C 9. C 10. A 11. C 12. B 13.

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1 NATINAL STANDARD EXAMINATIN IN CEMISTRY ANSWER KEYS SET No. C C 2. C 3. D 4. A & C 5. A 6. D 7. D 8. C 9. C 10. A 11. C 12. B 13. A 14. B 15. B 16. C 17. D 18. B 19. C 20. A 21. D 22. A 23. C 24. D 25. B 26. B 27. C 28. A 29. B 30. D 31. C 32. A 33. D 34. D 35. B 36. A 37. C 38. C 39. B 40. A 41. D 42. C 43. D 44. A 45. D 46. B 47. D 48. A 49. C 50. D 51. D 52. A 53. C 54. B 55. B 56. C 57. D 58. C 59. C 60. B 61. B 62. D 63. C 64. B 65. C 66. D 67. C 68. C 69. C 70. C 71. C 72. B 73. C 74. C 75. A 76. C 77. D 78. D 79. D 80. B

2 INTS & SLUTINS 1. C nly threshold energy and activation energy are changed by using catalyst. 2. C The configuration of alcohol as well as the substituted acid are fixed. ence no change in optically activity. 3. D C s 120 g 60 C g com o = kj mol 1 f o of C 2 = -393 kj mol = ( f o of C ) f o C 60 = 2390 kj mol 1 4. A & C Both ions 2 and S 2 are diamagnetic. The electron configuration of 2 is 1s 2 2s 2 2p 6 and that of S 2 is 2s 2 2s 2 2p 6 3s 2 3p 6 5. A K sp = s 2 = or, s mol L Mass of AgCl dissolved in one litre waer = = g = mg = mg ne litre can dissolve = 1.92 mg mg will be dissolved = 0.5L D The optical isomer are the non super imposable mirro images of each other. 7. D 2 Mg 2Cl MgCl Mg Na2C3 MgC3 2Na 2 Mg Na2S4 MgS4 2Na 8. C sol = yd + lattice = 0 5 = lattice lattice = E = kj mol C 2BF3 6Li 6LiF B A K K K f b

3 or K b K b 2 10 s C No. of geometrical isomers = 2 n = 2 4 = B V V0bP 1 bp 1 1 bp 1 bp or V V bp V bp V bp V0b P V0 y = mx + C positive slope and intercept 13. A = kj whenn 2 5 is a gas = = -220 kj mol B Equal fraction of reactants consumes in equal interval of time. ence it is a first order reaction. 15. B Salt p pk a log Acid Salt 2 log Acid 1 Acid 1 i.e. Salt i.e % C W = -P ext (V 2 V 1 ) W = -1(10 20) L atm W=-8L atm W= W=810.4 J(approx) 17. D It forms the most stable carbonium ion(benzyl) among the given compounds. 18. B C 9 6C r = kj mol Total energy needed = 7800 kj

4 19. C Energy released by 1 mole = Mass of glucose = = P P V P V V V kp 20. A Graphite is used as anode in the electrometallurgy of Al. 21. D Product of [ 3 + ][ ] is constant. 22. A 4 Sn 2e Sn 2 E V... G 0 2F Br 2e 2Br E 1.07 V... G 2F G G G G 2F F 1.07 G 177.6kJ 23. C 2KCl3 2KCl L 2 given by 2 moles L of 2 gives = mole a 24. D 3 C 2 5 C25Cl Na 25. B NA 0 atom B E hc longer Energy lesser required for rubidium 27. C

5 C 3 i 3 ii Zn/2 C 3 C 3 C 3 Cl 2 / Na 3 (adipic acid) G = TS At equilibrium G = 0 So, = TS T S = 97 T = K 97 S = PM d RT 2 44 d 4 gm / L (not in the option) or 4 Kg/m D Nucleophilic substitution reaction. 31. C Applying Nernst Equation o 2.303RT E E log 10 K 1F 1 o 2.303RT E E log K 1F Since they have asked magnitude of potential difference 25 K RT E E log F K = 82 mv 32. A e 1.23V Apply Nernst Equation

6 E 1.23 log 10 4 E = log E = V 33. D Explained on basis of rtho Effect. 34. D Use 2 n r 53 pm 1 pm = m 35. B Phase change occurs at constant temperature. 36. A Use electrophilic aromatic substitution. 37. C As hybridization is sp 3 d, so we have two different type of bonds (axial and equatorial). 38. C Change in oxidation number occurs. 39. B 2BrF 3 BrF 2 BrF4 2Cl Cl Cl A (a) Br (a) (b) (a) (b) Cl (a) (b) Chiral Quaternary 41. D Use EAS, also Na reacts with hydrogen attached to electronegative atom to release 2 (g). 42. C Each Ca 2+ will replace two K +. Ca has mass no. as 40 whereas K has D Follow either Ion electron/oxidation no. method to balance. 44. A Use aworth Projection rules. 45. D Use IUPAC Nomenclature rules. 46. B

7 Stronger the acid weaker is it s conjugate base. 47. D BrCl + 2 is bent shape due to sp 3 hybridisation. 48. A As after losing water, compound becomes Quasi aromatic species. 49. C Use rules for conversion of flying Wedge representation to Newman. 50. D Silicones have in general the chemical formula as [R 2 Si] n where R is alkyl or aryl group. 51. D yba2cu3 6.5 Use summation of oxidation no. = AB Find molecular formula by percentage composition given. The formula is C and option (A) & (B) both give positive Fehlings Test. 53. C Use IUPAC rules 54. B Explained on basis of Intermolecular ydrogen Bonding. 55. B Tl prefers +1 oxidation state. 56. C Cumulated alkene with even number of double bond is non planar. 57. D More resonance stabilized and more hyper conjugation. 58. C Due to TIN pest (Autocatalytic, allotropic transformation) 59. C Acidic hydrogen in conjugation with electron withdrawing group. 60. B As Boron is an electron deficit when compare with carbon. 61. B P is

8 3 C C 3 3 C C 3 excess C3 P.C.C C 3 MgBr 3 C C D N N is the most stable Lewis structure because octet of all the elements is complete and the +ve charge is present on less e.n and ve charge on more e.n atom. 63. C Cl / AlCl 3 N3 2S B Since both are isoelectronic species, but F has higher nuclear charge hence it is smaller than 2. N C Br Br NaN 2 2 /Pd/C/Quinoline 66. D Radial nodes = z = n l 1 Angular nodes = z = l = d-orbital ence n = 5 Therefore 5d orbital 67. C

9 C Cl C 2 dil.na Acidic K2Cr27 C C C C C C C I, II and III are correct statements. 69. C i & iii involve carbocation intermediate. 70. C due to positive charge on Mn, back bonding is not so effective(d - p) 71. C R 1 = K[X 0 ][Y 0 ] 72. B ence Y = 2[Y 0 ] 0.5 R 1 = k X Y 2 3 C C N 2 has highest dipole moment because of stable resonating structure 3 C C N 2 C C N 73. C milli eq of Cl = = 37.5 mili eq W Na 2 C W = 2 gram Na2C C [Fe(N 3 ) 6 ] 3+ has 1 unpaired and [FeF 6 ] 3 has 5 unpaired electron 75. A

10 Br - Mg Cl 1 mol C3 C C2 C C25 Cl C C2 CC C Zn 2+ conc n will be less than 1 M hence Zn 2+ will more spontaneously oxidized at anode E 1.1 log D M M D p = 2 [ + ] = 10-2 M 1 L = 10 3 g of water has 10-2 mole of + 1 ppm = 10 6 g of water has 10 mole of + = 5 mole of Ca 2+ ence weight of Ca = 5 40 = 200 ppm 79. D For compound Y 80. B.303 g of combine with 1 gm of Br g of Br = 48 g = 3 atm ence Y is Br 2 3 For Z.503g of '' 1g of Br = 80 g 160 g of Br ence no. of atoms = Z is Br 2 5