Unit-I - Physical Chemistry - Solution(Final)

Physical Chemistry1. Answer (2) [2C + O2 → 2CO] × 3 [3CO + Fe2O3 → 2Fe + 3CO2] × 2 6C + 3O2 → 2Fe2O3 → 4Fe + 6CO2 3 moles oxygen gives 2 mole Fe2O3 3y gm oxygen gives 2z gm Fe2O3 Q 2z gm Fe2O3 require 3y gm oxygen ∴ x gm Fe2O3 require = 2. Answer (2) Number of moles = weight molecular weight 2.56 256 UNIT 1 Section A : Straight Objective Type x × 3y 3 xy = 2z 2z = Number of moles = 10–2 Number of molecules = 10–2 N0 Q one molecule contain 16 lone pair electrons ∴ 10–2 N0 molecule will contain = 10–2 N0 × 16 = 0.16 N0 3. Answer (2) Moles of CaO = Mass of CaCl2 = %= 4. 1.62 = moles of CaCl2 56 1.62 × 111 = 3.21 gm 56 3.21 × 100 = 32.1% 10 Answer (3) 3O2(g) 1000 – 3x 2O3 2x 1000 – 3x + 2x = 888 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (1) Success Magnet (Solutions) Physical Chemistry x = 112 ml Volume of O3 at STP = 224 ml moles of O3 = 224 = 0.01 22400 O3 + 2KI + H2O → 2KOH + I2 + O2 moles of I2 = 0.01 weight of I2 liberated = 0.01 × 254 = 2.54 g 5. Answer (3) We know that N1V1 = N2V2 x1y1 = x2y2 x2 = x1y1 y2 x2 = final volume of solution Volume of H2O added = x2 – x1 = 6. Answer (1) 2IClx → I2 + moles of Cl2 = moles of IClx = x1y1 y2 ⎞ ⎛y − x1 = x1 ⎜ 1 − 1⎟ ⎟ ⎜y ⎠ ⎝ 2 x Cl 2 2 112 = 5 × 10 −3 22400 1.625 127 + 35.5 x −3 x⎛ ⎞ ⎜ 1.625 ⎟ = 5 × 10 2 ⎝ 127 + 35.5 x ⎠ ∴ moles of Cl2 = 1.625x = 10 × 127 × 10–3 + 10 (35.5 10–3 x) x (1.625 – 0.355) = 1.27 x= 7. Answer (4) 2KClO3 → KCl + KClO4 + O2 Molecular mass of KClO3 = (39 + 35.5 + 16 × 3) gm = 122.5 moles of KClO3 = 1.27 = 1 1.27 12.25 = 0.1 122.5 (2) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 Physical Chemistry Success Magnet (Solutions) moles of pure KClO3 = 0.1 × = 0.075 75 100 mass of residue = 9.1875 – 0.32 = 8.85 gm 8. Answer (2) Xe + x F2 → XeF x 2 1 mole 131 gm ∴ 2 gm 1 mole (131 + 19x) gm (131 + 19 x )2 gm of Xe Fx 131 2(131 + 19 x ) = 3.158 131 2(19x) = 131 × 1.158 ⇒ x=4 ∴ Formula of xenon fluoride is XeF4. 9. Answer (1) milli equivalent of HCl used with metal carbonate = 25 × 1 – 5 × 1 = 20 milli equivalent equivalents of metal carbonate = equivalents of HCl Mass = 20 × 10–3 equivalent mass equivalent mass = 10. Answer (4) Z2O3 + 1 = 1000 = 50 20 20 × 10 −3 3H2 → 2Z + 3H2O (2x + 48) gm 6 gm Q (2x + 48) gm metal oxide requires = 6 gm H2 gas ∴ 0.1596 metal oxide requires = 6( 0.1596 ) = 6 × 10 −3 2x + 48 6 × 0.1596 gm H2 gas. 2 x + 48 2x + 48 = 159.6 x = 55.8 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (3) New Delhi-75 Ph. of H2O2 Q In basic medium n factor of KMnO4 = 3 milli equivalent of H2O2 = milli equivalents of KMnO4 500 = 1 × 3 × volume (ml) Volume of KMnO4 = 12.5 × = 85. Plot No. Office : Aakash Tower. Answer (4) MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O equivalent mass of Cl2 = 71 = 35.Regd.02 mass of Mg = 0.01 22400 100 moles of Mg reacted = 0. Answer (3) n factor of HCl = 6 × 71 = 42.642 gm 13.02 × 24 = 0. Sector-11.: 45543147/8 Fax : 25084124 (4) . 4. Answer (1) K2Cr2O7 + 14 HCl → 2KCl + 2 CrCl3 + 7H2O + 3Cl2 Q 14 mole HCl produces = 3 moles Cl2 ∴ 1 mole HCl produces = 500 ml 3 3 moles Cl2 14 1 mole MnO2 + 4 HCl → MnCl2 + 2H2O + Cl2 1 mole ∴ moles of MnO2 = Mass of MnO2 = 3 14 3 × 87 gm 14 = 18.5 2 6NaOH + Cl2 → 5NaCl + NaClO3 + 3H2O equivalent mas of Cl2 = 14. Answer (2) KMnO4 + H2O2 → Product n=5 n=2 milli equivalents of KMnO4 = milli equivalents of H2O2 100 × 1 × 5 = milli eq.6 10 6 =3 14 7 7 3 equivalent mass of HCl = 36.48 gm Aakash IIT-JEE . Dwarka. Answer (3) 2Mg + O2 → 2MgO moles of O2 = 1120 × 20 = 0.16 15.Success Magnet (Solutions) Physical Chemistry 11. 08 3 0.08 1 mole 0. Answer (2) Number of equivalent of KMnO4 = 1 4 × 10 1000 = 4 × 10–4 Q 5 ml contains 4 × 10–4 equivalent of oxalate ion (equivalents of KMnO4 = equivalents of oxalate ion) ∴ 200 ml contains = 200 × 4 × 10 5 −4 = 16 × 10–3 weight of oxalate = 16 × 10–3 × 44 = 704 × 10–3 % of oxalate = 704 × 10 1. 4.05 Mass of NaHCO3 = 0.05 × 84 = 4.02 3 Mg + N2 → Mg3N2 3 mole 0. Sector-11. Plot No. Answer (1) Let the equivalents of Na2CO3 is X equivalents of NaHCO3 is Y Phenolphthalein indicator X = 2.5 × 0.: 45543147/8 Fax : 25084124 (5) .3 gm methyl orange indicator X +Y = 2.6 gm 3 3 ∴ mass of Mg3N2 = 16.1 24 moles of Mg reacted with nitrogen is 0.Physical Chemistry Success Magnet (Solutions) initial moles of Mg = 2·4 = 0. New Delhi-75 Ph.1 – 0.2 gm 17.5 × 10–3 = 0.08 8 × 100 = = 2.5 × 0.5 −3 × 100 = 47% Aakash IIT-JEE .Regd. Office : Aakash Tower.5 × 10–3 (in 10 mL) ∴ equivalents of NaHCO3 in 1 litre = 0.2 × 2 × 10–3 2 Y = 1 × 10–3 – 0.1 × 2 × 10–3 2 X = 1 × 10–3 (in 10 mL) ∴ In one litre = 1 × 10–1 mass of Na2CO3 = 5. Dwarka. 22. eq. 19. 8 4 3 4 63 3/ 4 equivalent mass of HNO3 = 63 = 4× 3 = 84 gm. Answer (2) Molecular weight of lewsite 1. Office : Aakash Tower.4 ml.: 45543147/8 Fax : 25084124 (6) . 2 HNO3 + 6H+ + 6e → 2NO + 4H2O 8moles of HNO3 exchange 6 moles of electrons 1 moles of HNO3 exchange n factor of HNO3 = 6 3 or mole of electrons. of NaOH = 20 × 0. moles of SO2 = 2 =1 2 volume of SO2 at STP = 22400 × 10–3 = 22. Answer (3) SO2 + H2O2 → H2SO4 m. Answer (3) XZ and YZ planes are nodal planes. Dwarka.1 = 2 m.66 × 10 −24 = 208. 4.66 × 10 −24 1. Plot No. Sector-11. of SO2 in air is 22. eq. 21. Answer (1) 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 2H2O In the above balance equation It is clear that only two of NO3– undergo change in oxidation state while six moles remain in same oxidation state. of SO2 = m. eq.25 × 10 −22 = 2 ×1+ 1.53 amu. New Delhi-75 Ph.78 × 10 −22 + 2 × 12 + 1. conc.4 ppm 20.Regd. of H2SO4 = m.Success Magnet (Solutions) Physical Chemistry 18. Answer (3) ∆X = ∆P (∆X)2 ≥ ΔX = h 4π h 4π Aakash IIT-JEE . 3 × 10 −19 19 4 = 6. Dwarka. Answer (1) 1 1 ⎞ ⎛ 1 = 109678 ⎜ 2 − 2 ⎟ λ ∞ ⎠ ⎝1 λ = 9.5h 2π π θ ν ν 0 tan θ = ⎛ ⎜e⎞ ⎟ ⎝h⎠ So the given graph will be a straight line with slope equal to e 1. Answer (2) Angular momentum (mvr) = = 24. Answer (4) Energy of infra radiation is less than the energy of ultraviolet radiation of the given transitions energy emitted in transition n = 5 → n = 4 is less than the energy emitted in transition n = 4 → n = 3.Regd. of photon ejected per second = 20 = 6. Answer (3) Angular momentum mvr = Angular momentum ∝ n n∝ r nh 2π Angular momentum ∝ r Aakash IIT-JEE .6 × 10 −34 × 3 × 10 8 = 3.06 × 1019 3. Answer (3) Light source is radiating energy at the rate 20 Js–1 hc 6.1176 × 10–6 cm = 911.6 × 10 −19 = = 2.3 × 10 −19 J Energy of single photon = λ = 600 × 10 −9 No.Physical Chemistry Success Magnet (Solutions) ∆X · ∆V = h 4πm h ΔV = h 4π 4πm ΔV = 1 h 2m π 23.76 Å 26. 27. 4.414 × 1014 h 6. New Delhi-75 Ph. Answer (3) hν = hν0 + eVstop ⎛e⎞ ν = ν0 + ⎜ ⎟ Vstop ⎝h⎠ nh 2π 3h = 1. Plot No.626 × 10 −34 Stopping potential 25. Sector-11.: 45543147/8 Fax : 25084124 (7) . Office : Aakash Tower.06 × 10 4 × 10 = 10 − 4 NAV 10 28. Answer (2) Let the no. Dwarka.626 × 10 −34 × 3 × 108 Aakash IIT-JEE . V P 4π 31. Answer (2) Let the electron be moving with momentum P its wavelength will be equal to ∆x = h P h P From Heisenberg’s uncertainty principle Δx · ΔP ≥ h 4π ΔP ≥ h P ΔP 1 × ⇒ ≥ 4π h P 4π Minimum percentage error in measuring velocity would be 100 × ΔV ΔP 100 = × 100 = = 7. l = 1. New Delhi-75 Ph. Answer (3) Cl(17) – 1s2. 3p5 n = 3. m = 1 33. Answer (2) KE = v2 = 1 mv2 = 4.96 ~ 8 . 32. Plot No. Answer (1) n(n − 1) = 10 2 n = 5th shell for visible spectrum transition must be n = 5 → n = 2 n = 4 → n = 2 n = 3 → n = 2 30.: 45543147/8 Fax : 25084124 (8) . 2p6. Office : Aakash Tower. Answer (4) It has highest number of orbitals among all mentioned ones hence maximum orientation is possible for f-orbitals.626 × 10 −34 = = 7.1× 10 −31 v = 103 m/s λ= λ 6.1× 10 −31 × 10 3 34.Success Magnet (Solutions) Physical Chemistry 29.55 × 10–25 2 2 × 4. 2s2. 3s2.Regd. 4.55 × 10 −25 = 1 × 106 9.6 = 28 photons hc 6.28 × 10–7 m mv 9. Sector-11. of photons required to be n nhc = 10 −17 λ −17 −17 × 10 −9 n = 10 λ = 10 × 550 = 27. 6 × 22 eV 12 = 54. Dwarka.Physical Chemistry Success Magnet (Solutions) 35.4 = 79 eV 37.: 45543147/8 Fax : 25084124 ⎛h⎞ ∴ p=⎜ ⎟ ⎝λ⎠ (9) . Sector-11.Regd.6 Z2/n2 eV For excited state n = 2 Z = 1 I.6 + 54. 3s2.E. 3s2. 3d 6 Cl– — 1s2. 2s2. Answer (2) hc =E λ ⎛E ⎞ c=⎜ ⎜p⎟ ⎟ ⎝ ⎠ Aakash IIT-JEE . Office : Aakash Tower.4 eV Energy required to remove both the electrons = binding energy + ionisation energy = 24. 2p6. 3p6 In Fe2+. Plot No.6 Z2/n2 eV = 13. 2s2. 36. = 13. Answer (4) KE = hν – hν0 1 = 3.2 × 1016 4 = 8 × 1015 Hz 40. 3p6. Answer (3) Ionisation energy of He = 13. 2p6. 4.6 × 39.4 eV 4 3 hν = hν − hν 0 4 ν0 = 1 ν 4 = 1 × 3. New Delhi-75 Ph. Answer (1) Ionisation energy = 13. Answer (4) Fe2+ –– 1s2. p electrons are 12 38. Answer (1) 1 = Rz 2 ⎡ 1 − 1 ⎤ ⎢ 2 2⎥ λ ⎢ ⎣ n1 n 2 ⎥ ⎦ 1 ∝ Z2 λ Since He+ Z = 2 ∴ its wavelength is one fourth of atomic hydrogen. d electrons are 6 while in Cl–. 4 = 20.6 U2 = U1 + 20.4 eV 1 22 Potential energy in the first excited level 2 U2 – U1 = 2 × 13. Sector-11.4 eV = 23. Answer (4) 9X 5 = 3. New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (10) . of electrons = 18 ∴ total no.27Å 32 Aakash IIT-JEE .8 eV. of spectrum is UV region = 6 – 1 = 5 43.6 − 2× 13. Dwarka.Regd.4 + 3.6 …(i) = 20. Answer (1) For M. Answer (2) For six energy level n = 6 No. Plot No.Success Magnet (Solutions) Physical Chemistry 41. Answer (1) 1 Shortest wavelength in Lyman series λ = = X R Longest wavelength in Balmer series for He+ = = 36 5RZ 2 36 5 × 4R (∴ Z = 2) = 9 5R = 44. Answer (2) 1 1⎤ ⎡1 = R × Z2 ⎢ 2 − 2 ⎥ λα 2 ⎦ ⎣1 …(ii) 1 1⎤ ⎡1 = RZ2 ⎢ 2 − 2 ⎥ λβ 3 ⎦ ⎣1 λβ λα = 3 9 × 4 8 = 27 32 λβ = 27 × 0. of orbitals = 9 42. 4. between n = 2 and n = 1 level Kinetic energy in first excited state = +13. n = 3 total no.4 eV If ground state is taken as zero potential level then U2 = 0 + 20.32 Å = 0.E. 45.4 eV 22 Difference in P.4 eV Then equation (i) and (ii) Total energy = 20. Office : Aakash Tower. after removing the first electron it occupy the noble gas configuration therefore it is not feasible to remove 2nd electron. Plot No. (2) will have lowest frequency as this falls in the Paschen series. 52.. 3d8. 47. 4s2 Total no. 48. 51. Answer (2) IE of Mg = 737 kJ/mol IE of Al = 577 kJ/mol IE of Na = 495. Answer (4) The lines in the Balmer series are emitted when the electrons jumps from n = 3.3 × 10–3 gm sodium requires = 49. 2s2.. The electrons will not be excited to the third allowed state and hence no line in the Balmer series will be emitted. 57. 3p6. Sector-11. i. 2p6. ∴ 2. 53. Answer (3) 1s2.: 45543147/8 Fax : 25084124 l(l + 1) h 2π (11) .e. 2p6. Answer (1) Radius in the third orbit = 9r (∴ rn ∝ n2) n3λ = 2πr3 3λ = 2π × 9r λ = 6πr 50. Answer (1) Since it is feasible to remove only one electron from the element therefore element belong group 1. 4. 3s1. Answer (1) Ni(28) – 1s2. Answer (3) 1 mole sodium = 23 gm sodium.. 56. 5. Answer (4) Inert gases has most stable electronic configuration therefore has least electron affinity. Dwarka. Office : Aakash Tower. 2s2. Answer (4) This is because in transition element the effect of increasing nuclear charge almost compensated by extra screening effect provided by increasing number of d-electrons.. electron shifts from lower to higher orbit out of (1) and (2). Q 23 gm sodium requires 495 kJ energy for ionisation. 3s2. New Delhi-75 Ph.Regd. orbits to the second allowed orbit. 4.Physical Chemistry Success Magnet (Solutions) 46. 54. Answer (3) BaO2 can exist in form of Ba2+ O22–. Answer (1) Orbital angular momentum = For s-orbital l = 0 ∴ Orbital angular momentum = 0.2 kJ/mol IE of Si = 786 kJ/mol Aakash IIT-JEE . Answer (2) Absorption line in the spectra arise when energy is absorbed.09 eV. orbitals = 15 49. 55.5 kJ.. Since the difference in energy between the third allowed state and the ground state is 12. 36 )2 + (0.5 × 1.732 = 2.5)2 1 2 = 1. 63.5 3 = 1. Answer (4) Nitrogen is chemically inert due to absence of bond polarity. New Delhi-75 Ph. 61.Regd. Plot No. Dwarka.36 × 3 = 0. 4. Answer (4) Alkali metal has low ionisation potential therefore can release an electron easily (oxidised) ∴ Good reducing agent. Answer (1) Bond angle 180º 120º 109º28′ < 109º28′ Molecules BeCl2 BCl3 CCl4 PCl3 Therefore BeCl2 > BCl3 > CCl4 > PCl3. 59.36 × 1.732 = 0. Answer (3) CH3 CH3 R = = 2 2 μ1 + μ1 + 2μ1μ1 cos 60 º (0.Success Magnet (Solutions) Physical Chemistry 58.36 )2 + 2(0.: 45543147/8 Fax : 25084124 (12) .6 D In fact it observed dipole moment is found to be much less due to bond angle diversion following ortho effect.5)2 + (1.62 D 60. Office : Aakash Tower.36)2 × 1 2 = 0. Answer (4) Cl Cl µ = = 2 2 2 μ1 + μ1 + 2μ1 cos 60 º (1.5)2 + 2(1. Aakash IIT-JEE . Answer (3) XeF4 is square planar in shape BrF4– also have square planar in shape. 62. Sector-11. Dwarka. among these given lewis acids back bonding is stronger in B – F. lone pairs = 6 – 4 = 2 2 2 CF4 XeF4 71. 65. 69. 2 2 BCl3 N = 3+3 = 3 → sp2 2 2 In [PtCl4]2– hybridisation is dsp2. Answer (4) In N2 there are pπ – pπ bonding itself and in CN– there is pπ – pπ bonding between C and N. Aakash IIT-JEE . Answer (1) Due to H-bonding H2O has higher boiling point than others. lone pairs = 5 – 4 = 1 2 2 N 4+4 = = 4.Physical Chemistry Success Magnet (Solutions) 64.: 45543147/8 Fax : 25084124 (13) . Answer (1) Due to back bonding BF3 is weaker acid. Plot No. 4. Answer (3) N = 8+2 = 5 2 2 Attached atoms = 3 ∴ T-shaped. 66. Sector-11. 67. N = 5 + 3 = 4 sp3 2 2 PCl5 N = 5+5 = 5 → sp3d. Answer (4) SF4 N 6+4 = = 5. New Delhi-75 Ph. Office : Aakash Tower.Regd. lone pairs = 4 – 4 = 0 2 2 N 8+4 = = 6. Answer (2) P 60º P P P 68. Answer (2) NH3 hybridisation is sp3. 70. Answer (1) Greater electronegativity when bonding through axial position. 275 × 4. 76. Na/NH3 conduct the electricity due to solvated ammonia electrons.03 × 10 −18 × 100 1.75 1. Answer (2) Species Cl – O– O = Cl – O– Bond order 1 1. Plot No. Office : Aakash Tower. 73. Answer (2) KO2 → K+ + O2– In O2– unpaired electrons = 1 → paramagnetic Na2O2 → 2Na+ + O22– In O22– – no unpaired electrons.03 × 10–18 = charge × 1. Answer (4) sp3d and dsp3 have same geometry but d-orbitals that takes part in hybridisation are different. Answer (3) Dipole moment µ = charge (q) × distance 1. Sector-11.03 × 10 −18 1. 4. 75. hence diamagnetic.: 45543147/8 Fax : 25084124 (14) .Success Magnet (Solutions) Physical Chemistry 72.8 × 10 −10 × 10 −8 Percentage ionic character = 74.66 O – O O = Cl – O O Bond length is inversely proportional to bond order. 77. attached atoms = 3 2 2 ∴ T-shaped.Regd. Answer (2) Diamond is sp3 hybridised Graphite is sp2 hybridised Acetylene is sp hybridised.275 × 10 −8 1. Answer (2) N = 6 + 3 + 1 = 5 .5 O Cl O 1. Dwarka. New Delhi-75 Ph. Aakash IIT-JEE .275 × 10–8 Charge = – 1. Answer (1) PV= E= 2 E 3 3 PV 2 For 1 mole gas PV = RT E= 3 RT therefore E represent here translational kinetic energy.01 = × P1 120 100 Number of molecules left = n2 × N0 = 6 × 1018.: 45543147/8 Fax : 25084124 (15) . Answer (2) P1 = P2 = n1 RT V n2 RT V n2 P2 = n1 P1 ⇒ n2 = n1 × P2 12 0.4 L 81. Answer (2) a ⎞ ⎛ ⎜ P + 2 ⎟( V − b) = RT V ⎠ ⎝ Z < 1. Sector-11. Office : Aakash Tower. Plot No.Physical Chemistry Success Magnet (Solutions) 78. Answer (2) At low pressure the volume is high a ⎞ ⎛ ⎜ P + 2 ⎟( V − b) = RT V ⎠ ⎝ a ⎞ ⎛ ⎜ P + 2 ⎟( V ) = RT V ⎝ ⎠ V −b ~ V PV + a = RT V 80. New Delhi-75 Ph.Regd. Answer (3) Volume is directly proportional to the number of molecules. 79. 2 Aakash IIT-JEE . Dwarka. 82. V – b ~ V Vr = RT a 1+ 2 V Vi = RT Vr < Vi since P = 1 ∴ Vr < 22. 4. Answer (1) P= dRT M d= PM RT d ∝ P. New Delhi-75 Ph.51 atm. 84. 86.Regd. d ∝ 1 T 87. Sector-11. 4. Office : Aakash Tower. Answer (2) Solubility of gases in liquids increases on increasing the pressure.Success Magnet (Solutions) Physical Chemistry 83. Answer (3) Since in adiabatic process there is no exchange of heat between system and surrounding therefore in expansion temperature falls down and pressure will be less than the pressure in isothermal process. Answer (4) PV = nRT 10 × V = 1 R × 320 32 V = R litre After leakage 5 10 × R = n × R × 320 8 n= 10 × 5 1 = 8 × 300 48 ∴ mass of gas = 32 2 = gm 48 3 Mass of gas leaked out 1 – 2 1 gas = gm = 0. 3 3 Aakash IIT-JEE . Dwarka. 85.: 45543147/8 Fax : 25084124 (16) . 88. Plot No. Answer (2) PV = nRT 2 × 3 = nAR × 273 nA = 6 273 R for vessel B 4 × 1 = nBR × 300 nB = 4 300R After the connection 4 ⎞ ⎛ 6 + ⎟ R × 300 P×7= ⎜ ⎝ 273R 300R ⎠ P = 1. Answer (4) Because intermolecular force of attraction in NH3 is high.33 gm. Sector-11. Office : Aakash Tower. Answer (2) Since PV = K(constant) 92. Plot No.: 45543147/8 Fax : 25084124 (17) .2 × R × 600 …(i) …(ii) 2NO2(g) 0 2α dividing equation (i) by (ii) then P = 2. Answer (1) rA = rB MB MA volume diffused 50 = time t rate of diffusion = 50 t = MB 40 MA t 5 = 4 MB 64 25 MB = 16 64 MB = 100 93. Since 8 mole O2 produces 4 moles of gaseous product. Therefore pressure reduced to half. Dwarka.4 atm 91. Answer (2) N2O4(g) 1 1–α α = 0. Answer (4) 4A3O4 → 3A4 + 8O2 for the 3 moles of A4 8 moles of O2 required.2 1 × V = 1 × R × 300 P × V = 1.Physical Chemistry Success Magnet (Solutions) 89.2 Total number of moles after equilibrium = 1. Answer (3) 2 4 HCOOH ⎯⎯ ⎯ ⎯ → H2O(g) + CO(g) H SO X moles X moles H2 SO4 ⎯ ⎯→ CO2 (g)+ CO(g)+ H2 O(g) COOH ⎯⎯ Y moles Y moles COOH y moles Aakash IIT-JEE . 90. 4.Regd. New Delhi-75 Ph. Success Magnet (Solutions) Physical Chemistry Y 1 = X + 2Y 6 6Y = X + 2Y X = 4Y X = 4 :1 Y 94.6 63 − X 57 = = 0. 98.: 45543147/8 Fax : 25084124 (18) . Plot No. Answer (2) 1 mHe = 2 1 mH2 4 rH2 rHe = nH2 nHe 4 = 2 2 :1 2 97. Sector-11. X = 6 mm fraction of methane = 95. Answer (2) CH4 + 2O2 → CO2 + 2H2O volume of C2H2 is X mL Pressure (63 – X)mm C 2H2 + Pr essure X(mm ) (63 – X)mm 5 ⎯→ 2CO 2 + H2 O ⎯ 2 2 2 X( mm ) total pressure of CO2 = (63 –X) + 2X = 63 + X 63 + X = 69.72 × 5 = 48. New Delhi-75 Ph.9 63 63 ⇒ 1.6 kcal ∆H = ∆E + ∆nRT ∆E = 48. Answer (2) PH2 = n H2 nH2 + nHe + nCH4 × 2. Answer (3) The expression for standard heat of formation of gaseous carbon is C(graphite) → C(gas) ∆H = 725 kJ/mol As graphite is thermodynamically more stable than diamond so heat required to convert graphite to gaseous carbon should be more.87 kcal. 4. Office : Aakash Tower. Dwarka.6 – (5 × 2 × 10–3 × 373) = 44. Answer (3) Change in enthalpy = Heat of evaporation × Number of moles = 9. Aakash IIT-JEE .6 atm 96.Regd. Aakash IIT-JEE . 105. 7 The rest is used to do work against external pressure 0. Cv = R 2 2 Only 5 = 0. Answer (1) For an adiabatic process PVγ = constant log P = –γ log V + constant Thus slope of log P versus log V graph is –γ. 100. Answer (3) For a diatomic gas Cp = 7 5 R . Thus curve C should respond to helium. Answer (4) At constant temperature ∆T = 0 ∆E = 0. Sector-11. 103.6 kcal. Office : Aakash Tower.71 × 60 = 42. ∆H = 0 and at constant temperature.Physical Chemistry Success Magnet (Solutions) 99.71 of energy supplied increases the temperature of gas.Regd.: 45543147/8 Fax : 25084124 (19) . Answer (1) 3O2(g) → 2O3(g) is endothermic 4EO 3 − 3EO 2 < 0 E O3 < 3 O2 4 This in equality valid only E O2 > E O . Answer (1) γ −1 T1V1γ −1 = T2 V2 γ −1 ∴ T2 ⎛ V1 ⎞ ⎟ =⎜ ⎟ T1 ⎜ ⎝ V2 ⎠ = 2 γ −1 Since γ is more for the gas X. PV = K(constant) – Boyle’s law When temperature is constant PV is constant ∆H = ∆E + ∆(PV) = 0. 4. The value of γ is maximum for helium monoatomic gas. 3 102. Plot No. Answer (2) 2HgO(s) → 2Hg(l) + O2(g) As the reactant from its solid state is converting to liquid and gas phase heat is required for this decomposition ∆H > 0 further more entropy increases ∆S > 0. 101. Dwarka. 104. Answer (3) Heat of formation of a compound is defined as the change in enthalpy when one mole of the compound has been formed from its constituent elements. New Delhi-75 Ph. The temperature will also be more for it. Answer (3) S(g) + 6F(g) → SF6(g). Answer (1) N≡N+ + – 1 (O = O) → N = N = O(g) 2 1 ⎛ ⎞ ∆Hf = ⎜ 946 + × 498 ⎟ − ( 607 + 418 ) = 170 kJ mol–1 2 ⎝ ⎠ Resonance energy = observed heat of formation – calculated heat of formation = 82 – 170 = –88 kJ/mol.95 = 7. 1 atm-litre = 101.95 2 2 H2C2O4 → C2O42– + 2H+. ∆H = –53. Dwarka. 1 1 H2C2O4 + NaOH → Na2C2O4 + H2O. ∆H = 50 kJ/mol. ∆H = 3. Sector-11. ∆H = –57. 109.Regd.8 = 180 × 4. 4. 2 2 Therefore heat of formation = Bond energy of reactants – Bond energy of products –1100 = (275 + 6 × 80) – 6 × (S – F) Thus bond energy of 6 × (S – F) = 309 kJ/mol. ∆H = –1100 kJ mol–1 ∆H = +275 kJ/mol ∆H = 80 kJ/mol 1 F (g) → F(g).Success Magnet (Solutions) Physical Chemistry 106. New Delhi-75 Ph. ∆H = –53.35 kJ 2 2 1 1 H2C2O4 + OH– → C2O42– + H2O. Answer (4) By definition heat of neutralization we have.3 kJ Subtracting eq (1) from eq (2) we get 1 1 H2C2O4 → C2O42– + H+.35 kJ 2 2 …(1) …(2) H+ + OH– → H2O. ∆H = –150 kJ/mol On adding both equations we get BaCl2 + aq → Ba2+(aq) + 2Cl–(aq).184 × ∆T ∆T = 0.3 J = 607. Office : Aakash Tower. 110.8 J Let ∆T be the change in temperature P∆V = mS∆T 607. Answer (2) BaCl2·2H2O + aq → Ba2+(aq) + 2Cl–(aq) + 2H2O.81 K Tf = Ti + ∆T = 290.9 kJ. Plot No. S(s) → S(g).3 J Work done = 6 × 101.8. 108. ∆H = 2 × 3. Answer (2) Work done in expansion = P × V = 3(5 – 3) = 6 atm-litre We have. Aakash IIT-JEE . ∆H = 200 kJ/mol BaCl2 + 2H2O → BaCl2·2H2O.: 45543147/8 Fax : 25084124 (20) . 107. 1 [H+] = 0. Plot No.0095 atm 2 Since atmospheric pressure is 1 atm so percentage of CO2 in air = PCO2 Ptotal × 100 = 0.Physical Chemistry Success Magnet (Solutions) 111. 114. 200 after reaction moles 0 Since CH3COOH is weak acid therefore we assume that H+ ions from CH3COOH is so less than can be neglected ∴ HCl 0. Answer (3) CH3COONa + HCl initially moles 0. Aakash IIT-JEE . the NaOH will use 18 m. Thus to prevent the decomposition of CaCO3 at 100°C the % of CO2 in air must be greater than 0.1 → CH3COOH + NaCl 0 0.mol of NaOH 2NaOH + CO2 → Na2CO3 + H2O The resulting solution contains 18 m mol of NaOH and 1 m.1 Kp = PCO = 0.2 0.1 + Cl– 0.095 N . 4. mol of acid and Na2CO3 will use 1 m. On titration upto phenolphthalein end point.mol of acid.1 M pH = – log[H+] = – log 0.1 =1 113.mol of NaOH 1 m.mol of CO2 reacts with 2 m.95%.95% . Hence Normality = 112.1 = 20 m.3.1 0.: 45543147/8 Fax : 25084124 (21) . Answer (1) The solution contains = 200 × 0. Dwarka. Answer (1) For CaCO3 (s) CaO(s) + CO2(g) → H+ 0. Office : Aakash Tower. New Delhi-75 Ph. pKa + pKb = 14 pKa = 14 – 4. Answer (1) C Csalt ⇒ pH = pK + log salt pH = pK a + log C a C acid acid ∴ Csalt = Cacid.3 pH = pK a pH = 9.1 0 0.7 = 9.Regd.1 (18 + 1) = 0.mol of Na2CO3. Sector-11. 118.2 B+ 0.1 x = 1 × 10 −5 = 5 × 10 −6 2 ∴ degree of dissociation = 116.1 0. 4.1 OH– C2α OH– C1 C2 = [NH4OH] = 0.2 ~ 0. So this pressure is not sufficient to maintain the system in equilibrium therefore total pressure in the chamber would be equal to 30 atm. 2) x = 10 −5 0.x kb = k f (a − x ) x (22) Aakash IIT-JEE . at t = 0 conc at equi. Answer (2) C conc. Sector-11.1 – x ~ 0.: 45543147/8 Fax : 25084124 .2 B+ 0 (x + 0.2 BOH initially moles after equilibrium Kb = 0. a a–x Kf Kb D 0 x when equilibrium is achieved kf (a – x) = kb.1 – x + Cl– 0.2 0. Answer (2) BCl 0. Plot No.2) + OH– 0 x ( x + 0.PNH3 = (20) (10)2 = 2000 atm3 Kp = 2020 atm3 Since Qp < Kp. Office : Aakash Tower. New Delhi-75 Ph.Success Magnet (Solutions) Physical Chemistry 115. 1 − x ∴ x + 0.15 M NH4+ C2α + + Kb [NH4 ][OH− ] C 2 α(C1 + C 2 α ) = = C1α = [NH4OH] C 2 (1 − α ) α Kb −4 = C = 1. 0.8 × 10 1 117. Answer (2) NH4OH C2(1 – α) NaOH –→ Na+ C1 + 5 × 10 −6 = 5 × 10 −5 . Answer (3) 2 Qp = PCO 2 .Regd. Dwarka. 44 × 10 –4 = 1.2 × 10 –4 M Solubility of PbSO4 = 1.5 mL. 123. Answer (3) PCl5(g) COCl2(g) PCl3(g) + Cl2(g) CO(g) + Cl2(g) If some amount of CO has been added into the vessel at constant volume then the second equilibrium will move for backward direction.36 mg litre–1.2 × 10–4 = 1. Therefore to attain the new equilibrium first reaction will move to forward direction and the conc of PCl5 present at new equilibrium will be less 120. 4.36 = 27. Dwarka. 121. = 36. Answer (2) [S2–] = 5 × 10 −21 = 1.8 × 1.50.05 [H+ ][S 2− ] = 1× 10 −7 × 1× 10 −14 [H2S] 1× 10 −21 × 0. As a result the equilibrium concentration of Cl2 will be less. New Delhi-75 Ph. Plot No. Answer (2) CO(g) 1 1–x + NO2(g) 1 1–x CO2(g) + 1 1+x NO(g) 1 1+x t=0 at equilibrium CO2 + Ba(OH)2 → BaCO3 ↓ +H2O white Aakash IIT-JEE .Physical Chemistry Success Magnet (Solutions) 119.0 × 10 –19 0.24 × 10 −10 10 −14 Its equilibrium constant Keq = = = 1. Answer (3) K a × Kb Kw 3. So the equilibrium constant of the first reaction will also be disturbed and reaction quotient will be less than the equilibrium constant.2 × 10–4 × 303 × 103. Volume of water needed to dissolve 1 mg of PbSO4 = 1000 36.Regd.1 1× 10 −19 [H+] = pH = 1. Sector-11. Office : Aakash Tower. Answer (2) Solubility of PbSO4 = K sp = 1.8 × 104 122.: 45543147/8 Fax : 25084124 (23) . Sector-11.0457 Min.0821× 600 Min.8 × 10 −5 [NH3 ] [NH3 ] [NH3] = 0.2 − x ) x = 1.082 × 363 = = 0.25 Kc = ⎜ ⎟ ⎜ ⎟ ⎝ 1− x ⎠ ⎝ 0 . Answer (2) Kp = PCO2 = 2.8 ⎠ 2 2 124.5 × 1) Cl2 Cl2 T = 0.57 gm. moles of CaCO3 required = 0.8 × 10 − 2 = [H2 ] (0.38 atm V 1 (24) Aakash IIT-JEE .Regd. Plot No. Answer (2) H2(g) + At t = 0 At eq.2 197 moles of CO2 at equilibrium ⇒ 1 + x = 1. 125.27 × 10 −2 × 0.2 ⎞ = 2.50 2 PCl ( X Cl × PT )2 (0.125 + 1.1265 M 126. 4. weight of CaCO3 required = 0. Answer (2) NH3 + H2O + NH4+ + OH– Kb = [NH4 ][OH − ] 1.2 0.27 × 10–2 moles/litre PH2 S = nRT 1. New Delhi-75 Ph.0457 × 100 = 4. Dwarka.5 × 10 −3 = = 1. Answer (1) X Cl2 = XCl = 0.25 Number of moles of CO2 = 2. Office : Aakash Tower.5 M 127.25 × 1 0.5 × 10–3 = 0.5 × 1)2 Kp = P = X × P = (0.2 – x [H2S] x = 6.5 × 10 −3 × 1.2 ⎛ 1 + x ⎞ = ⎛ 1.: 45543147/8 Fax : 25084124 .2 x = 0.Success Magnet (Solutions) Physical Chemistry moles of BaCO3 = 236.4 = 1.125 M total [NH3] required = 0. Kc = S(s) 1 1–x H2S(g) 0 x 0. 322 × 10 −3 )2 = = 1. Dwarka.Regd.314 × T log2000 = 20700 + 12T T = 404.37 × 10 –12 Kb = (Cα )2 (1.4 [OH–] = antilog(–3.4) = 3.5 [H3BO3 ] 40 K= 1. New Delhi-75 Ph. Answer (1) M(OH)2(s) pH = 10.9 [Glycerine] 1.322 × 10 −3 5.9 [H3BO3 ][Glycerine] [Complex ] 60 = = 1.303 × 8.Physical Chemistry Success Magnet (Solutions) 128. Office : Aakash Tower.27 [H+] = 5.6 = 3.: 45543147/8 Fax : 25084124 (25) .1× 10 –15 = 1.5 = 0.6 pOH = 14 – 10.1 129.9 [Glycerine] = 131.98 × 10 −4 = 1.7 M 0.98 × 10–4 M M2+ + 2OH– x 2x x= 3.37 × 10–12 [OH–] = 7.99 × 10 − 4 M 2 Ksp = [M2+][OH–]2 = 4x3 = 4 × (1. Answer (2) K= [Complex ] = 0.15 × 10–11 Aakash IIT-JEE .75 × 10 – 5 C 0. Answer (3) Since PCO < PCO2 Kp = PCO2 PCO = 1 500 × 10 −6 = 2000 Since 2. 4.5 = 1. Answer (1) [In–] % of = [In − ] [In ] + [HIn] − × 100 = 10 × 100 = 91% 10 + 1 132. Answer (3) pH = 11.99 × 10–4)3 = 3.303 RTlogKp = − ΔGo p 2.27 – log[H+] = 11. Sector-11. Plot No.3 K 130. 13 V – (–0. Sector-11.04 = 4. Answer (2) Equivalents of Cr deposited = equivalent of O2 evolved 2 .1 × 10–5 = 5 – log 1. of sulphur in hypo are –2 and +6 136.Regd.Success Magnet (Solutions) Physical Chemistry 133.44) = + 0. 4.96 = 9. Dwarka.1 × 10–5 pOH = – log 1. Plot No.994 V Aakash IIT-JEE .1 = 5 – 0.2 × 10–10 [OH–] = 1. Cr 2 + ⎯ 137.6 0.2 × 10–11 [OH–]2 = 1. The electrode with higher reduction potential (Pb2+/Pb) acts as a cathode while other electrode (Fe/Fe2+) with lower reduction potential acts as anode At anode At cathode Fe ⎯ ⎯→ Fe 2+ + 2e Pb 2+ + 2e ⎯ ⎯→ Pb Net reaction Pb 2+ + Fe ⎯ ⎯→ Fe 2+ + Pb º º Eº cell = E cathode − E anode = –0.04 134.769 = 0.: 45543147/8 Fax : 25084124 (26) . New Delhi-75 Ph. Answer (3) Na2S2O3 S Na + — –1 +1 +2 –2 –1 – O — S+2 — O Na +1 –2 + O Since total charge in the sulphurs are –2 and +6 each ∴ Oxidation no.4 ⎯→ Cr n = 2 i. Answer (4) Higher the reduction potential greater is tendency for reduction. Answer (2) E°cell = E°cathode – E°anode = E°H2O2 ( aq)| H2O( l) − E°Fe3+ ( aq)| Fe2+ ( aq ) = 1. Hence more of Pb and Fe2+ are formed.) = 1.763 – 0.96 pH = 14 – 4.56 ×n = ×4 52 22. Answer (4) Mg(OH)2(s) [Mg2+][OH–]2 Mg2+ (aq) + 2OH– (aq. Office : Aakash Tower.e.31V Since the standard emf to the cell is positive the reaction is spontaneous. 135. HO S HO O HO O HO S O HO O HO S O O Aakash IIT-JEE .Regd. Sector-11.Physical Chemistry Success Magnet (Solutions) 138. Answer (2) As the surface area of contact of an electrode with electrolyte increases. Office : Aakash Tower. Dwarka. Answer (3) As the Cu2+ ions lost from the solution are compensated by copper anode therefore concentration of the solution remain same At anode At cathode 139. 4.35 days −3 i ×E 2 × 10 × 87 × 3600 × 24 143. Answer (2) 2MnO2 + Zn2 + + 2e ⎯ ⎯→ ZnMn2O 4 EMnO2 = 87 1 t= w × 96500 8 × 96500 = = 51. 141. Answer (2) Half cell reactions of the given electrodes are First electrode Second electrode Third electrode + + AmO 2 ⎯→ AmO 2 2 +e⎯ + + AmO 2 ⎯→ Am 4 + + 2H2O 2 + 4H + 2e ⎯ Cu(s) ⎯ ⎯→ Cu2 + + 2e Cu2 + + 2e ⎯ ⎯→ Cu(s) Am 4 + + 2e ⎯ ⎯→ Am2+ Thus it is evident that half cell reaction of only second electrode involves H+ ions so its reduction potential will change with varying pH value.059 log = 0. Answer (1) Anode H2 ⎯ ⎯→ 2H+ + 2e 10–6 M Cathode 2H+ + 2e ⎯ ⎯→ H2 Ecell [H+ ] 2 cathode 0. 140.118 = 2 (10 − 6 )2 On solving we get [H+]cathode = 10 –4 M 142. Conductance of electrolyte increase thereby time rate of electrolysis increases.: 45543147/8 Fax : 25084124 (27) . Plot No. Answer (4) In pure state sulphuric acid makes cyclic ring type of structure in the absence of water so it cannot give off H2 gas to react with metals. New Delhi-75 Ph. The reactions occuring at two electrodes are Cathode Anode Net reaction (Hg ) n+ n C + ne ⎯ ⎯→ nHg + nHg ⎯ ⎯→ Hg n n ( ) A + ne − (Hg ) n+ n C + ⎯ ⎯→ Hgn n ( ) A Thus the given cell is electrolyte concentration cell E cell = + 0.0295 = log ⎝ ⎠ n ⎛ 1 ⎞ ⎜ ⎟ ⎝ 20 ⎠ n=2 + It means mercury ion exists as Hg2 2 Aakash IIT-JEE . 4. Plot No. Dwarka. Sector-11. Answer (1) For the following electrochemical cell Pt | Cl2 (1 atm) | Cl− (C1 ) || Cl− (C2 ) | Cl2 (1 atm) | Pt The half cell reactions are Anode Cathode The Ecell is given by 1 Cl− ⎯→ Cl2 (g) + e − A ⎯ 2 1 − Cl2 (g) + e ⎯ ⎯→ Clc 2 Ecell = − C C [Cl − ] 0.0591 2 0.059 0.059 100 log + 2 2 [H ] 0.059 1 log + 2 2 = E°H+ / H − log + 2 2 2 2 [H ] [H ] 0. New Delhi-75 Ph.Regd. Answer (1) 2H+ + 2e ⎯ ⎯→ H2 E = E°H+ / H − 2 PH 0. Answer (2) + Let the formula of mercury ion is Hgn n then the formula of mercury nitrate would be Hgn(NO3)n.Success Magnet (Solutions) Physical Chemistry 144.059 V + 2 2 [H ] E′ = E°H+ / H − 2 ∴ E H+ / H − E ′ = H+ / H 2 2 145.059 0.059 100 log × [H+ ] 2 = 0.0591 [Hgn n ]C log + n [Hgn n ]A ⎛ 1⎞ ⎜ ⎟ 0.059 log 2 = 0.: 45543147/8 Fax : 25084124 (28) . Office : Aakash Tower.059 log 1 log − C = − 1 C1 C2 1 [Cl ] A For Ecell to be positive C1 > C2 146. of H+ be x.059 [Mn 2+ ] log − 0. New Delhi-75 Ph. Answer (1) Q + 2H+ + 2e Ecell = E°cell − QH2 0.0591 1 log + 2 2 [H ] Ecell = E°cell − 0. AgCl is consumed.46 mV Aakash IIT-JEE . 2 2 o o ΔG3 = − n F EH O/H 2 − 2 .699 V ⎛ 0.02846 8 5 [MnO − 4 ]x o 2+ − = E MnO− 4 / Mn Electrode potential decreases by 28.Physical Chemistry Success Magnet (Solutions) 147. o ΔG1 = −RT ln K w ⎯→ H+ + e ⎯ o ΔG 2 = − nF Eo + H / H2 ⎯→ H2O + e– ⎯ 1 H + OH– . OH− Eo H O/H 2 2 .492 ⎞ ⎟ = 3. OH o o Since E H+ / H2 = 0 So ΔG 2 = 0 o Thus ΔG1 = ΔGo 3 o –RTln Kw = − n F EH2O / H2 . Answer (4) o − EMnO− / Mn2 + = EMnO − / Mn2 + 4 4 [Mn 2+ ] 0.Regd.059 log − 5 [MnO 4 ][H+ ]8 Let the initial conc. Answer (1) H2O H+ + OH– 1 H .059 − log 28 log − 8 5 5 [MnO 4 ]x 0. Office : Aakash Tower.: 45543147/8 Fax : 25084124 (29) . Plot No. 2 2 .699 − 0. Answer (1) Above half cell is metal-metal insoluble salt-anion electrode when it acts as cathode half cell reaction will be AgCl(s) + e − ⎯ ⎯→ Ag(s) + Cl− (aq) i. When it is reduced to x/2 the electrode potential is given by E′ MnO − / Mn2 + 4 o 2+ − = E MnO− 4 / Mn 0. So during reaction quantity of AgCl decreases. 5 pH = ⎜ 0.059 0. 148. Sector-11.0591 ⎝ ⎠ 149.0591 pH intercept = Eo cell = 0.e. 4.059 [Mn 2 + ][2]8 log 8 5 [MnO − 4 ]x o − − = EMnO4 / Mn2 + [Mn2+ ] 0. Dwarka. OH − = RT ln K w sin ce (n = 1) F 150. ΔG2 = –1F(0. 154. Answer (2) [ Tl+ ]2 [Sn 2+ ] 0. Plot No. Answer (2) Λ= ⎛ 24 ⎞ =⎜ ⎟ = 0.05 ⎟= ⎠ 390 ⎛ Λ Degree of dissociation = ⎜ ∞ ⎝Λ 156.23 − ( −0.01 ⎞ 19.44 = 1.47 2Tl + Sn4+ → Sn2+ + 2Tl+ Ecell = 0.5 N 0.: 45543147/8 Fax : 25084124 . Dwarka. Answer (3) o Cu2+ + 2e → Cu.23 + 0. 4.47 − 0.67 –3 ∆G° = – nFEo cell = –2 × 96500 × 1.25) = 0. (30) Aakash IIT-JEE . Answer (1) o o Eo cell = E RP cathode − ERP anode = 0. Office : Aakash Tower.337) o Cu+ + e → Cu.5 × 10 −5 = = 19.059 log log[10] 2 = 0.411 V = 0. ΔG3 = –1FE° o o ΔGo 3 = ΔG1 − ΔG 2 –FE° = –2F(0.153 = 0.47 – 157. Answer (4) Ni(s) + Cu2+(aq) → Cu(s) + Ni2+(aq) o o Eo cell = ERP cathode − ERP anode = 0. Answer (3) ∞ ∞ Λ∞ = Λ∞ NH4NO3 + Λ KOH − Λ KNO3 = 128 + 239 − 125 = 242 Degree of dissociation = 155.153) E° = 2 × 0. Answer (4) Since oxidation potential of Cu is more than Ag therefore Cu will go to solution as Cu2+ and Ag+ will go as Ag.59 V Q Eo cell is positive therefore reaction will be spontaneous. but Mg will not be deposited because H+ preferentially discharge.47 – 4+ 2 [Sn ] 2 Eo cell = 1. New Delhi-75 Ph.059 0.34) = 0.5 = 0.059 = 0.34 − ( −0. Sector-11.337 – 0. Λ ⎝ 242 ⎠ ∞ Λ 1000 K 1000 × 19.153) o Cu2+ + e → Cu+.521 volt.10 . ΔG1 = –2F(0.Regd.Success Magnet (Solutions) Physical Chemistry 151. 152. Answer (4) The metal which has high reduction potential reduce first ∴ Sequence of deposition of metal Mg < Cu < Hg.44) = 1.337) + F(0.67 × 10 kJ = –322 kJ 158.13 – (–0. 153. 44) Fe3+ + 3e → Fe. New Delhi-75 Ph.3 ⎢ Cu ⎦ ⎥ ⎣ 163. 4.771 V 162.: 45543147/8 Fax : 25084124 (31) .059 ⎡ Zn2 + ⎤ log⎢ 2+ ⎥ 2 ⎢ Cu ⎦ ⎥ ⎣ ⎡ Zn 2+ ⎤ log⎢ 2+ ⎥ = 37.44) E = –3 × 0. ΔGo 2 = –3F(–0. Pt 160.44 = 0.059 ⎛ 1 ⎞ log⎜ ⎟ 2 ⎝ 0. Office : Aakash Tower. ΔG1 = –2F(–0.059 log ⎢ 2+ ⎥ 2 ⎢ ⎣ Cu ⎥ ⎦ 1. Answer (4) When the cell is completely discharged Ecell = 0 o = E cell ⎡ Zn2+ ⎤ 0. Plot No.1 = 0.01 M E2 = E o cell − 0. ΔGo 3 = –1FE o o ΔGo 3 = ΔG 2 − ΔG1 –FE = –3F(–0. Sector-11.036) Fe3+ + e → Fe2+.01 ⎠ It is clear from both equation E1 > E2 161. Dwarka.01 ⎞ log⎜ ⎟ 2 ⎝ 1 ⎠ when the [Zn2+] = 1 M[Cu2+] = 0.Regd.036) – 2F(0. Answer (2) Since the cell reactions proceed in the standard condition and E o cell is negative therefore the electricity cannot be produced. Answer (4) o Fe2+ + 2e → Fe. Answer (2) Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) E1 = E o cell − 0.3 ⎢ ⎣ Cu ⎥ ⎦ ⎡ Zn2+ ⎤ ⎢ 2+ ⎥ = 1037.Physical Chemistry Success Magnet (Solutions) 159.036 + 2 × 0.059 ⎛ 0. Answer (1) Since oxidation potential of Zn is high therefore Zn will be oxidised ∴ Zn(s)|Zn2+(aq)||H+(aq)|H2(g). Aakash IIT-JEE . T.Success Magnet (Solutions) Physical Chemistry 164.303 RT log Ksp log Ksp = ∴ −0. Sector-11. cathode 2OH– → 1 O + H2O + 2e 2 2 2H+ + 2e → H2 7200 I 1 × 96500 4 7200 I 1 × 96500 2 Moles of O2 released at anode = Moles of H2 released at cathode = Volume of H2 + volume of O2 = 672 (at S.: 45543147/8 Fax : 25084124 . Answer (2) Let the I amp current is passed for 2 hrs. 4. From the given equations ∆G3° = ∆G1° + ∆G2° E° Cl− / AgCl / Ag = E° Ag+ / Ag + 2.Regd.) 22400 × 7200 3 I × = 672 96500 4 I = 0. Dwarka.303 RT log Ksp –0. Answer (1) 2Cl– → Cl2 + 2e (anode) Cu2+ + 2e → Cu (cathode) ∴ 1 mole of copper deposited at cathode (32) Aakash IIT-JEE . Office : Aakash Tower. AgCl(s) + e → Ag(s) + Cl–(aq) .80 + 2.15 = 0. 165. anode At.303 RT log Ksp ∆G2° = −1F E° Ag+ / Ag ∆G3° = −1F E°Cl− / AgCl / Ag Ag+(aq) + e → Ag(s) . 166. Answer (2) AgCl(s) Ag+(aq) + Cl–(aq) .95 = −16. Plot No. Answer (3) Copper is oxidised to Cu2+ 167. Charge = 2 × 60 × 60 × I = 7200 I Moles of electrons passed = 7200 I 96500 At.P.101 0. New Delhi-75 Ph.92 × 10–17. ∆G1° = –2.059 Ksp = 7.536 amp. 2 gm / Cm 3 6. 172.023 × 5.32 gm/Cm3.023 × ( 4. 173. Sector-11.46 169. a = 2 2r 3 3 3 Volume of cube = a = (2 2r ) = 16 2 r 4 3 16 3 πr = πr 3 3 16 3 πr π volume of sphere = 3 = Packing fraction = 3 volume of cube 3 2 16 2 r 174. Dwarka.N. Answer (4) Z×M d= N AV × a 3 = 4 × 62 6. Office : Aakash Tower. Answer (3) d= Z×M N AV × a 3 = 4(78 ) 6. effective one ion of B) and 4 ions of A are removed from edge centres (effective one ion) ∴ New formula effective atoms A+ B– 3 3 Since equal number of cations and anions are missing therefore defect is Schottky detect. Answer (2) 3 face centre + 1 corner atom forms tetrahedral void in fcc. 4. Answer (1) Total volume of sphere = 4 × For fcc unit cell.e. Aakash IIT-JEE . 177.6 Å Z×M = 4 × 78 6.023 × 10 (5.6)3 = 5. 176. Answer (3) When one face plane is removed then four corners ions and one face centre ion of B are removed (i. Z = 4).023 × 10 23 × (10 −8 )3 = 4 × 620 = 411. 175. 170. New Delhi-75 Ph. = 7. Answer (4) Total number of effective atoms in unit cell = 4 + 4 = 8.: 45543147/8 Fax : 25084124 (33) . Plot No. Answer (3) Triclinic is most unsymmetrical crystal system a ≠ b ≠ c and α ≠ β ≠ γ = 90°. 171.Physical Chemistry Success Magnet (Solutions) 168.3 × 2 = 4.e.023 In Frenkel defect the density remains unaltered.75 g / Cm 3 6.023 × 10 23 × ( 4.46 × 10 23 −8 3 ) = 4 × 780 = 3.Regd.6)3 × 10 − 24 4 × 780 d= N AV × a 3 = 6. Answer (4) All have same number of formula unit (i. Answer (1) r+ + r– = a 2 a = 2. Answer (3) The centre atom is surrounded by six atom and one atom lies over this therefore C. : 45543147/8 Fax : 25084124 (34) .09 × 10 −8 )3 Unit cell is fcc. Office : Aakash Tower.732 0. Answer (3) In hexagonal crystal system a = b ≠ c. Answer (3) Orthorhombic exist in body centred. Sector-11.Regd. 2a = 4r 4r a= 2 = 2 2r = 2 2× 3 × 297 4 = 3 × 297 2 = 363. end centred. Aakash IIT-JEE .732 r– = 182. Answer (2) r+ r− = 0. γ = 120°. 184.79 pm.5 = Z=4 ∴ 6. α = β = 90°. face centred as well as primitive unit cells. 179. 4.5 = Effective number of ions of A = 4 − ∴ 1 7 = 2 2 5 2 New formula of compound A7/2B5/2.732 r+ 100 = = 136 . 0. 183. Plot No. New Delhi-75 Ph. 3a = 4r r= 3a = 4 3 × 297 4 In fcc. Answer (1) Co-ordination number of each sphere is 6. 180.023 × 10 23 × (4. 181. Answer (4) When all the atoms touching body diagonal then 2 face centred + 4 corners ions will be removed of B and 2 edge centre ions of A will be removed therfore effective number of ions of B = 4 – 1.Success Magnet (Solutions) Physical Chemistry 178.61 pm. Answer (1) In bcc. Dwarka. Answer (2) Z ×M d= NAV × a 3 Z × 108 10. Office : Aakash Tower. Plot No.303 log (a − x ) t 2. a + 3x = Pt x= a (a – x) a = P0 B (g) + 0 x 2C(g) + 0 2x D (g) 0 x Pt − P0 3 For first order reaction. At t = ‘t’. Sector-11. Answer (2) − 1 d [H2 ] 1 d [NH3 ] =+ 3 dt 2 dt d [H2 ] 3 − = × 2 × 10 − 4 = 3 × 10 − 4 dt 2 So. Answer (3) In the rate of reaction. Dwarka. 187. reciprocal of coefficient is written not in the rate of appearance of product. k= k= a 2.: 45543147/8 Fax : 25084124 (35) . the formula of compound per unit cell is A1/2BC2. 189. −d[H2 ] d[NH3 ] × = 3 × 10 − 4 × 2 × 10 − 4 = 6 × 10–8 dt dt 188. New Delhi-75 Ph. simplest formula of compound is AB2C4. Answer (3) A (g) ⎯ ⎯→ At t = 0. Number fo A atoms = Number of B atoms = Number of C atoms = 1 1 ×4 = 2 8 185. 3 ⎝6⎠ (2r ) 186.303 3 P0 log t ( 4P0 − Pt ) Aakash IIT-JEE .Regd.Physical Chemistry Success Magnet (Solutions) 4 3 πr ⎛ π⎞ In simple cubic unit-cell the packing fraction = 3 = ⎜ ⎟. At t = 0 At t = ‘t’. 4. Answer (1) 1 ×2 =1 2 1 × 4 +1= 2 4 So.303 log t P0 ⎛ Pt − P0 ⎞ P0 − ⎜ ⎟ ⎝ 3 ⎠ k= 2. Answer (3) Number of alternate corners = 4 Number of alternate edges = 4 Number of alternate faces = 2 Hence. 7 1. d [ A 2B5 ] 1 d [ AB2 ] d [B2 ] =+ = +2 dt 2 dt dt 1 k1[ A 2B 5 ] = × k 2 [ A 2B 5 ] = 2 × k 3 [ A 2B 5 ] 2 the relation becomes − 2k1 = k2 = 4k3 193. Answer (2) A 2B5 ⎯ ⎯→ 2AB 2 + ½ B2 So. Answer (1) Total time = n × t 1 2 69.303 N log 0 t Nt N0 λ×t = Nt 2.2 = n × 138. Answer (3) λ= 2.303 (36) . New Delhi-75 Ph.3 g 4 He −2 206 210 84 Po ⎯⎯ ⎯→ 82 Po Volume of helium accumulated = 22400 × 0. Answer (3) Lower is the activation energy. Office : Aakash Tower. Answer (3) 238 92 U is disintegrated through (4n + 2) series.7 = 0.Regd.: 45543147/8 Fax : 25084124 λ 2. 4.Success Magnet (Solutions) Physical Chemistry 190. Activation energy for reverse reaction = ∆H (only magnitude) + Activation energy for forward reaction = 20 + 30 = 50 Aakash IIT-JEE . Plot No. Sector-11.4 n =½ n 1 ⎛ 1⎞ ⎛ 1 ⎞2 N = N0 ⎜ ⎟ = 1 ⎜ ⎟ 2 ⎝ ⎠ ⎝2⎠ 1 N= 2 = 1 = 0.414 Disintegrated amount = 1 – 0. Dwarka. Answer (3) For exothermic reaction. 192. 191.3 g 210 = 32 ml. 195. higher is the rate of reaction.303 log Comparing with y = mx + C then Slope = + 194. Answer (2) 2A(g) + B(g) ⎯ ⎯→ 2C(g) + 1 d [C] 120 − 100 = =2 2 dt 10 + d [C] = 4 mm / min dt 197.Physical Chemistry Success Magnet (Solutions) 196. New Delhi-75 Ph. Answer (3) ⎯→ 4C + 5D 2A + 3B ⎯ − 1 d [B] 1 d [C] =+ 3 dt 4 dt 4 d [B] d [C] =+ 3 dt dt − 201. Plot No.303 = × 0.5 × 109 years. Answer (2) 238 ⎛ ⎞ × 0.303 206 ⎝ ⎠ k= log t 1 0. Dwarka.: 45543147/8 Fax : 25084124 (37) . 4. Answer (2) Overall order of reaction is 2.5 × 10 t = 1. 200.693 2.Regd. Answer (4) For ‘B M= 1000 × K b × 1 (M = molar mass of B) 1× 100 M 10 Kb = For ‘A’ M1 = 1000 × M × 2 1× 10 × 100 (M1 = molar mass of A) M1 = 2 M Aakash IIT-JEE . Answer (1) t½ ∝ 1 (a)n−1 (n = order of reaction) 198.1 9 t 4. Sector-11. 199.2 ⎟ ⎜1 + 2. Office : Aakash Tower. : 45543147/8 Fax : 25084124 …(i) …(ii) (38) . Aakash IIT-JEE .1 For Ba(NO3)2. 4. Answer (2) Freezing will start at –1. Answer (3) ∆Tf = Kf × m ∆Tb = Kb × m By adding (i) and (ii) ∆Tf + ∆Tb = Kf × m + Kb × m = m(Kf + Kb) (Q ∆Tf + ∆Tb = 2. Plot No.5 204. o o ·X A + PB ·XB Ps = PA o o Ps = PA ·X A + PB (1 − X A ) (Q XA + XB = 1) o o o Ps = PA ·X A + PB − PB ·X A o o o Ps = PB − X A (PB − PA ) …(i) Comparing the equation (i) with Ps = 210 – 120XA then we get o o PB = 210 and PA = 90 203.5 56. Sector-11.93°C. answer is (3).52) m = 1 for non electrolyte solute.38) 2. freezing point of solution = 0 – 0.86°C not 0°C because ∆Tf = 1. Glucose doesn’t freeze.86 and as the value of ∆Tf increases then the molality also increases due to the freezing of water. m = 0. Hence. New Delhi-75 Ph.558 Total depression in freezing point = ∆Tf for NaCl + ∆Tf for Ba(NO3)2 = 0.5 18 nurea 1 1 = = nurea + n water 1 + 55.1 = 0.38 = m(1. nwater = xurea = 1000 = 55.86 × 0.93 Hence.86 × m m=1 It shows that 1 mole of urea dissolved in 1000 g of water.372 + 0.86 × 0. m = 0.1 = 0. nurea = 1. Answer (4) For NaCl.1 ∆Tf for NaCl = i × Kf × m = 2 × 1. 205. Dwarka.86 + 0. Answer (2) From Roult’s law.Regd.93 = –0.372 ∆Tf for Ba(NO3)2 = i × Kf × m = 3 × 1. 206. Answer (1) ∆Tf = Kf × m 1. Office : Aakash Tower.558 = 0.Success Magnet (Solutions) Physical Chemistry 202.86 = 1. 29 g Ice separated = 200 – 161.2) = 0.78 Hence. 212.5 = 2.71 g 209.45 F. Plot No. Answer (2) A negatively charged sol is formed due to adsorption of I–. Office : Aakash Tower. Answer (1) M= 62 = M Ca(NO 3 )2 and 0. Dwarka.78 – (–2.79 = –2.79°C B. i = 3) ∆Tb = i × Kb × m = 3 × 0.Physical Chemistry Success Magnet (Solutions) 207.5 = 0. Coagulation value ∝ 1 . 210.P.78 = 100. effective molarity are same. B. Answer (2) In case of 208. F. Answer (1) Emulsifying agent stabilised the emulsion of oil in water by forming an interfacial film between suspended particles and the medium.2 × 1.79 Hence. of solution = 100 + 0. 213. Sector-11. of solution = 0 – 2.2 0.: 45543147/8 Fax : 25084124 (39) . 2 4 20 1000 × K f × w ΔT × W 1000 × 1.Regd.8 i= H+ + X– 0 0 before dissociation after dissociation 0.2 1 ∆Tf = i × Kf × m = 1. Answer (3) 9.2 = 1.P.78°C ∆Tf = i × Kf × m = 3 × 1.3 × W W(unfreezed water) = 161.52 × 0. coagulating power Aakash IIT-JEE .2 1. Answer (2) If same masses of same type of electrolytes are taken then lower is the molecular mass higher is the molarity and higher is the colligative properties.57°C. of solution = 0 – 0. New Delhi-75 Ph.P.86 × 50 9. Answer (1) HX 1 (1 – 0. 4.45 = –0.45°C 211.5 m 200 1000 (For MgCl2.86 × 0.79) = 103.P. – F.86 × 0. = 100.5 m = 95 = 0.P.2 = 0.29 = 38.05 M Na SO . B : Multiple Choice Questions 1.1× number of moles of NaF formed = 0.01 [NaF] = 0.1× 100 = 0. 219. Answer (1) Gold number is equal to the number of milligram of substance required to prevent the coagulation of 10 ml gold sol before adding 1 ml of 10% NaCl solution. Answer (1) The potential required to stop electro-osmosis is known as Dorn potential. 216. 217. Office : Aakash Tower. 218. Plot No. Answer (2) K is highly reactive towards water. Answer (2. Sector-11. 220. Answer (3) As2S3 sol is negatively charged. Answer (2) van der Waal’s adsorption occurs at low temperature and high pressure. New Delhi-75 Ph. Answer (2. Answer (1. Dwarka.05 M 200 1000 Aakash IIT-JEE .01 1000 moles of NaOH = 0. 3) %C in C2H5OH = 24 × 100 = 52% 24 + 5 + 16 + 1 72 × 100 = 40% 72 + 12 + 96 %C in C6H12O6 = %C in CH3COOH = 24 × 100 = 40% 12 + 3 + 12 + 32 + 1 %C in C2H5NH2 = 3. Answer (3) Fe(OH)3 is positively charged sol due to adsorption of Fe3+ ions. 4) 24 × 100 = 53% 24 + 5 + 14 + 2 HF + NaOH ⎯ ⎯→ NaF + H2O moles of HF = 0.: 45543147/8 Fax : 25084124 (40) . 2. 4.Success Magnet (Solutions) Physical Chemistry 214. Answer (1) Higher is the value of van der Waal constant ‘a’ more is the adsorption on charcoal.Regd.01 = 0. Section . 215.01 1000 100 = 0. 4) The substances having same composition of atoms and similar crystal structure are isomorphous. equivalents of NaOH = 20 × 0. 4.Regd. New Delhi-75 Ph. Cl is present in lowest oxidation state In HClO4. 2. range of oxidation number is from –1 to 7. Sector-11.Physical Chemistry Success Magnet (Solutions) 4. Concentration of SO2 in air is 22. Office : Aakash Tower. equivalents of SO2 = m. 3) SO2 + H2O2 → H2SO4 m.1 = 2 n factor of SO2 = 2 =1 2 Volume of SO2 at STP = 22400 × 10–3 = 22. Plot No. Answer (1. Dwarka. Answer (1. equivalents of H2SO4 = m. Answer (2.025 40 Normality = mass / equivalent mass vol (litre) 1 × x molecular mass of NaOH = 40 40 40 x = 1 gm 8. 3) KCrO3Cl = 1 + x + 3 × –2 + (–1) = 0 x = +6 O O CrO5 O Cr O O Oxidation number of Cr = +6 5. Answer (1.4 pm Aakash IIT-JEE .: 45543147/8 Fax : 25084124 (41) .4 ml. In HCl. For Cl. 3) Resultant normality of solution N1V1 + N2V2 + N3V3 = N4V4 5 × 1 + 20 × n=1 equivalents of MnO2 = equivalents of (NH4)2SO4 1 1 + 30 × = N4(1000) 2 3 25 = N 4 1000 N4 = 1 N 40 Resultant [H+] = 1 = 0. 3) Those substance can be oxidised and reduced. Answer (2. in which central element is neither in lowest nor in highest oxidation state. 3) MnO2 + (NH4)2SO4 → MnSO4 + (NH4)2S2O8 n=2 1 × 2 = 1x x=2 7. 2. 6. 2. Cl is present in highest oxidation state. Sector-11. Answer (1. Answer (2.Regd. 4) 6 3 or mole of electrons 8 4 3 4 63 = 84 gm. 3) Ca(OH)2 + H2SO4 → CaSO4 + 2H2O n=2 n=2 equivalent mass of H2SO4 = 98 = 49. Dwarka. 2 (42) Aakash IIT-JEE .Success Magnet (Solutions) Physical Chemistry 9. Office : Aakash Tower. 12. 2) molecular mass . New Delhi-75 Ph. 4. 4) 3Cu + 8HNO3 ⎯→ 3Cu(NO 3 )2 + 2NO + 9H2 O 0 +2 In the above balance equation is it clear that only two moles of NO3– undergo change in oxidation state while six moles remain in same oxidation state 2HNO3 + 6H+ + 6e → 2NO + 2H2O Total 8 moles of HNO3 exchange 6 mole of electrons 1 mole of HNO3 exchanges ∴ n factors of HNO3 = Cu is oxidised of Cu2+ equivalents mass of HNO3 = 10. 3.: 45543147/8 Fax : 25084124 . 3/ 4 MnO 2 + 4HCl → MnCl2 + Cl2 + 2H2 O n factor of HCl = 2 1 = 4 2 +4 −1 +2 0 n factor of MnO2 = 2 equivalent mass of MnO2 = 11. 2 Ba(MnO 4 )2 ⎯ ⎯→ Mn2+ ( n=10 ) +7 +2 milliequivalents of Ba(MnO4)2 = 100 × Fe2+ → Fe3+ + e n=1 1 × 10 = 100 10 milliequivalents of FeSO4 = 100 × 1 = 100 Fe C 2 O 4 → Fe 3 + + CO 2 (n = 3 ) +2 +3 +3 +4 milliequivalents of FeC2O4 = 100 × 3 = 100 3 equivalents of Ba(MnO4)2 = equivalents of FeSO4 = equivalents of FeC2O4. Plot No. Answer (1. Answer (1. 3. 1. l = 2 Radial nodes = 3 – 2 – 1 = 0 For 4f. 3) Number of scattered α-article in Rutherford experiment is inversely proportional to square of kinetic energy of incident α-particles.Physical Chemistry Success Magnet (Solutions) 13. 4) Radial node for 1s ⇒ n = 1. 4) For mth line n2 = (m + 1) ⎡1 1 1 ⎤ = Rz 2 ⎢ 2 − ⎥ λm (m + 1)2 ⎦ ⎢1 ⎥ ⎣ for nth line n2 = (n +1) ⎡1 1 1 ⎤ = Rz 2 ⎢ 2 − ⎥ λn (n + 1)2 ⎦ ⎢1 ⎥ ⎣ λ m (m + 1)2 = λn (n + 1)2 2 ⎧ ⎪ (n + 1) − 1 ⎫ ⎪ ⎨ ⎬ 2 ⎪ ⎩ (m + 1) − 1⎪ ⎭ 16.: 45543147/8 Fax : 25084124 (43) . 4. Answer (3. Plot No. 3. 17. 3. 4) n = 4. Answer (1. Answer (1. 2. 3. Dwarka. 2. Office : Aakash Tower. Answer (3. Answer (1. m = –2. 1. 2. +2 s= + 1 (correct) 2 −∞ ∫ ψ dxdydz = 1 2 n = 4. 2. l = 0 Radial node = (n – l – 1) = 1 – 0 – 1 = 0 For 3d.e. 2. 4) X – [Ar]4s1 Y – [Ar]5s1 Since in the Y electron is in higher state therefore energy required to change (X) to (Y) Since in X element there is 19 electron which represent the K-atom. Answer (1. New Delhi-75 Ph. l = 0. –10 + 1 + 2 s= − 1 (correct) 2 Aakash IIT-JEE . 18.Regd. –1. l = 0. 3 for l = 2. 4) An acceptable solution of schrodinger wave equation must satisfy the following condition (i) It should be single valued ∞ (ii) It should be continuous (iii) The function should be normalised i. n = 3. m = –2. l = 3 Radial nodes = 4 – 3 – 1 = 0 14. 3 l = 2.0 + 1. or n = 4. Sector-11. 15. New Delhi-75 Ph. Aakash IIT-JEE . Answer (1. Office : Aakash Tower. Transition corresponding 4 → 1.: 45543147/8 Fax : 25084124 (44) . ∴ Hybridisation = sp ⎪ 2 2 ⎭ 25. Dwarka. 26. 4. 2) rn = n2r1 2 ⎛ πrn ⎛ An ⎞ ⎜ ⎟ = ln ⎜ ln ⎜A ⎟ ⎜ πr 2 ⎝ 1⎠ ⎝ 1 ⎞ ⎟ = ln n 4 = 4 ln(n) ⎟ ⎠ ln An A1 ln(n) 22. 24. 2 → 1 belongs to lyman series. Answer (1. 3. 3. 23. 21. 2. 4) H3 O + N 6 + 3 −1 ⎫ = = 4 lone pair = 1 . Plot No. 4) (1) KF combines with HF and forms KHF2 with exists as K+ + [F -----. 4) Since only six different wavelengths are emitted therefore highest excited state is n = 4 therefore (1) is correct. 4) Kinetic energy of the ejected electron depends on the frequency of incident radiation not on the intensity. 3) Energy of orbital of hydrogen depend upon n and not on l. 2. 2) Cu and Al. Answer (2. (2) Due to smaller size of Li+ ions it has high polarising power therefore predominantly covalent in nature. 3 → 1. So 1s – 1s3. Sector-11.H —— F]–. In the emitted radiation two wavelength are shorter than λ0 it means that initially atoms were in excited state therefore (2) is also correct. 3. ∴ Hybridisation = sp 3 ⎪ 2 2 ⎪ 4 + 3 +1 − N 3⎪ CH3 = = 4 lone pair = 1 . 3. 20. Si and Ge donot show inert pair effect. Answer (1. Intense and weak beam are having more or less number of photons. (3) CsBr3 exists as Cs+ + Br3–. Answer (2. first period would have 3 vertical columns. 4) There are three possible values of spin quantum number it means an orbital can accommodate 3 electrons. (4) Sodium sulphate is soluble in water but BaSO4 is sparingly soluble because hydration energy of BaSO4 is less than its lattice energy.Regd.Success Magnet (Solutions) Physical Chemistry 19. Answer (1. ∴ Hybridisation = sp ⎬ trigonal pyramidal 2 2 ⎪ N 5+3 ⎪ 3 NH3 = = 4 lone pair = 1 . Answer (1. Answer (2. 2. 2) Rb+ – ionic radius 1. 3) Alkali metal have lowest I. 2.48 Å O2– – ionic radius 1. 4. 29. F– is weakest reducing agent among halide due to maximum stability due to highest hydration energy. Answer (2.68 Å Mg2+ – ionic radius – 0. 3) Low ionisation energy.Regd. high electron affinity and high lattice energy favours the ionic bond formation. Answer (1. Plot No. energy because after releasing one electron these acquires noble gas configuration. 4) PCl5 in solid form exists as [PCl4+][PCl6–] and PBr5 exists as [PBr4+][Br–] in solid state. Office : Aakash Tower. Answer (1. 31. Aakash IIT-JEE . 3) Electron affinity of O(g) and S(g) are negative therefore involve emission of energy.: 45543147/8 Fax : 25084124 (45) . 4) Electronegativity. Electron affinity of nitrogen is less than oxygen because nitrogen has half filled p-orbital therefore it is more stable.40 Å and Li+ – ionic radius – 0. 4) XeOF2 NH2 − N 8+2 = = 5 attached atoms = 3. 3. 3) ∴ Angular F B–C≡C–B F sp2 sp sp sp2 SiH3 SiH3 N – SiH3 F F ∴ planar Lone pair of nitrogen is involve in pπ-dπ bonding there fore delocalised therefore nitrogen is sp2 hybridsed and planar. Answer (2. ionisation energy and oxidizing power increases from iodine to fluorine. Answer (1. New Delhi-75 Ph.Physical Chemistry Success Magnet (Solutions) 27. 34. Sector-11. 4) Increasing metallic character increase the electron donating tendency of metal. 33. 32. Answer (1. 30. Dwarka. Answer (1. 2.60 Å 28. Answer (2. Answer (1.E. 4) Tl+ is more stable than Tl3+ Ga3+ is more stable than Ga+ Pb4+ is less stable than Pb2+ Bi3+ is more stable than Bi5+ These are due to inert pair effect. ∴ T-shaped 2 2 N 5 + 2+1 = = 4 attached atoms = 2. Answer (1. 35. 2 2 36. 3. 4) Above critical temperature gas cannot be liquified. 3. Answer (1. 4) N O O (μ = 0) Cl (μ ~ – 0) H C2H5 (μ ≠ 0) Statement 2 and 3 are facts n2a V2 Unit of P = Unit of ∴ Unit of a = atm L2mol–2 42. 3. 45. 43. Dwarka.5° Bond angle of F2O – 103° 39. 2. Sector-11. 2. 4. 4) Electron affinity of anion is positive and non spontaneous due to electron –2 repulsion. Answer (1. 4) Z= PV RT 3 RT 2 Average KE = 44. 3) CH3 O O N O Br CH3 C C H O CH3 (μ ≠ 0) 41. 2. 3. New Delhi-75 Ph. 4) Z= Vmolar ( Vm ) Videal ( Vid ) Aakash IIT-JEE . Answer (1. Office : Aakash Tower. 3.Regd. Answer (3. 4) Mutual attraction of molecules known as van der Waal intermolecular force. Answer (2.: 45543147/8 Fax : 25084124 (46) . Answer (2. Answer (2. Answer (1. 4) (a) On decreasing the electronegativity of central atom bond angle decreases (b) Bond angle of NH3 – 107° Bond angle of H2O – 104. 40. Answer (3.Success Magnet (Solutions) Physical Chemistry 37. 4) – N = C = O – linear S = C = S – linear 38. Plot No. 3. Answer (1.Physical Chemistry Success Magnet (Solutions) When Vm > Vid Vm = Vid Vm < Vid 46. N2O5 forms nitric acid H2O + N2O5 → 2 HNO3 48. Plot No. Answer (1. Answer (1. 4. therefore these are reducing in nature. 3. Answer (2. 4) Cl Cl (1) Cl Br (2) Br P Br Cl Br P dipole moment μ ≠ 0 Br Cl dipole moment μ = 0 (3) CH3 O O CH3 H (4) N H O H dipole moment is not zero dipole moment is not zero due to following structure 50. 4) Order of acidic strength H3PO2 > H3PO3 > H3PO4. H3PO2 there is P—H bond present.Regd. hybridisation of phosphorus in all acids are not sp3. 3) Z>1 Z=1 Z<1 If force of attraction dominates then Z < 1 Mg2C3 → 2 Mg2+ + C34– 2 Mg2+ + [2–C C C2–] Two sigma and two pi bonds CaCN2 ⇒ Ca2+ + CN22– [–N C N–] Two sigma and two pi bonds 47. Answer (2. 3. 2. New Delhi-75 Ph. Sector-11. Office : Aakash Tower. In H3PO3. 4) KK (σ2s2) (σ*2s2) (π 2px2 = π 2py2) Four electrons are present in 2π molecular orbitals that’s why double bond contains both π bonds. + − O In gaseous state N2O5 exists as O N O N O therefore hybridisation of each nitrogen is sp2. Dwarka. 51. Aakash IIT-JEE . O N2O5 is called anhydride of nitric acid because in reaction with H2O. 3) In solid state N2O5 exist as [NO 2 ] [NO 3 ] therefore hybridisation of each nitrogen is sp and sp2 respectively. 2. 49.: 45543147/8 Fax : 25084124 (47) . 4) Under the critical condition gases does not follow ideal behaviour for Z is not equal to 1 at absolute zero temperature the kinetic energy of gas molecules will be zero. Answer (1. 56. Office : Aakash Tower. 3. 3.Success Magnet (Solutions) Physical Chemistry 52. Answer (1. Sector-11. Answer (1. 4) All are well known relations. 4) ump 2RT M 1 : uaverage : urms : 8RT πM : 3RT M : 1. 2) Pc = a 27b 2 Vc = 3b. 3. 61. Dwarka.128 : 1. Answer (1. 4) van der Waal’s constant a measure the intermolecular force of attraction b is called excluded volume Vc = 3b. 60. 4) Absolute values of entropy and internal energy cannot be calculated. Aakash IIT-JEE . 4. 53.Regd. PQ are intensive variables and property. ∆H < 0 and ∆E < 0 58. 59. Answer (2. New Delhi-75 Ph. 3. 2. Answer (2. 2) On the expansion the volume increases therefore pressure decreases as the temperature is constant therefore kinetic energy of gas molecule remain same.: 45543147/8 Fax : 25084124 dP is intensive dQ (48) . 3) Kinetic energy for 1 mole gas = 3 RT 2 1 mole of gas has molecules = Nav. 2. 4) If P. 55. 4) For spontaneous process ∆G < 0. 4) Statement of IInd law of thermodynamics. Answer (1. Answer (1. Answer (1. Q are arbitrarily chosen intensive variables then P/Q.224 57. 54. Plot No. Answer (2. Answer (1. 68. Gibbs free energy are state function. Answer (1.5 × 2 2 = ×V 5 15 V = 7.: 45543147/8 Fax : 25084124 (49) . 4) Standard heat of formation of all elements in their standard states is zero ∆Hf(O) ≠ 0 and ∆Hf (diamond) ≠ 0 because these are not standard state. 2. 4) State function depends only on initial and final position therefore Enthalpy. Dwarka. 3. 67. New Delhi-75 Ph. 3) During the streching of rubber band the long flexible macromolecules get uncoiled the uncoiled arrangement has more specific geometry and more order thus entropy decreases. Answer (2. Plot No.Physical Chemistry Success Magnet (Solutions) 62.(a). Office : Aakash Tower. 3. Answer (1. 68. 63.5 ml ∴ Milli equivalents of salt = 1 Aakash IIT-JEE .Regd. Entropy. 4) Work done in reversible process is more than work done in irreversible process at equilibrium ∆G is zero. Answer (3) At equivalence point (IIT-JEE 2008) N1V1 = N2 V2 (base) (acid) 2. 4) In isothermal process T = constant P1V1 = P2V2 (Boyle’s law at constant T) ∆U = 0 ∆H1 = ∆H2 64. Answer (1. Sector-11. Answer (1. 4) BOH + x x− 3 x 4 HCl 3 x 4 → BCl 0 ⎛ 3x ⎞ ⎜ ⎟ ⎝ 4 ⎠ + H2O 0 ⎛ 3x ⎞ ⎜ ⎟ ⎝ 4 ⎠ 0 pOH = pK b + log [Salt ] 3x × 4 = 5 + log [BoH] 4× x pH = 14 – 5 – log 3 = 8. 2) ∆E is a state function ∴ ∴ It is zero is cyclic process internal energy of ideal gas depends only on temperature In isothermal process ∆E = 0. 4. 66. Answer (2. Answer (3. 65.523. Sector-11.5 = 3. 3) At equilibrium ∆G = 0 ∆G° = – 2. Answer (1. 4. Answer (1.303 RT log K At 25°C E° cell = 0.303 RT Log K – nF E ° cell = – 2. 2. 2) N2(g) + 3H2(g) 2NH3(g) ∆H = negative Na+ + CN– HCN + OH– (basic) CH3COO– + Na+ CH3COOH + OH– (basic) 2Na+ + CO32– H2CO3 + 2OH– (basic) Since the no.0591 log K . 3) The aqueous solution of NH4Cl and CuSO4 are acidic.2 × 10–2 M 69.5 = 1. Office : Aakash Tower. Answer (1. 2. Ag+ are electron deficient SnCl4 can expand its octet due to vacant d-orbital therefore behaves as a lewis acid. Answer (1. therefore NH3 formation will be fast. The temperature at which atmospheric pressure is equal to vapour pressure is called boiling point if pressure boiling point will be increased . of moles of gases decreases in product sides therefore on increasing the pressure forward reaction favours.: 45543147/8 Fax : 25084124 (50) . Answer (1. 4) NaCN CN– + H2O CH3COONa CH3COO– + H2O Na2CO3 CO32– + 2H2O 73. n 70. 3. 4) Pressure favours the forward reaction.Success Magnet (Solutions) Physical Chemistry pH = 7 – 1 1 pkb – log C 2 2 1 1 log 10 2 =7–6– = 7 – 6 + 0.5 (H+) = 10–1. Catalyst increases the rate of reaction. 3) On increasing the ammonia the partial pressure of NH3 increases where as increasing the temperature favours the dissociation of NH4HS therefore more NH3 will be formed. 74. Answer (2. 72. Aakash IIT-JEE . New Delhi-75 Ph. Plot No.Regd. Answer (1. Dwarka. 3) Electron deficient species are called lewis acid therefore BF3. 2. 75. 71. New Delhi-75 Ph. 4) N2(g) + 3H2(g) initial moles at equilibrium 3 3–x 9 9 – 3x 2NH3(g) 0 2x (51) Aakash IIT-JEE . Answer (1. Sector-11. Answer (2. 4) NO3– is a conjugate base of HNO3 which is strong acid HSO4– is a conjugate base of H2SO4 which is strong acid 78. 4) CH3COOH is weak acid its concentration of H+ ions is less than 10–6 M therefore pH > 6 CH3COOH + NaOH → CH3COONa + H2O initial m. Plot No. 2. 81. 4. Answer (1. Answer (3.52 for 75% red ⎛ 25 ⎞ 1 ⎟ = × 10 −5 = 5. 2) Hln Ka = [H+] = H+ + In– [H+ ][In − ] [HIn] K a [acid] [base] ⎡ 75 ⎤ [H+] = 10 −5 ⎢ ⎥ = 3 × 10 −5 ⎣ 25 ⎦ pH = 4. The aqueous solution of CH3COONH4 is generally neutral ∴ pH > 6.Regd. mol after reaction 2 0 6 4 0 2 0 2 Since solution is basic therefore pH > 7. 82. 77. Dwarka. 3. AgNO3 and CaCl2 solution.47 for 75% blue [H+] = 10 −5 ⎜ ⎝ 75 ⎠ 3 80. 4) K ΔH ⎡ 1 1⎤ log 2 = ⎢ − ⎥ K1 2. Answer (1.: 45543147/8 Fax : 25084124 . 4) Due to common ion effect solubility of AgCl will be less than water in NaCl. 3. 2. Answer (1. Office : Aakash Tower.Physical Chemistry Success Magnet (Solutions) 76. 3) On increasing the temperature ionic product of water increases so pH and pOH decreases but water will remain neutral. 79. Answer (2.303R ⎣ T1 T2 ⎦ Since in the option (3) ∆H = 0 because (Eaf = Eab) ∴ Therefore only this reaction is independent of temperature and (K2 = K1) on the other hand there is not ∆H = 0 therefore equilibrium constant depends on temperature. 84.Regd. 3 2 W (N2 ) (3 − x )28 ⎛ 14 ⎞ = = ⎜ ⎟ remain same at t = t1 as well as t = t1 .: 45543147/8 Fax : 25084124 (52) . 3) Due to smaller size of Li+ is more solvated than Na+ ion therefore conductivity is less than Na+ ion. W (H2 ) ( 9 − 3 x )2 ⎝ 3 ⎠ 2 3 83. New Delhi-75 Ph. 3) Ecell = E°cell – 0.e.Success Magnet (Solutions) Physical Chemistry t = t1 the reaction attains the equilibrium therefore the amount of NH3 remins constant after t1 time 2x = 2 x = 1 mole Moles of N2 = 2. t1 t1 and because initial molar ratio is 1 : 3. Sector-11.. Dwarka. So y ml should be less than x ml that required for phenolphthalein end point. 85. 3) Pb(s) (Pb2+)c(aq) (Pb2+)A(aq) Pb(s) for this cell E°cell = 0 Ecell = − 0. Answer (2.059 [Cd2+ ] log 2 [Cu2+ ] on increasing the concentration of [Cu2+] and decreasing the concentration of [Cd2+] Ecell will be more positive. Answer (1. Plot No. 4.059 ⎦ ⎣ log = 1 2 [K sp (PbSO 4 ] 2 1 86. 2. In next step NaHCO3 coming from Na2CO3 neutralised by HCl using methyl orange indicator. Answer (2. 4) Upto phenolphthalein NaOH is fully neutralised and Na2CO3 will be converted to NaHCO3.059 [Pb 2 + ] A log 2 [Pb 2+ ]C Ksp = [Pb2+][SO42–] = s2 s= K sp (PbSO4 ) Ksp = [Pb2+][I–]2 = 4s3 ⎤3 ⎡ K sp (PbI2 )⎥ s= ⎢ ⎦ ⎣ 4 1 Ecell ⎡ K sp (PbI2 ) ⎤ 3 ⎥ ⎢ 4 0. Answer (1. Office : Aakash Tower. moles of H2 = 6. moles of NH3 = 2 W(N2) + W(H2) + W(NH3) = 2 × 28 + 2 × 6 + 2 × 17 = 102 g at t = 2t1 Molar ratio of N2 and H2 same at two time i. Aakash IIT-JEE . 3) –2 O –1 +2 +1 +1 –1 +2 H — O — S — O — O — H total charge at s = +6 –2 O –2 –1 O +1 +1 +1 +2 +1 –1 O CrO5 –1 O total charge on Cr = +6. Answer (2. 89.. 3) Ecell = 0 − (10 −5 )2 0. Answer (1.e.Regd.2)3 0.1)2 6 (0. charge on cation and anion produced on the dissociation of the electrolyte in the solution. Answer (1. Office : Aakash Tower.5 −1 90.059 (0. 3) The value of the constant A for a given solvent and temperature depends on the type of electrolyte i. 2. –1 Cr O O 88.059 log = positive 2 (10 −3 )2 E°H+ / H = 0 2 3+ Cr ⎯ ⎯→ Cr (aq) + 3e Cu2+(aq) + 2e ⎯ ⎯→ Cu 3+ 2Cr + 3Cu2+ ⎯ ⎯→ 2Cr + 3Cu Ecell = E°cell – (Cr 3 + )2 0.059 3 log 2 6 = E°cell + 91. Answer (1.059 log 6 (Cu2+ )3 = E°cell – 0. 2) 2Na2 S 2 O3 + I2 ⎯ ⎯→ Na2 S 4 O6 + 2N aI n =1 n=2 n=2 +2 0 +2. Dwarka. Plot No. Aakash IIT-JEE . 4. Answer (1. 2) Molar conductance of an electrolyte depends upon its degree of dissociation with increase in dilution the molar conductance increases due to increase in dissociation specific conductance decreases upon dilution because number of current carrying ions per unit volume of solution decreases.: 45543147/8 Fax : 25084124 (53) . New Delhi-75 Ph. Sector-11.Physical Chemistry Success Magnet (Solutions) 87. 3) Mole of Fe3+ = 0. Ag+. New Delhi-75 Ph.100 mol electron required to reduce all the Fe3+ to Fe2+ 0. 2. Au3+ these ions will be discharged ahead of H+. 95.1 × 1 = 0. 3) Cu S ⎯ ⎯→ Cu+ SO 2 n= 4 +2 −2 0 +4 O O H – O – S – O – O – S – OH O O (peroxy linkage) equivalent mass = M 4 M 4 n= 4 O2 ⎯ ⎯→ H2 O 0 −2 equivalent mass = Aakash IIT-JEE . 3) E A / A n+ = E° A / A n+ − 2. Answer (1.1 Mole of electron = Fe3+ + e → Fe2+ 0. Office : Aakash Tower. Plot No.: 45543147/8 Fax : 25084124 (54) . If metal ion with negative potential is coupled H-electrode the hydrogen half cell should function as cathode thus metal electrode will be negative half cell (anode). 4) As cell proceeds Ecell tend to zero to attain equilibrium state reaction quotient also increases to reach the state of equilibrium. 2. 3) If a given metal ion has negative reduction potential H+ will be reduced by metal. Na+ and Mg2+ the H+ will get preferentially reduced while in aqueous solution of Cu2+. 4. 2.049 = 0.Regd.Success Magnet (Solutions) Physical Chemistry 92. 94.049 mol electron to reduce the Fe2+ to Fe Fe2+ + 2e → Fe Mole of iron formed = 96. Answer (1.303 RT log[ A n + ] from this equation it is clear that nF E A / A n + decreases with increasing [An+] E A n + / A = E° A n + / A − 2. Answer (1. Answer (2. Similarly if reduction potential is positive metal will be reduced by H2.149 mole 96500 1 × 0. In aqueous solution containing Zn2+. Answer (1. Dwarka. 2) 3600 × 4 = 0. Answer (1. Sector-11.303 RT 1 log nF [ A n+ ] 93.025 mole Fe 2 O H–O–S–O–O–H O (peroxy linkage) 97. Answer (1. New Delhi-75 Ph. 2 edge atoms. formula of compound is A 1 BC 4 2 1 2 or AB4 C10 100. 101. Answer (3.5% tetrahedral voids. formula of compound is A 1 C 5 or AC10 4 2 Possibility II : Suppose 2 face atoms are not removed. Aakash IIT-JEE . Answer (1. On placing body diagonal plane. 4) Original Formula 1 1 ×4= 2 8 Number of A atoms = Number of B atoms = 1 × 2 =1 2 1 × 12 + 1 = 4 4 Number of C atoms = Hence. Sector-11. 2. 4) In spinel structure (MgAl2O4). 1 1 1 – 2× = 2 8 4 Number of A atoms = Number of B atoms = 1 – 1 = 0 ⎛1 ⎞ 1 Number of C atoms = 4 – ⎜ × 2 + 1⎟ = 2 ⎝4 ⎠ 2 Hence. two Al3+ occupy 50% octahedral voids and one Mg2+ occupies 12. Dwarka. 1 body atoms are removed but 2 face atoms may or may not be removed. Office : Aakash Tower.Regd. 4) Frenkel defect is shown by those ionic solids in which the difference in the size of cation and anion is very large. 2 face atoms are removed. 4. 2 corner atoms. Plot No. 2. the formula of compound is A1/2 BC4 or AB2 C8.: 45543147/8 Fax : 25084124 (55) . 3.Physical Chemistry Success Magnet (Solutions) 98. O2– ions occupy ccp lattice. Answer (1. 1 1 1 – ×2= 2 8 4 Number of A atoms = Number of B atoms = 1 – 0 = 1 1 ⎛1 ⎞ Number of C atoms = 4 – ⎜ × 2 + 1⎟ = 2 4 2 ⎝ ⎠ Hence. Possibility I : Suppose. 4) Factual type 99. 4) Square close packing coordination number = 4 Hexagonal close packing coordination number = 6 (56) Aakash IIT-JEE . 103. IInd Combination : Two face Cl– ions are removed Number of Na+ ions = 4 Number of Cl– ions = 4 – 1 ×2 = 3 2 Hence. 4) a= Body diagonal touches corner & body centre.69 + 1. formula per unit cell is Na4Cl3 IIIrd Combination : One corner and one face Cl– ions are removed.375. 4) The given plane represents face plane in fcc. coordinate no. Cl– ions are removed.: 45543147/8 Fax : 25084124 . Plot No. 4) Fluoride structure (CaF2) has cation constituting ccp whereas anions are present in all tetrahedral voids whereas for antifluorite structure anion constitute lattice of cation are present at tetrahedral voids. 108.Success Magnet (Solutions) Physical Chemistry 102. 3) r+ 1 . Answer (3. 2. Number of Na+ ions = 4 Number of Cl– ions = 4 – 1 1 × 1 + × 1 = 3. Answer (1. 4) In F-centre. Number of Na+ ions = 4 Number of Cl– ions = 4 – 2 × 1 = 3. 105. = 8 r− 1 . Answer (1. 2. formula per unit cell is Na4Cl3. Ist Combination : Two corner Cl– ions are removed. Dwarka. 86 i. Office : Aakash Tower.732 104. 69 = = 0 . formula per unit cell is Na4Cl3. 106.75 8 Hence. New Delhi-75 Ph.95) 1.Regd. 3) Octahedral voids form at the edge centre as well as the body centre at fcc. Answer (1. 95 ∴ bcc structure ∴ ∴ 2 (r+ + r− ) = 3a 2 (1.375 8 2 Hence. Answer (1. 107.75. Answer (1. 4. Answer (1.e. Sector-11. 116. New Delhi-75 Ph. 110. Answer (1. 3) Doping of group 14 elements with group 15 elements produces excess of electrons and doping of group 14 elements with group 13 elements produces holes in the crystals. n/p ratio decreases. So.303 Comparing with y = mx + C then we get (1) and (4) t½ does not depend on concentration for first order reaction. Plot No. 123. 2) A is called pre exponential factor. temperature. 3. 4) Maxwell and Ostwald theories are exclusively related to chemical kinetics. (2) and (3) options. Answer (2. 4) log [R0 ] k·t = [R] 2. nature of catalyst and surface area of reactants. 113. 4) Effective molarity of Ba3(PO4)2 is more than MgSO4. 115. Answer (3. Answer (1.r. Answer (1. 4) In α-decay.Regd. Answer (1. SPM allow the movement of solvent only not the solute. 2.: 45543147/8 Fax : 25084124 (57) . Dwarka. 4) Rate of reaction depends on nature of reactants. Answer (1. the osmotic pressure of Ba3(PO4)2 is more. Answer (1. no ppt. 3) Zn2+ is present in alternate tetrahedral voids therefore its C. boiling point of solutions is same. 3) Statement 4 is incorrect because of B decreases then C increases hence there must be a –ve sign. 4) In all options. 114. 120. 2. 121. Answer (1.t. Answer (1. Answer (1. 2. overall order of reaction is two & first order w. In rock salt structure there is 4Na+ and 4Cl– ions present in a unit cell. Sector-11. 3. 122. 4) t½ ∝ 1 (n = order of reaction) and the unit of frequency factor ‘A’ is equal to the unit of ‘k’’ (a)n−1 118. 2. 2. 3. Answer (1. effective molarity are same. 3) A has units of rate constant of reaction not rate of reaction. So. Answer (2. 3.2 M.e. 2. 2. positron emission & k-electron capture. Answer (1. 124. 4. of BaSO4 is formed in right side. Office : Aakash Tower. [I–] 119. 112. 3) Statement 4 is incorrect because K = A e–Ea/RT. 111. Aakash IIT-JEE .N is 4. 2. 4) From the rate expression. So. 2. n/p ratio increases while in β-decay.Physical Chemistry Success Magnet (Solutions) 109. effective molarity are same i. 117. 4) The alkaline hydrolysis of ethyl acetate is second order while others are first order reaction. Answer (1. Answer (1. 3) In (1). Size of true particle < 10–7 cm in diameter. the colloidal system is called aerosol. Answer (2. Answer (2. Answer (1. 129. New Delhi-75 Ph. 3) Acetone and chloroform shows negative deviation from Raoult’s law while ethanol and water shows positive deviation from Raoult’s law. Sector-11. CuSO4 + 4NH4OH → [Cu(NH3)4]SO4 Hence. 133. greater is the penetrating power. Answer (1. Dwarka. 2) When dispersion medium is gas. 4) Statement 1 and 2 are incorrect because sol particles neither move toward anode nor cathode. dust are example of aerosols of solids whereas fog.: 45543147/8 Fax : 25084124 Fe(s) Pt(s) 2NH3 (Haber’s process) 2SO3 (Contact process). 127. Sulphur in water is an example of ryophobic sol. the physical state of reactants and catalyst are same N2(g) + 3H2(g) 2SO2(g) + O2(g) 135. (58) .Success Magnet (Solutions) Physical Chemistry 125. Answer (1. 128. 134. Sulphur in water is an example of lyophobic sol. 126. 2) When CuSO4 is dissolved in NH4OH then association takes place instead of dissociation. electrophoresis is the property of colloids. 4) Chemisorption is specific in nature and it is shown by the gases which can react with adsorbent. 131.Regd. 3) Size of suspension particles are > 10–5 cm in diameter. Chemisorption is unimolecular not multimolecular and favourable at high temperature not at low temperature. 3. Plot No. 3) An anion caused the precipitation of a positively charged sol and vice versa. 4. Answer (1. 3) Scattering of light can’t done by water and CaCl2 is more effective coagulant than NaCl because As2S3 is negatively charged sol. Answer (3. smoke. Answer (2. Office : Aakash Tower. 4) Starch. 132. Answer (2. clouds are example of aerosol of liquids. Answer (1. 4) Peptization is the preparation method of colloids. Aakash IIT-JEE . Answer (2. 3) In homogeneous catalysis. Size of colloidal particles are 10–7 –10–5 cm in diameter. freezing point of solution is raised and boiling point of solution is lowered. The higher the valency of the effective ion. While ultrafilteration and electrodialysis are the purification method of colloids. 130. gum and protein in water are examples of lyophilic sols. Dwarka.003 M 1000 C3.5 4 = 23. Sector-11. of neutrons.11 + 18 106 gm water contains = 300 gm CaCO3 ∴ Molarity = = 300 / 100 mol / L 1000 3 = 0. Y2– and Z3– are isoelectonic therefore Number of electrons in increasing order will be X > Y > Z X–. Answer (3) 11.11 + 55.: 45543147/8 Fax : 25084124 (59) .2 100 10X + 1100 – 11X = 1020 X = 80 % of B–10 = 80 2. Office : Aakash Tower. 1. Answer (1) Δx ⋅ Δp = Δx = h 4π (∆x = ∆p) h 4π Aakash IIT-JEE .167 1000 11. increasing order will be Z<Y<X C2. Answer (2) Let the % of B-10 = X then % of B-11 = (100 – X) Average atomic mass = 10 X + (100 − X )11 = 10.55 11. Plot No.Regd.07 2. Mass percentage of water = 18 × 100 18 + 60 = 1800 78 3 × 35 + 1× 37 = 35. Y2– and Z3– all have same no. Answer (1) 1 kg water has 11. 4. Answer (3) Average atomic mass = 3. New Delhi-75 Ph. 1. Answer (2) Since X–.11 moles of solute mole fraction of solute = 3.11 11.C : Linked Comprehension C1.11 = = 0. Therefore atomic no.Physical Chemistry Success Magnet (Solutions) Section . Answer (2) Let the moles of water = 1 mole Moles of urea will also be 1. 1. of electrons = 2n2 = 2 × 42 = 32 Aakash IIT-JEE . Answer (2) ∆x = h 4 πm Δ v 6. Answer (2) ⎡1 1⎤ 13. Sector-11.14 × 10 −3 × 2 × 10 −2 = = 2. Answer (1) λ = v (given) λ = v = v2 = h mV h mv h m v= h m C4.1 = 8 × 1012 m/s 2.92 n2 = 0.14 9.64 × 10–30 m 3.6Z 2 ⎢ − ⎥ ⎣16 9 ⎦ = 2.529 × n2 Z 16.626 × 10 −34 4 × 3.: 45543147/8 Fax : 25084124 (60) .Regd. Plot No. New Delhi-75 Ph. 4. Office : Aakash Tower.6⎢ − ⎥ Z 2 E3 − E2 ⎣9 4 ⎦ = ⎡ 1 1⎤ E 4 − E3 13. 1. Dwarka. Answer (2) 5 144 20 × = 36 7 7 rn = 0.Success Magnet (Solutions) Physical Chemistry Δx ⋅ m Δv = Δv = h 4π h 4πm ⋅ Δx = = h 4π × 4πm h h 1 × 4π m = 6.626 × 10 −34 10 31 × 4 × 3.529 × 2 1 n=4 Maximum no. New Delhi-75 Ph. C5. 4. So the wavelength will be maximum. Answer (2) For the maximum wavelength energy should be minimum. Sector-11.529 Å 3. l = 1 Radial nodes = 3 – 0 – 1 = 2 Radial nodes = 2 – 1 – 1 = 0 Z2 n2 1 2 2 = −13 . 3s2. 1. Energy emission is minimum in this transition. Answer (1) For the first excited state n = 2 Energy = − 13. For the Balmar series the electron should jumps from higher level to 2nd energy level.529 Å rBe 3 + = 0. 3p6. Answer (1) Radial nodes = (n – l – 1) For 3s . If it emit a photon it will drop into nucleus. Office : Aakash Tower.529 n2 1 = 0.: 45543147/8 Fax : 25084124 (61) .4 eV 2.Regd. l = 0 n = 2. Answer (2) Energy electron for H-atom depends only on the value of n not the value of l. 3d5 3d 5 1 5 Total spin = 5 × = 2 2 2. Therefore 4s > 3d = 3p = 3s 3. Answer (4) Spin quantum no. that is not possible. Answer (3) Mn2+ – 1s2.6 eV 4 = –3. 2p6. Dwarka. Plot No. 2.Physical Chemistry Success Magnet (Solutions) 3.6 × = −13 .6 × n = 3. In the option (2) there is no shell change. 3. For 2p . C7.529 × = 0. Answer (4) 22 4 1s electronic level allow the H-atom to absorb a photon but not to emit a photon. Answer (4) For the visible region transition belongs to Balmar series.529 × 1 Å = 0. 1. Answer (3) Maximum number of electrons in any subshell having same value of spin quantum number is (2l + 1) C6. Answer (4) rH = 0 . Aakash IIT-JEE . is not derived from Schrondiger wave equation. 1. 2s2. Success Magnet (Solutions) Physical Chemistry C8. 1. Answer (1) N(7) – 1s2, 2s2, 2p3 – Half filled (more stable) O(8) – 1s2, 2s2, 2p4 – Less stable ∴ Electron Affinity of Nitrogen is less than Oxygen. 2. Answer (4) Electron Affinity of Br is less than Chlorine. 3. Answer (1) Na(g) → Na+ (g) + e Na+(g) + e → Na ∆H = ionisation energy ∆H = Electron affinity …(1) …(2) Since (1) and (2) process are opposite therefore ionisation energy of Na is equal to electron affinity of Na+. C9. 1. Answer (4) In He+ ions the electrons are tightly held up by the nucleus therefore its ionisation energy is more than He. 2. Answer (1) Be (4) → 1s2, 2s2 – full filled (more stable) B (5) → 1s2, 2s2, 2p1 – Less stable ∴ Be has more first I.E. than I.E. of B. 3. Answer (1) Due to half filled P-orbital nitrogen electronic configuration is more stable. Therefore its ionisation energy is more. C10. 1. Answer (2) Since IE of element B is less therefore it is most reactive metal amongst given elements. 2. Answer (4) Element D has high IE but less than IE of A. Therefore (D) is non metal. 3. Answer (1) The first ionisation potential is highest for element A therefore A is noble gas. C11. 1. Answer (2) F B—F F μ=0 .. N H H H Addition .. N F F F Subtraction ∴ BF3 < NF3 < NH3. 2. Answer (3) Cl Cl Cl μ=0 3. Answer (2) % ionic character = 16 (∆EN) + 3.5 (∆EN)2 = 16(0.2) + 3.5(0.2)2 = 3.34 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (62) Physical Chemistry Success Magnet (Solutions) C12. 1. Answer (2) .. .. .N = O . . .. N .. .. .. O ..O 2. Answer (2) O – O—C O – 1 Bond order = 1 + = 1.33 3 3. Answer (3) CO molecule Total no. of electrons = 14. ⎡π2p 2 ⎤ y Arrangement – σ1s2 σ*1s2 σ2s2 σ* 2s2 ⎢ σ2px2. 2⎥ π 2 p ⎢ ⎥ z⎦ ⎣ B.O. = 1 (10 – 4) = 3. 2 In CO+ ion, one electron is released from antibonding molecular orbital. ∴ B.O. = 1 [10 – 3] = 3.5. 2 C13. 1. Answer (2) 2 2 2 μ R = μ1 + μ1 + 2μ1 cos 104.5º 2. Answer (3) µ=q×d 1.03 × 3.33 × 10–30 = charge × 1.27 × 10–10 Charge = 2.7 × 10–20 Percentage ionic character = 2.7 × 10 −20 × 100 = 16.87% 1.6 × 10 −19 ~ 17% 3. Answer (1) O=C=O + C14. 1. Answer (4) µ=0 2⎤ ⎡π2 p y 2 2 2 2 2 − ⎢ ⎥ → 1 s * 1 s 2 s * 2 s 2 p σ σ σ σ σ O2 x 2 2 ⎢ ⎣π2 pz ⎥ ⎦ 2⎤ ⎡ π * 2 py ⎢ ⎥ 2 ⎢ ⎣ π * 2 pz ⎥ ⎦ There is no unpaired electrons ∴ paramagnetism is not shown. 2. Answer (2) Bond order of N2 is 3. Bond order of O2 is 2. Bond order of F2 and Cl2 are 1. ∴ 1 Bond length ∝ Bond order Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (63) Success Magnet (Solutions) Physical Chemistry 3. Answer (1) In NO molecule there are total no. of 15 electrons. ⎡π2p 2 ⎤ y 1 Arrangement - σ1s2 σ* 1s2 σ2s2 σ*2s2 σ2px2 ⎢ 2 ⎥ π* 2py . π 2 p ⎢ ⎥ z⎦ ⎣ ∴ Unpaired electrons = 1. C15. 1. Answer (3) rH2 rmix 4= = Mmix MH2 Mmix 2 16 = Mmix 2 Mmix = 32 Let the molar ratio of C2H4 and CO2 is a : b. Mmix = 28a + 44b = 32 a+b 32a + 32b = 28a + 44b 4a = 12b a 3 = = 3 :1 b 1 2. Answer (2) rN2 rHe = 2 4 = 0.252 3 28 Let the moles of He is coming out to be x Moles of N2 coming out is 0.252x nN2 n total = 0.252x = 0.2 1.252x nHe = 0.8 n total Mmix = 0.2 × 28 + 0.8 × 4 = 8.8 rO2 rmix = 8.8 = 0.52 32 3. Answer (1) Cl2(g) Initial moles At equilibrium 1 1–x 2Cl(g) 0 2x Total no. of moles at equilibrium = 1 – x + 2x = 1 + x = 1 + 0.14 = 1.14 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (64) Physical Chemistry Success Magnet (Solutions) The composition of gas mixture is coming out would be rCl2 rCl = 0. Answer (3) Dalton’s law of partial pressure is not valid for reacting gases NH3 (g) + HCl(g) ⎯ ⎯→ NH4 Cl(g) ↑ white fumes 3.32 Mmix.4 = = 0.8 = 1.68 3.64 rmix MKr = = rkr Mmix 83.5) = 59. Coming out would be 2. Office : Aakash Tower. 1. 4.2 Mole fraction of O3 = volume of O 3 0.8 Total volume of gaseous mixture = 1 – 3x + 2x = 0.Regd.68 × 71) + (0.8 Aakash IIT-JEE .5 = 2.86 35.64 C16.18 59. Plot No. New Delhi-75 Ph. = (0.28 71 Let the moles of Cl atom coming out be x. Answer (4) Volume percent of H2 = = Pressure of H2 × 100 Total pressure of mixture 150 × 100 150 + 300 + 200 + 100 = 150 × 100 = 20% 750 C17. 1.17 0.32 × 35. Answer (2) Partial pressure of hydrogen = w 2 w w + 2 16 × Ptotal w 2 = × Ptotal 8w + w 16 = 8w × Ptotal 9w ⎛8⎞ = ⎜ ⎟ × Ptotal ⎝9⎠ 2.17 x XCl = 0.17 x X Cl2 = 2.5 Total volume 0. then moles of Cl2.: 45543147/8 Fax : 25084124 (65) . Sector-11.17 x = 0. Dwarka. Answer (1) 3O2(g) Initial volume After reaction 1 litre 1 – 3x 2O3(g) 0 2x x = 0. Regd. Dwarka. Answer (2) PV = K(constant) 2. Answer (2) turpentine oil (O 3 + O 2 ) ⎯⎯ ⎯ ⎯⎯ ⎯→ O 2 100 mL 40 mL Therefore volume of O3 is 60 mL 2O3 2 mL 60 mL 3O2 3 mL 90 mL = 130 mL Change (increase) in volume = 130 – 100 = 30 mL 3.: 45543147/8 Fax : 25084124 (66) . Answer (4) At point A near low pressure region volume is very high thus (V – b) ~ V a ⎞ ⎛ ⎜ p + 2 ⎟ V = RT V ⎠ ⎝ Aakash IIT-JEE . Answer (2) M(average) = 16x + 28y = 20x + 20y 4x = 8y Maverage = = 16 y + 28 x x+y 16 y + 28(2y ) 72 = = 24 3y 3 C19. Office : Aakash Tower. Sector-11. Answer (1) Let the volume of NH3 be x mL Volume of H2 = (50 – x) mL Total volume now = 40 + 90 1 3 NH3 ⎯ ⎯→ N2 + H2 2 2 Since 40 mL of O2 is added and sparked it must have reacted with H2 to form liquid water. New Delhi-75 Ph. 1. Answer (4) PV = nRT n is not directly proportional to T ∴ Its graph will not be straight line. Moreover since 6 mL contraction occurs with alkaline pyrogallol. 34 mL is the volume of O2 is used up. Total volume of H2 is 68 (Q 2H2 + O2 → 2H2O) (50 − x ) + 3 x = 68 2 x = 36 % NH3 = 72 C18.Success Magnet (Solutions) Physical Chemistry 2. 1. 4. Plot No. 16 x + 28 y = 20 x+y 3. 4. Answer (1) Z= a ⎞ ⎛ ⎜ P + 2 ⎟( V − b) = RT V ⎠ ⎝ PV + a ab − − Pb = RT V V2 PV = RT + Pb − = RT + a V RTb a − V V ⎡ a ⎞⎤ 1⎛ = RT ⎢1 + ⎜ b − ⎟⎥ RT ⎠⎦ ⎣ V⎝ Comparing to equation of state B ⎛ ⎞ PV = R + ⎜1 + + .: 45543147/8 Fax : 25084124 1L atm = 0. ⎟ V ⎝ ⎠ a ⎞ ⎛ B = ⎜b − ⎟ RT ⎝ ⎠ 19(a).. Answer (3) At point B at high pressure volume is low a ⎞ ⎛ ⎜P + 2 ⎟ ~ p V ⎠ ⎝ P(V – b) = RT PV Pb = 1+ RT RT 3.1 kJ (67) .. therefore volume is constant –56 = ∆U + 1(40 – 70) × 0.Physical Chemistry Success Magnet (Solutions) PV + a = RT V PV a = Z = 1− RT RTV 2. Real gas exert lower pressure than the same gas behaving ideally due to intermolecular force of attraction. so in Van der Waal’s equation ⎜ P + 2 ⎟ (V – b) = RT. Coefficients depends on the identity of the gas but are independent of the temperature. Office : Aakash Tower. New Delhi-75 Ph. 3. Plot No. 4) (IIT-JEE 2008) a ⎞ ⎛ a Because V is very large. Answer (3) ∆H = ∆U + ∆(PV) = ∆U + V(P2 – P1). C20. 2 and b are V ⎠ ⎝ V neglected and equation becomes PV = RT.1 ∆U = –56 + 3 = – 53 kJ/mol Aakash IIT-JEE . Answer (1..Regd. 1. Sector-11... Dwarka. 303 nRT log V 1 (IIT-JEE 2008) = − 2. of moles of ice needed to melt to absorb this heat = 3. Answer (2) Entropy change in isothermal process. Answer (2) ΔH ∆S = T b Tb = 30 × 1000 = 400 K 75 C21(a).314 log = 38.303 × 2 × 8.2 × 500 × 20 = 42000 J =7 6 × 103 Since each ice cube contains one mole water so at least 7 ice cube are required. Dwarka.Success Magnet (Solutions) Physical Chemistry 2. Answer (4) For spontaneous process ∆G should be negative. ∆S = 2.303 × 30 × 0.: 45543147/8 Fax : 25084124 (68) . New Delhi-75 Ph. ∴ ∆H < 0 and ∆S > 0 2.303 nR log V2 V1 10 1 42000 since process is isothermal ∆T = 0 = 2. Plot No.45 atm × litre Aakash IIT-JEE . Sector-11. 1. 4. =0 C21. Answer (3) Enthalpy change ∆H = nCp∆T. Office : Aakash Tower.5 = 34.54 = –25. No.Regd. Answer (4) Work done in AB process = –P∆V W1 = 3 (30 – 10) = –60 atm×litre Work done in B → C process = 0 Work done in C → A process V2 W2 = –2.29 JK–1 mol–1 3. Answer (4) ∆G = ∆G° + RT ln Q at equilibrium ∆G = 0 Q = Keq ∴ ∆G° = –RT ln Keq C22.54 atm × litre Net work done in the process = –60 + 34. 1.303(1× 30 ) log 10 30 = +2. Answer (2) Amount of heat required to lower the temperature of 500 g of water from 20º to 0ºC = 4. Sector-11.1739 = 4. Plot No.Regd.1 = 4. Answer (2) ∆Q = nS∆T Molar heat capacity = ΔQ nΔT Since in ice water equilibrium there is change in temperature is zero. 1. Answer (2) Since heat of ionisation for HCN is more than CH3COOH. Therefore Ka(HCN) will be less than Ka(CH3COOH) ∴ pK a (HCN) > pK a (CH3 COOH) 2.8) = 0 ΔHf (Na+ ) = −241. 4. C23.74 + log 0.Physical Chemistry Success Magnet (Solutions) 2.9 kJ 3. Answer (3) Na+(aq) + OH– (aq) → NaOH (aq) Heat of ionisation of strong base = 0 ΔH = ΔHf (NaOH) − ΔHf (Na + ) − ΔHf (OH− ) = 0 ⇒ − 470. Office : Aakash Tower. Answer (1) For acidic buffer pH [ salt ] = pKa + log [acid] ⎡ 2 1000 ⎤ ⎢ 59 × 500 ⎥ ⎣ ⎦ = 4. Answer (1) At point A → PV = nRT T= 3 × 10 = 375 K 0. 1. Answer (2) pV 30 = = 375K nR 1× 0. It means HCN is weaker acid than acetic acid.57 Aakash IIT-JEE .7 − ΔHf (Na+ ) − ( −228.08 Since process is cyclic therefore ∆H and ∆U will be zero because in cyclic process state functions will be zero. New Delhi-75 Ph. Dwarka.74 + log 40 59 = 4.08 × 1 Temperature at point C = T = 3. ∴ Molar heat capacity = ∞ C24.74 – 0.: 45543147/8 Fax : 25084124 (69) . 12 × 10–3 Kf = 6. (70) Aakash IIT-JEE .3.01) = 1. Answer (4) [OH–] = 2 × 10–7 pOH = – log (2×10–7) = 7 – log 2 = 6. Answer (1) CH3COOH 1–α Ka = CH3COO– α + H+ α + 0. Answer (3) pH = pKa + log [salt] = [acid] pH = pKa = 4.74 + log [acid] [ salt ] [acid] 6.74 = 4. Sector-11.16 ×10–3 × 57 = 66.74. 2. 3. Plot No. 3.8 × 10 – 5 1− α on solving we get pH = 1. Answer (4) Kf ⎞ ⎛ Since reaction is endothermic on increasing the temperature equilibrium constant increases ⎜ K c = K ⎟ r ⎠ ⎝ but Kf increases more than Kr. Dwarka.01 α( α + 0.74 + log [ salt ] [acid] [ salt ] 100 = [acid] 1 % of acid in the mixture = 1 × 100 101 =1% ∴ % dissociation of acid = 99% C25. Answer (3) Keq Kf = K r Kf = Keq × Kr = 1.937. 1. Answer (3) [ salt ] pH = 4. 1. Answer (1) Kf > Kr.7 pH = 14 – 6.612 × 10–2 2.Success Magnet (Solutions) Physical Chemistry 2. C26. New Delhi-75 Ph.Regd.: 45543147/8 Fax : 25084124 . Office : Aakash Tower. 4.7 = 7. Dwarka. ∆H = – 136. Office : Aakash Tower.01 MNaNO2 < 0. 3.01 MH2SO4 3. Plot No. Sector-11. 1.Regd. of moles decreases in forward direction therefore on increasing the pressure forward reaction favours. New Delhi-75 Ph. 2.01 MNaCl < 0.Physical Chemistry Success Magnet (Solutions) 3. Aakash IIT-JEE .018 × 10–14 = 18 × 10–17. ⎝ ⎠ C28.01 MH2S < 0. Answer (3) There is no effect at equilibrium at constant volume.55 = = 0.8 Since reaction is exothermic therefore decrease in temperature favour forward reaction. Since no.moles (initial) After reaction 10 6 4 0 buffer solution CH3COONa + H2O 0 4 0 4 [ salt ] pH = pKa + log [acid] pH – pKa = log 4 6 ⎛ 2⎞ = log ⎜ 3 ⎟ . C27. Answer (3) Kw Ka = [H+] [H–] = 10–14 [H+ ][OH − ] 10 −14 = [H2O] 55. Answer (2) C2H4(g) + H2(g) C2H6(g). 4. Answer (2) CH3COOH + NaOH m. 1.: 45543147/8 Fax : 25084124 (71) . Answer (4) CH3COONa → CH3COO– + Na+ CH3COO– + H2O 2. Answer (4) NaNO3(s) NaNO2(s) + 1 O (g) 2 2 ∴ since no. Answer (3) H2S NaCl NaNO2 H2SO4 H+ + HS solution is acidic [H+] > 10–7 m → → → solution is neutral [H+] = 10–7 solution is basic [H+] < 10–7 solution is strong acidic CH3COOH + OH– (basic). of gaseous moles decrese in backward direction thererfore on increasing the presure reverse reaction favour. 0. 059 log 2 (PH2 ) A 0.059 log Zn2+ = E° cell − 2 2 H+ 0.059 log 2 H+ [ ] [ ] 2 A 2 C ⎛ C1 ⎞ = – 0.12 = log⎛ C1 ⎞ ⇒ log⎛ C1 ⎞ = −36 ⎜ ⎟ ⎜ ⎟ .Success Magnet (Solutions) Physical Chemistry C29. 0.059 × 0. Answer (1) Zn → Zn2+ + 2e 2H+ + 2e → H2 Ecell 0.28 = – 0. 4. 1. Sector-11.90 mV 2 2.78 – (KC ) a 2 C1 2 – 2.059 log⎛ 1 ⎞ = – ⎜ ⎟ 2 ⎝2⎠ = + 0.059 log 2 [ ] [ ] 0.059 log ⎜ C ⎟ ⎝ 2⎠ for non spontaneous process C1 > C2.Regd. 3.: 45543147/8 Fax : 25084124 (72) . Answer (1) H2(g) → 2H+ + 2e (anode) 2H+ + 2e → H2 (cathode) Ecell = – (PH2 )C 0. Office : Aakash Tower.059 ⎝ K aC2 ⎠ ⎝ K aC2 ⎠ C1 = 10−36 K aC2 Ka = C1 × 1036 C2 Aakash IIT-JEE . Answer (1) H2 → (2H+)A + 2e (2H+)C + 2e → H2 Ecell H+ = 0 − 0. Dwarka.3010 = 8. Plot No.059 log 2 2 = 0. New Delhi-75 Ph. Answer (1) Λm = 1000 × K M 27.9.Regd. Aakash IIT-JEE .001028 = 48. Office : Aakash Tower.95 × 10 –5 N 0. 2. Specific conductance is directly proportional to concentration.1 = Λ 3.9 × 0.1 1000 Λ 279 = 27. K = = 2.5 ⎟ ⎠ ⎝Λ ⎠ ⎝ = 0. New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (73) . Answer (3) 1 ⎛l⎞ K = R ×⎜a⎟ ⎝ ⎠ 1 ⎛l⎞ 1. it decreases 8 times. Answer (4) On doubling the edge length.15 ⎞ degree of dissociation = ⎜ ∞ ⎟ = ⎜ 390. 3. 1.Physical Chemistry Success Magnet (Solutions) C30. volume of cube becomes 8 times and the concentration of solution decreases 8 times. C31. Hence.79 × 10–3. Plot No. Sector-11. 1.15 ⎛ Λ ⎞ ⎛ 48.2 × 55 = 66 m −1 ⎝a⎠ or 66 × 10–2 cm–1. Answer (3) Λ= 1000 × K = 1000 × 4. 4. Answer (3) Λ∞ m = 119 + 160 = 279 α Λ = Λ∞ m degree of dissociation 0.2 = 55 × ⎜ a ⎟ ⎝ ⎠ ⎛l⎞ ⎜ ⎟ = 1. Dwarka.123. Answer (2) Specific conductivity is directly proportional to the concentration. 2. 23 pm 0. of oh voids = 4 and td voids = 8 Out of 4 oh voids 3 are filled (atoms are present at edge centre) and 4 td voids are filled ∴ fraction occupied = 7 = 0.Success Magnet (Solutions) Physical Chemistry C32. Plot No.: 45543147/8 Fax : 25084124 (74) . Office : Aakash Tower. Answer (2) A 8× 1 8 B 1 1 2 × + 12 × 2 4 C 4× 1 +4 2 ⇒ AB4C6 2. formula of left compound = AB4C5 3.294 (i. the basic lattice is ccp ∴ No.21 2.Regd. formula of left compound = AB3C6 If C is removed. Answer (4) Q Anion is bigger Q it will constitute lattice r + 0. Answer (1) A at corner B at 2 face centre and C at rest 4 face centres i.e. range of coordination number = 4) r − 2. 4. Answer (1) Tetrad axis will remove 2 atoms at face centre and 1 of body centre (which is not present in this compound) 2 face centre atoms can be B or C. New Delhi-75 Ph.414 (for maximum packing efficiency) r− ∴ r− = 80 = 193.58 12 Aakash IIT-JEE . 1. Answer (2) oh voids in fcc is situated at all edge centres and body centre and td voids are present at each body diagonal. Answer (1) r+ = 0.e. ∴ distance = 3 a 4 td void √3 a 4 oh void body √3a diagonal td void C33. Nearest oh and td void pair is oh void at body centre and td void at edge centre.414 3. If B is removed.65 = = 0. 1. Dwarka. Sector-11. Dwarka. Answer (3) body diagonal plane (Bcc) Aakash IIT-JEE . 2. New Delhi-75 Ph. 1. 4. 1. Answer (2) V = (100)3 × 10–30 = 10–24 cm3 1 cm3 = 10 g ∴ ∴ 10–24 cm3 = 10–23 g 10–23 g = 2 atoms (Bcc) 100 g = 2 × 100 × 1023 g = 2×1025 atoms C35.48 3 ρbcc (2) 2 3. Answer (3) body diagonal plane of fcc.Physical Chemistry Success Magnet (Solutions) C34. Answer (1) ρfcc 4 (2.8)3 = × = 5.: 45543147/8 Fax : 25084124 (75) . Plot No. Answer (3) Density of CsCl Z = 1 (1 formula unit is present) For bcc structure 3a = 2(r+ + r− ) a= 2(r+ + r− ) 3 pm Volume = [2(r+ + r− )]3 ( 3) 3 × 10 −30 cm3 ∴ ρ= [2(r+ + r− )]3 × NA × 10 −30 MCsCl × ( 3 )3 2.Regd. Answer (1) Rectangular plane (fcc) 3. Office : Aakash Tower. Sector-11. : 45543147/8 Fax : 25084124 (76) . Dwarka.314 ⎢ ⎣ 700 × 800 ⎦ 2 (oh) + 1( td) × 100 = 25% 12 log 4 = − By solving. 4. p p Aakash IIT-JEE . Answer (2) % of voids occupied = C37.303RT C38. Answer (2) O2+ = 4 (ccp) Mg2+ = 1 (Q 1/8 of td voids) Al3+ = 2 (Q 1/2 of oh voids) ∴ formula = MgAl2O4 2. Answer (2) O2– replaced = 6 (face centred) Charge replaced = 6 × 2 = 12 (negative units) ∴ ∴ Y to be doped = 4 formula = MgAl2Y2O 3. 1. Office : Aakash Tower.Success Magnet (Solutions) Physical Chemistry C36. Answer (4) According to Arrhenious equation.303 × 8. ratio increases. Answer (4) 0 13 +1 e ⎯→ 13 7 N ⎯⎯ 6 C unstable stable n n ratio decreases while in K-electron capture. Sector-11. Ea = 64 kJ 3. Plot No. 2.303RT When Ea becomes zero then K = A. Answer (1) On increasing temperature. Answer (3) On β-emission. 2. 1. log K = log A − Ea 2. K increases but Ea and A remains same. α-emission and positron emission.Regd. 1. New Delhi-75 Ph. Answer (3) log K2 E a ⎡ T1 − T2 ⎤ =− ⎢ ⎥ K1 2.303R ⎣ T1 ⋅ T2 ⎦ Ea ⎡ 700 − 800 ⎤ ⎥ 2. 2. Answer (3) Rate ∝ [A]x 2 ∝ (4)x (4)1/2 ∝ (4)x x= 1 2 2. C39. 2. Answer (1) In the 1st experiment. of neutrons Elements with higher atomic number are more stable if they have slight excess of neutron as this increase the attractive force and also reduces repulsion between protons.: 45543147/8 Fax : 25084124 (77) . more is the radioactivity. p (IIT-JEE 2008) Z (At no. 2. Office : Aakash Tower. New Delhi-75 Ph. Answer (4) K= 2. 4.Physical Chemistry Success Magnet (Solutions) 3. Sector-11.Regd. Answer (3) Lesser is the half life. Answer (4) The order w.t. 1. Dwarka. Rate ∝ [A]x (3)3 = 27 ∝ (3)x x=3 In the IInd experiment. Plot No.e. C40. Answer (2) n ratio i.r. 1. Answer (1) Tritium (1T3) has highest 38(a).) 45o no. B is zero. Rate ∝ [A]x [B]y 8 ∝ (2)3 (2)y y=0 3.303 a log t (a − x ) Aakash IIT-JEE . 693 2.2. Dwarka.303 = log 69. Office : Aakash Tower.8 PT = 140 + 480 = 620 mm 2.Regd.303 log 60 10 …(1) K2 = 100 2. ntoluene = 78 92 Total moles = x x 170 x + = 78 92 78 × 92 x 78 = 92 xbenzene = 170 x 170 78 × 92 xtoluene = 1 − 92 78 = 170 170 o o PT = Pbenzene × x benzene + Ptoluene × x toluene PT = 700 × 92 78 + 600 × 170 170 PT = 378.82 + 275.: 45543147/8 Fax : 25084124 (78) . Sector-11. 1.303 100 2.303 log = log 10 60 t 10 t = 45 minutes C41.3 t 20 t = 161 minutes 3. x g benzene and x g toluene are mixed nbenzene = x x . New Delhi-75 Ph.Success Magnet (Solutions) Physical Chemistry 100 0.11 mm Aakash IIT-JEE . Plot No.2 + 600 × 0.8 5 5 According to Roult’s law o o PT = Pbenzene × x benzene + Ptoluene × x toluene PT = 700 × 0.29 PT = 654. xtoluene = = 0. Answer (2) Suppose. 4. Answer (1) K1 = 100 2. Answer (3) xbenzene = 1 4 = 0.303 log t 10 …(2) K1 = K2 100 2. 1 m = 60 = 500 60 500 1000 Pbenzene 140 = = 0. Office : Aakash Tower. Hence. Answer (2) In benzene.Physical Chemistry Success Magnet (Solutions) 3.Regd. 1. 3. Plot No.1 i = 0.5 It shows that acetic acid is 100% dimerised in benzene.6 × 0. Answer (4) The ratio of effective molarity of 0.5α = 0 . 2 M urea and 0.: 45543147/8 Fax : 25084124 . Hence molecular weight of acetic acid will be 120.225 PT 620 In beaker ‘A’ ∆Tb = i × Kb × m 0.8 1 2.6 = 2. the ratio of osmotic pressure is also 2 : 2 : 1.6) 1 1 + 3 × 0. Answer (3) K3[Fe(CN)6] 1 (1 – α) 3K+ + [Fe(CN)6]3– 0 3α 0 α before dissociation after dissociation i= i= 1 + 3α (Q α = 0.5 M AlCl3. 1. Sector-11.2 M K4[Fe(CN)6] is 2 : 2 : 1. acetic acid dimerises 100%.17 = i × 1.8 1 (79) Aakash IIT-JEE . Answer (2) 3 3 1000 × = 0. 2. Dwarka.7 × 0.1 i=1 It shows that acetic acid remains normal molecule in acetone. New Delhi-75 Ph. Answer (2) Mole fraction of benzene in vapour phase = C42. C43. In beaker B ∆Tb = i × Kb × m 0. Hence. 4. Answer (2) 2CH3COOH 1 (1 – α) (CH3COOH)2 0 α 2 before association after association Here. 3. molecular weight of acetic acid will be 60. acetic acid remains the normal molecule.13 = i × 2. Answer (1) In acetone. α = degree of dimerisation i= 1 − 0. Office : Aakash Tower.Success Magnet (Solutions) Physical Chemistry α= 0. x aP = m 1 + bP 1 b . O2–. Answer (2) According to Langmuir adsorption isotherm. Sector-11. Answer (2) In this reaction Br2 is oxidised as well as reduced.5 Percentage of dimerisation = 40%. 4. intercept = . 2.D : Assertion . C44. Answer (1) According to Langmuir adsorption isotherm.Reason Type 1.4 0.6 × normality 3. 2. log 1 x = log K + log P m n 1 .: 45543147/8 Fax : 25084124 (80) . a a At very low pressure. 1.2 = 0. Dwarka. n Comparing the above equation with y = mx + c then we get. Answer (3) As2S3 is negatively charged sol. m Section . Answer (3) Fe(OH)3 sol is positively charged. C45. m b (1) = + x a aP Comparing the above equation with y = mx + c then we get. F– all have 10 electrons therefore isoelectronic. Na+. Answer (3) According to Freundlich adsorption isotherm. Then we get x = aP . Answer (3) Mg2+. New Delhi-75 Ph. 3. Answer (2) Volume strength = 5. 1. Answer (1) Number of milligram of lyophilic colloid required to protect 10 ml gold sol in 1 ml 10% NaCl solution is equal to gold number of lyophilic colloid. Plot No. slope = 3. intercept = log K and slope = 2.Regd. Aakash IIT-JEE . bP is negligible in comparison to 1. 2p6. 2s2. 3d 5 – (Half filled d-orbital more stable) unpaired electrons = 5 11. Answer (4) In KClO3 the Cl is reduced and oxygen atom is oxidised both are different atom therefore it is not example of disproportionation reaction. Aakash IIT-JEE . Answer (2) KMnO 4 + FeC2 O 4 ⎯ ⎯→ Mn2 + + Fe3 + + CO2 n=5 n=3 equivalents of KMnO4 = equivalents of FeC2O4 equivalents of FeC2O4 = 2 × 3 = 6 equivalents of KMnO4 = 1. Answer (2) He(2) – 1s2 ∴ diamagnetic 1s Maximum number of electrons in an orbital is two. 3p6. 4s1 Cu+ → 1s2. 10. 2s2. It is in highest oxidation state therefore can acts as a oxidising agent.023 × 1024. 3s2. 2s2. 2p6. 3p6. 3p6. New Delhi-75 Ph. 2p6. Sector-11. Answer (3) Oxidation number of nitrogen in HNO2 is +3 therefore N can gain higher oxidation state +5 as well as lower oxidation state –3. 2p6. Plot No. 3d 10. 13. 3s2. 4. 2s2. 3p6. 3s2. 6.2 × 5 = 6 9. 2p6. 5.Regd. 3d 6 – unpaired electrons = 4 Fe3+ → 1s2. Answer (1) Heisenberg’s uncertainty principle is applicable for microscopic particle. 3d 6. 4s2 Fe2+ → 1s2. 3d 10 – No unpaired electrons Cu2+ → No. Dwarka. Answer (2) Na 2 S2 O 3 + I2 ⎯ ⎯→ Na 2S 4O 6 + 2NaI−1 n=2 0 8. Office : Aakash Tower.Physical Chemistry Success Magnet (Solutions) 4. Answer (3) Fe(26) → 1s2. of unpaired electrons = 1 (paramagnetic) 12. 2s2. 3s2. Answer (2) Cu(30) → 1s2.: 45543147/8 Fax : 25084124 (81) . Answer (2) Oxidation number of sulphur in H2SO4 is +6. 3s2. 3p6. 7. Answer (3) 18 ml water = 18 gm water (d = 1 gm/ml) ∴ ∴ ∴ Number of molecules of water = NAV 1 molecule of water contains = 10 electrons Total number of electrons = 10 NAV = 6. 22. In other atom the energy depends on n as well as value of l. Plot No. Office : Aakash Tower. 19. 4. Answer (3) Orbital angular momentum = l ( l + 1) For s-subshell l = 0 Aakash IIT-JEE . 15. 4s2 Zn2+ → 1s2. 20. 3d 10 Since Zn2+ has no unpaired electrons therefore it is diamagnetic. 2p6. Answer (2) Zn(30) → 1s2. 3s2. Answer (4) Electrons are filled in the orbital from lower energy level to higher energy level 3d orbital has more energy than 4s. Therefore 4s orbital will be filled first. 3s2. 16. 2 (because value of l ranging from 0 to (n – 1) Value of m depends on l and ranging from –l to +l. 3d 10. mV y x dx2 – y2 In d x 2 − y 2 the electron density lies along the x and y axis.: 45543147/8 Fax : 25084124 h 2π (82) .6 Z2 n2 It is clear the energy of electrons depends on principal quantum number. 2s2.Regd. New Delhi-75 Ph. For n = 3 l = 0. Answer (4) Fixed circular path around nucleus is not possible because it violates the Heisenberg’s uncertainty principle. 3p6. 1. Dwarka.Success Magnet (Solutions) Physical Chemistry 14. Answer (4) h mV h . Answer (3) Since the wavelength for the electron and proton is same but mass of electron is less than proton so its velocity will be more than proton. Answer (4) de Broglie wavelength λ = 18. It is clear from this expression λ = 17. Answer (1) The value of l depends on the n. Sector-11. Answer (3) Energy of entron in nth orbit for H-atom or H-like ions = −13. 21. It is applicable to moving particle. 2p6. 3p6. 2s2. Answer (3) :P P : P P: Hybridisation of p is sp3 and each p atom is liked with three other p atoms. 2p1 . 31. Dwarka. 2s2 (full filled s-orbital) .Physical Chemistry Success Magnet (Solutions) 23. Answer (2) Oxidising power of fluorine is more than oxygen. Answer (2) Aakash IIT-JEE . 30. due to greater reduction potential. New Delhi-75 Ph. 4. Plot No. Answer (4) : : N F F F Subtraction H : N H H Addition therefore dipole moment of NH3 is more than NF3 32. Answer (3) Due to smaller size of fluorine atom there is great electron-2 repulsion therefore electron affinity of fluorine is less than chlorine.more stable B(5) → 1s2. Answer (4) Be and Al shows diagonal relationship. Answer (3) Atomic volume is a guide to the size of atoms. 28. Answer (1) OF2 oxidation number of oxygen x + 2(–1) = 0 x = +2 The electronegativity of fluorine is highest in periodic table. Ionic radius of Mg2+ is smaller than Ba2+.less stable 2p orbital has higher energy than 2s-orbital. Office : Aakash Tower. Answer (2) BaSO4 is insoluble in water. Sector-11.: 45543147/8 Fax : 25084124 : ⎡ ⎢ ⎣ I I I Linear ⎤ :⎥ ⎦ : Hybridisation of central atom is sp3d (83) .Regd. 26. 24. 25. 29. 2s2. 27. Answer (3) Be(4) → 1s2. If atomic radius increases atomic volume also increases. Answer (2) Due to non available vancant d-orbital NCl5 does not exist down the group atomic size increases therefore N is smaller in size than p.O. of N2 2 is 2.O. Answer (3) He2 ( 4) − σ1s 2 σ * 1s 2 B. = He + 2 (3) bond order = Since B. 36. of He2 is zero therefore it does not exist. Answer (2) Hydrogen in non polar molecule due to absence of electronegativity difference between the two H-atoms. Answer (3) Due to available vacant d-orbital only SiCl4 reacts with water both are covalent compound. The electronegativity of Li and Cl is not small. New Delhi-75 Ph. Office : Aakash Tower.O.Regd. Dwarka.O. of N+ 2 − But B. 39. Answer (2) − B.: 45543147/8 Fax : 25084124 : ⎡π * 2 p1 ⎤ y ⎢ ⎥ 1⎥ ⎢ π * 2 p z ⎣ ⎦ 10 − 6 =2 2 Number of unpaired electrons = 2 2−2 =0 2 2 −1 1 = 2 2 (84) .O.Success Magnet (Solutions) Physical Chemistry 33. Aakash IIT-JEE . ⎡π2p 2 ⎤ y 2 2 2 2 ⎥ σ2p 2 N2(14) – σ1s σ * 1s σ2s σ * 2s ⎢ x 2⎥ ⎢ 2 p π z⎦ ⎣ 37. 38. 4.5 = B. Sector-11. Plot No. 40. = 35. Answer (3) : N : : : : N O O : O:– : O : – N 5 +1 = = 3 attached atoms = 2 2 2 ∴ shape is angular 34. Hence it is least stable. Answer (3) Bond order of O2 2⎤ ⎡π 2 p y ⎥ O2(16) — σ1s 2 σ * 1s 2 σ2s 2 σ * 2s 2 σ2 px 2 ⎢ 2⎥ ⎢ π 2 p z ⎣ ⎦ : : B.O. of N2 = 2. Answer (3) LiCl is covalent in nature due to smaller size of Li+ ion. Physical Chemistry Success Magnet (Solutions) 41. Answer (2) H2O is liquid due to hydrogen bonding but no H-bonding present in H2S. Oxygen is more electronegative than sulphur. 42. Answer (2) Intermolecular force of attraction of inert is very low. Since inert gases have highly stable electronic configuration therefore ionisation energy is quite high. 43. Answer (3) PV = nRT P ∝ T (when n and V are kept constant) 44. Answer (2) Heat absorbed during isothermal expansion of an ideal gas against vaccum is zero. Volume of the gas molecules is negligible compared to the volume occupied by gas. 45. Answer (4) Gases molecules have different types of speed, (i.e. Vrms, Vmp, Vav) molecules of ideal gas neither attract nor repel to each other. 46. Answer (2) Molar heat capacity at constant pressure is more than that of molar heat capacity at constant volume. The average kinetic energy of ideal gas depends only on temperature (i.e. 47. Answer (1) Due to great intermolecular force among the NH3 molecules the value at ‘a’ is more for NH3 than that of N2. 48. Answer (4) Compresibility of non ideal gas is not equal to 1 i.e. (Z ≠ 1) it may be greater than 1 or less than 1. Due to intermolecular force of attractions pressure of non ideal gas is lower than expected. 49. Answer (1) Due to the free valancies the transition metals adsorb the gases. 50. Answer (3) Due to stronger inter molecular force of attraction SO2 is easily liquified. Since critical temperature is directly proportional to the value of ‘a’ ∴ Critical temperature of SO2 will be more than H2. 3 RT) 2 51. Answer (3) ⎛ a ⎞ ⎜ 2 ⎟ represents the inter molecular force of attraction. ⎝V ⎠ It can not be neglected because in real gas the molecules are closer to each other. 52. Answer (4) 2 ⎞ ⎛ ⎜ p + n a ⎟( V − nb ) = nRT ⎜ V2 ⎟ ⎝ ⎠ Unit of V = unit of nb Unit of b = litre mol–1 ‘a’ represents inter molecular force of attraction Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (85) Success Magnet (Solutions) Physical Chemistry 53. Answer (4) V1 V2 = T1 V2 V2 = 4 V 3 ⇒ V V = 2 300 400 At constant pressure gases follows the Charle’s law V ∝ T (at constant P) 54. Answer (1) ∆E is a state function because it depends only on initial and final position only. 55. Answer (1) In cyclic process the change in enthalpy (∆H), change in entropy (∆S) and change in free energy (∆G) are zero because these are state function. 56. Answer (1) Heat of neutralisation of HF with NaOH is more than 13.7 KCal because heat of hydration of F– ion is very high due to smaller size of F ion. 57. Answer (2) For spontaneous process ∆G must be negative. ∆G = ∆H – T∆S 58. Answer (1) Enthalpy of formation of H2O(l) is more than H2O(g) because some extra heat evolved when the water vapour is condensed. 59. Answer (4) Pressure and temperature are intensive properties whereas volume is extensive property. Extensive property depends on the mass of substance. 60. Answer (2) ∆G = ∆H – T∆S If ∆S is negative then at high temperature T∆S will be more than ∆H therefore ∆G will be positive and reaction will be non spontaneous at low temperature. ∆H will be more than T∆S therefore ∆G will be negative and reaction will be spontaneous. 61. Answer (3) ∆G = ∆H – T∆S Since ∆S = negative ∴ T∆S is positive. If ∆G = (–) then ∆H must be negative. 62. Answer (2) Solubility of HgI2 is more in KI solution due to complex formation 2KI + HgI2 → K2[HgI4] due to bigger size I– ion is highly polarizable. 63. Answer (1) Kc = [CO2] on increasing the volume the concentration of CO2 decreases. So to maintain the concentration of CO2 equilibrium shifts in forward direction. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (86) Physical Chemistry Success Magnet (Solutions) 64. Answer (4) CH3COOH CH3COO– + H+ on addition of CH3COONa the concentration of CH3COO– increases and equilibrium shifts in backward direction so pH will increases because [H+] decreases. 65. Answer (1) For neutral solution [H+] = [OH–] ∴ pH + pOH = pKw ∴ pH = pH = pOH pKw 2 ∴ ∆H = –1 66. Answer (1) CuO + H2 Cu + H2O at 75ºC the water in form of liquid ∴ Kc > kp ∴ ∆H = 0 Kp = Kc(RT)–1 But at 175º water in forms of gaseous Kp = Kc 67. Answer (4) The aqueous solution of salt of strong base with weak acid is basic due to cationic hydrolysis the solution becomes acidic. ∴ pH < 7 ∴ pH > 7 68. Answer (3) Solubility of salt increases on dilution the solubility product (Ksp) depends only on temperature. 69. Answer (3) H3PO 4 H+ + H2PO − 4 − H+ + HPO2 4 Ka1 Ka2 H2PO− 4 Ka1 > Ka2 pKa1 < pKa2 70. Answer (1) ∴ H3PO4 is stronger acid then H2PO− 4 Catalyst speeds up the rate of forward reaction as well as backward reaction. 71. Answer (2) AT the equilibrium rate of forward reaction is equal to the rate of backward reaction so concentration of reactants as well as product does not change with time. 72. Answer (4) Equilibrium constant depends only on temperature. A(g) Initial moles at equilibrium 1 1–x B(g) + C(g) 0 x 0 x ⎛ x ⎞⎛ x ⎞ ⎜ ⎟⎜ ⎟ V V x2 K c = ⎝ ⎠⎝ ⎠ = (1 − x )V ⎛ 1− x ⎞ ⎜ ⎟ ⎝ V ⎠ Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (87) Answer (2) NH+ 4 H + H—N—H H 77. O O O +2 –1 O oxidation number of chromium is +6. 81. 80.Success Magnet (Solutions) Physical Chemistry 73. –1 Cr +1 +1 +1 O –1 78. Answer (1) + NH4 + OH− on addition of NH4Cl the concentration of NH+ 4 increases and equilibrium will be – shifted in backward direction therefore [OH ] decreases as a result pOH increases and pH will decreases. 4. New Delhi-75 Ph. Office : Aakash Tower. NH4OH 75. Answer (3) Reduction potential of Na is very less than water therefore water reduced first and H2 gas evolved at cathode. Answer (2) SO 2 + 2H2S ⎯ ⎯→ 3S + 2H2O Bleaching action of SO2 is due to reduction. 76. Answer (3) –1 –2 +1 form of charge on nitrogen is +1. For spontaneous process ∆G must be negavite. Answer (1) FeC 2O 4 ⎯ ⎯→ Fe 3 + + CO 2 n=3 − FeC 2O 4 + MnO 4 ⎯ ⎯→ Mn2+ + Fe3 + + CO 2 ( n= 3 ) (n = 5 ) Equivalents of FeC2O4 = equivalent MnO− 4 1 × 3 = 0.Regd.6 × 5 79. Answer (1) − Blood is a basic buffer solution of (H2CO3 + HCO3 ). Liberation or deposition depends on the reduction potential. Aakash IIT-JEE . Answer (2) In electrochemical cell the chemical energy is converted to electrical energy for this process the cell reaction should be spontaneous. Sector-11. Dwarka.: 45543147/8 Fax : 25084124 (88) . Plot No. Answer (1) CH3COONa + HCl 2M 1M 1M 0 Acidic buffer solution CH3COOH + NaCl 0 0 1M 1M 74. Answer (2) NaCl ⇒ coordination no. Answer (1) When Zn electrodes are used in ZnSO4 solution then it acts as attacking electrodes. of atoms = 1 bcc unit cell has no. Answer (1) KCl is used in salt bridge because K+ and Cl– ions have nearly same ionic mobility. Answer (3) When CuSO4 is electrolysed then Cu deposited at cathode and O2 gas is evolved at anode. 86.732 size of r+ is maximum to be fitted in oh void which would cause expansion of lattice. molar conductivity of any electrolyte (strong or weak) is equal to the sum of molar conductivities of the ions produced by it.: 45543147/8 Fax : 25084124 (89) . Answer (2) At in finite dilution. Sector-11. of atoms = 2 fcc unit cell has no.414 size of r+ is least to be fitted in oh void r− r+ = 0. Plot No.Physical Chemistry Success Magnet (Solutions) 82. 88.0591 ⎛ C1 ⎞ log⎜ ⎜C ⎟ ⎟ 2 ⎝ 2⎠ For spontaneous process C1 < C2 85. Answer (1) r+ = 0. ratio = 8 : 4 Inverse simplified ratio = 1 : 2 Molar ratio = 1 : 2 ∴ Statement (1) and (2) both are correct 90. New Delhi-75 Ph. Answer (3) 1 Faraday electricity is required to deposit 1 equivalent of substance not one mole 89.414 to 0. Dwarka. Office : Aakash Tower. of atoms = 4 Aakash IIT-JEE . 87. 84. = 6) r− When r+ = 0. r− but when 91.Regd. Answer (2) Cubic unit cell has no. Answer (2) º For concentration cell E cell =0 º E cell =− 0. Answer (3) Equivalent conductance is directly proportional to the dilution because mobility of ions increase. 83.732 (for coordination no. ratio = 6 : 6 Inverse simplified ratio = 1 : 1 Molar ratio = 1 : 1 For CaF2 ⇒ coordination no. 4. Answer (1) The end product of (4n + 2) series is 105. Answer (2) ccp has ABC ABC. Answer (2) Each tetrahedral void is at ¼th distance from corner at body diagnol. Dwarka. 95. They turn to paramagnetic substance when heated above curies temperature. Cl– will constitute lattice. 106. pattern of stacking. Office : Aakash Tower.. 4. 94. Answer (1) Co-ordination number in bcc is 8. Answer (3) Common salt has F electrons. 93. Answer (1) Distance between 2 nearest spheres in fcc = Distance between 2 nearest spheres in bcc = 96. Answer (1) For second order.. 104. Answer (4) The value of Arrhenious constant is not affected by temperature. Answer (1) For LiCl. 97.Molar ratio = 1 : 1 ∴ Simple coordination number ratio = 1 : 1 Q Simple coordination number of Zn2+ = 4 (present in tetrahedral void) ∴ Simple coordination number of S2– = 4 103. Plot No. Answer (2) Ferromagnetic substances are those paramagnetic substance which persists their magnetic moment (i. Sector-11. New Delhi-75 Ph. 98.: 45543147/8 Fax : 25084124 (90) .Success Magnet (Solutions) Physical Chemistry 92. Answer (1) Ions come closer in crystals suffering from frenkel defects. t1/2 is inversely proportional to concentration. spin alignment) even in absence of magnetic field.. 99.Regd. Answer (4) Inverse and normal spinel structure have same packing efficiency. 101.. responsible for colour. Answer (4) 10–3 mole of SrCl2 will develop 10–3 NA cationic vacancies. Answer (2) The rate law is − 1 d[ A ] d[B] d[C] =− =+ = Rate of reaction 2 dt dt dt 206 82 Pb .e. 102. 2 a 2 3 a 2 Aakash IIT-JEE . 100. Answer (1) ZnS :. : 45543147/8 Fax : 25084124 (91) . Answer (4) In benzene. neutron is converted into proton and the atomic number increases by one. Hence. Answer (2) For coagulation. Answer (1) The effective molarity of KCl is twice the sugar because KCl gives two ions. Sector-11. the van’t Hoff factor is less than 1. Answer (3) Alkaline hydrolysis of ester is known as saponification and it is second order reaction. 111. 112. Answer (3) The effective molarity of 1 M CuSO4. acetic acid dimerises. greater will its coagulating power. 120. 109. HgI2 + 2KI → K2HgI4 potassium mercuric iodide In aqueous solution. 114. 116. the vapour pressure of KCl is less than sugar. Answer (1) In 1 M aqueous solution. Answer (4) Micelles is formed at above CMC and at above Kraft temperature. Answer (2) During β-decay. 108. 110. solvent is less than 1000 g while in 1 m aqueous solution. Plot No. solvent is 1000 g.P. Answer (3) Rate constant of any order reaction is directly proportional to temperature. 115. Answer (4) When mercuric iodide is added in KI solution then association takes place and freezing point is raised. higher is the charge on oppositely charged ions.Physical Chemistry Success Magnet (Solutions) 107. 119. HgI2 dissociates as HgI2 → Hg2+ + 2I– 113.5 M AlCl3 and 2 M urea solution are same and elevation in B. So. 1 M is more concentrated than 1m. Boiling point is not a colligative property. New Delhi-75 Ph. Answer (3) In presence of more volatile solute. is a colligative property. Answer (1) The extent of adsorption of CO2 is much more higher than H2 because critical temperature and van der Waal’s force of attraction of CO2 is much higher than H2. Aakash IIT-JEE . Answer (3) Molality and mole fraction is independent of temperature and molality is the number of gm moles of solute dissolved in 1 kg of solvent.Regd. Dwarka. 118. 0. 117. Office : Aakash Tower. Answer (4) Solution of starch in water is a lyophilic sol it is reversible sol. vapour pressure of solution is developed by solute and solvent both. So. 4. s). Answer .A(p.7 gm NH3 moles = 1. B(q.1 17 molecules = 0. q.1 N0 atoms = 0. of electrons = 0. r).Regd. Answer . Office : Aakash Tower. Dwarka.4 N0 volume = 2.3 N0 volume = 2.4 gm SO2 moles = 6 .6 = 0 .6 gm C2H2 moles = 2 . D(p) (A) 60% metal in metal oxide Q 40 gm oxygen reacts with 60 gm metal ∴ 8 gm oxygen reacts with = 8 × 60 = 12 40 ⇒ Equivalent weight of metal = 12 Equivalent weight of oxygen = 8 (B) 64.24 L 2.4 = 0 .2 gm oxygen moles = 3. Sector-11.4 N0 volume = 2. r). New Delhi-75 Ph.1 32 molecules = 0.2 N0 volume = 2. C(r.1 26 molecules = 0.: 45543147/8 Fax : 25084124 (92) .A(p. q).2 = 0. Plot No.1 N0 atoms = 0.4% metal in metal oxide Q 35.4 gm metal Aakash IIT-JEE . s). 4.Success Magnet (Solutions) Physical Chemistry Section . C(p.24 L (D) 6.1 N0 × 10 = N0 (B) 3.E : Matrix-Match Type 1. q).1 N0 atoms = 0.1 64 molecules = 0.6 gm oxygen reacts with 64.24 L (C) 2.1 N0 atoms = 0. D(q) (A) 1. B(q.7 = 0.24 litre no. 5 71 ⇒ Equivalent weight of metal = 14.A(r. B(p. C(s). Since it is in highest oxidation state therefore reduction is possible hence acts as a oxidising agent HNO2 ⇒ Oxidation No. Sector-11. r).5 Equivalent weight of oxygen = 8 (C) 29% metal in metal chloride Q 71 gm chlorine reacts with 29 gm metal ∴ 35. Plot No. D(s) (A) Fe C2 O 4 ⎯ ⎯→ Fe 3+ + CO 2 +2 +3 +3 +4 +5 Oxidation +3 Reduction –3 n factor of FeC2O4 = 1 + 1 × 2 = 3 (B) Cu2 S ⎯ ⎯→ Cu2+ + SO 2 +1 −2 +2 +4 n factor of Cu2S = 1 × 2 + 6 = 8 (C) Fe S 2 ⎯ ⎯→ Fe3+ + SO2 n factor of FeS2 = 1 + 5 × 2 = 11 +2 −1 +3 +4 (D) Fe(NO3 )3 ⎯ ⎯→ Fe + 2 + NO +3 +5 +2 +2 n factor of Fe(NO3)3 = 1 + 3 × 3 = 10 Aakash IIT-JEE . r). 4.5 (D) Equivalent weight of Mg = 24 = 12 2 3.A(p. Answer . C(q).Regd.5 gm chlorine reacts with = 35. q.4 = 14. New Delhi-75 Ph.Physical Chemistry Success Magnet (Solutions) ∴ 8 gm oxygen reacts with = 8 × 64. Answer . B(p). s). (temperature dependent) ∴ Volume of solution depends on the temperature. of Nitrogen = +1 N3H is called hydrazoic acid 5. B(q). D(r) HNO3 ⇒ Oxidation number of nitrogen is +5. r) D(r) Molarity = No. of gram formula mass of solute (Temperatu re dependent) Volume (litre) Formality = Strength of solution – It is defined as the amount of solute in grams present in one litre of solution. C(p. 4. of Nitrogen = +3 ∴ acts as a reducing as well as oxidising agent N2O – Oxidation No. Dwarka.5 × 29 = 14.6 ⇒ Equivalent weight of metal = 14.: 45543147/8 Fax : 25084124 (93) . of moles of solute (Temperatur e dependent ) Volume (litre) Molality = No. of moles of solute (Temperatur e independen t ) Mass of solvent (kg) No.A(r). Answer . Office : Aakash Tower.5 35. 3p6.A(r. 3d3 No. of spectrum = = (n2 − n1)(n2 − n1 + 1) 2 (6 − 3)( 6 − 3 + 1) = 6 Infrared region 2 ∴ Diamagnetic ∴ Paramagnetic ∴ Paramagnetic ∴ Paramagnetic Aakash IIT-JEE . B(p. of spectrum = (A) (B) (C) (D) No. Answer . Plot No. of spectrum = No. Dwarka. s). D(p) (A) N2 = No. of spectrum = No. 3s2. 3s2. of electron = 6 + 8 = 14 e– CN– = No. 3d10 4s0 No. 2s2. of electron = 6 + 7 + 1 = 14 e– (B) 238 92U ⎫ ⎪ ⎬ (isoelectronic ) ⎪ ⎭ no. s). r) B(p.A(p. Answer . hence isodiaphers 15 14 16 7N 6N 8O 14 14 7N 6N — — all have 8 no. D(p. C(q.: 45543147/8 Fax : 25084124 (94) . C(r. of unpaired electrons = 3 Cu2+ — 1s2. of spectrum = No. 3p6. of unpaired electrons = 3 Co2+ — 1s2. of neutrons = 234 – 90 = 144 (n – p) = 144 – 90 = 54 ∴ (C) (D) (n – p) is same in both. of unpaired electrons = 0 Cr3+ — 1s2. of spectrum = (n2 − n1 )(n 2 − n1 + 1) 2 (5 − 2)(5 − 2 + 1) = 6 belongs to visible region 2 ( 6 − 3)( 6 − 3 + 1) = 6 belongs to infrared region 2 ( 4 − 2)( 4 − 2 + 1) = 3 belongs to visible region 2 (8 − 4)(8 − 4 + 1) = 10 belongs to infrared region 2 8. Sector-11. 3p6. 2p6.Success Magnet (Solutions) Physical Chemistry 6. 3s2. r). Answer . 2s2. D(s) Zn2+ — 1s2. B(r). C(q.Regd. of electron = 2 × 7 = 14 e– CO = No. 3d 9 No. q). s). 2p6. q) (A) No. 4. 3p6. of neutrons therefore isotones since mass no. C(q). Office : Aakash Tower. of unpaired electrons = 1 9. r). s). 2p6. 3s2. s). of neutrons = 238 – 92 = 146 (n – p) = 146 – 92 = 54 234 90U no. 3d 7 No. D(s) No. New Delhi-75 Ph.A(p. Answer . 2s2. 2p6. is same therefore isobars 7. 2s2.A(s). B(q. Answer . D(r) (A) (B) (C) (D) Alkali metals are group 1 elements K belongs to group one Alkaline earth metals are group 2 elements Ba belongs to group 2 Fr is radioactive element As is metalloid. s). Answer . Dwarka. 4. r.A(q).: 45543147/8 Fax : 25084124 . D(p. B(r). 2.A(r). 2s2. r.A(p). Plot No. New Delhi-75 Ph. Answer . r). Answer . B(p). But in case of nitrogen multielectronic species energy depends on both principal and azimuthal quantum number N(7) – 1s2. C(p. B(p. 1. s) (A) Orbital angular momentum = for p-orbital l = 1 ∴ Orbital angular momentum = 2 h 2π l(l + 1) h 2π (B) (C) (D) Angular momentum mvr = nh where n is principal quantum number 2π p. Office : Aakash Tower. 3 Therefore in N-shell d and p-subshell will be present and number of waves = principal quantum number = 4 12.A(q). d-subshell has 3 and 5 degenarate orbitals respectively For N-shell n = 4 ∴ l = 0. 2p3 Here energy of valence electrons depends on the exchange energy and symmetry also. 11. of spectrum = (D) No. C(s). q) Hydrogne atom Z = 1. C(s). Sector-11. D(p. –1s1 Hydrogen and He+ are monovalent therefore energy deciding factor is only principal quantum number.Regd.Physical Chemistry Success Magnet (Solutions) (B) No. of spectrum = 10. C(p). (95) Aakash IIT-JEE . of spectrum = (7 − 3 )(7 − 3 + 1) = 10 – Infrared region 2 (5 − 2)(5 − 2 + 1) = 6 – Visible region 2 ( 6 − 2)( 6 − 2 + 1) = 10 – Visible region 2 (C) No. B(q). D(p) (A) (B) (C) (D) Fluorine has maximum electronegativity Chlorine has maximum electron affinity Fe is transition element therefore has variable valency He is inert gas (most stable configuration) therefore has maximum ionisation energy 13. q. s) (A) (B) (C) (D) Acid has pH < 7 Base has pH > 7 Acidic buffer is a mixture of weak acid + salt of this acid with strong base Basic buffer is a mixture of weak base + salt of this base with strong acid 18.A(p). Cl2 Metallic radius is bigger than covalent radius Stevenson equation is used to calculate actual bond length of polar covalent molecules.A(r. C(s). C(p. H2SO4 Phosphorus exist in allotropic form and forms oxyacids like H3PO4. Sector-11. s).5 2 10 − 8 =1 2 10 − 5 = 2. s). C(p. Answer . D(q. 17. B(s). s) (A) (B) (C) (D) Chlorine has highest electron affinity and forms oxy acids like HClO. New Delhi-75 Ph. 16. r). D(p.Success Magnet (Solutions) Physical Chemistry 14. r). C(p. Answer . s). Office : Aakash Tower.5 2 2 unpaired electrons 1 unpaired electrons ∴ Paramagnetic ∴ Paramagnetic zero unpaired electron ∴ Diamagnetic one unpaired electron ∴ Paramagnetic Bond order of O+ 2 Aakash IIT-JEE . H3PO3 15. Answer . HClO2 Fluorine is most electronegative elements Sulphur exists in allotropic forms oxyacids like H2SO3. B(q. µR = 0. Answer .A(q. q. B(r). C(p). µR = 0 H C=C CH 3 Additive. Answer . D(q. µR ≠ 0 (both have different bond moment) F Substractive. Dwarka. D(p. s). 4. D(p) (A) (B) (C) (D) van der Waal’s radius is bigger the covalent radius and used for gaseous molecules Covalent radius is used for covalent molecules HCl. r) O=C=O + F H I F B F Subtractive. r). B(p. s).Regd. µR ≠ 0 F Substractive. s).: 45543147/8 Fax : 25084124 (96) . B(q). Plot No. s) 1⎤ ⎡ 2 2 ⎤⎡ 2 2 2 2 2 ⎢ π p y ⎥ ⎢ π * 2p y ⎥ 1 s * 1 s 2 s * 2 s p σ σ σ σ σ O2 (16) — x ⎢π 2p 2 ⎥ ⎢π * 2p1 ⎥ z ⎦⎣ z⎦ ⎣ Bond order of O2 Bond order of O− 2 − Bond order of O2 2 = = = = Nb − Na 10 − 6 = =2 2 2 10 − 7 = 1.A(q.A(p. Answer . D(r) (A) + NO2 N 5 −1 = =2 2 2 N 5 +1 = =3 2 2 N 5 + 4 −1 = =4 2 2 N 5+3 = =4 2 2 ∴ hybridisation sp (B) − NO 3 ∴ hybridisation sp2 ∴ hybridisation sp3 ∴ hybridisation sp3. (B) NH4NO3 can be written as NH 4 + NO 3 + + − − ∴ hybridisation of nitrogen in NH 4 is sp3 and hybridisation of nitrogen in NO 3 is sp2 Both ionic and covalent bond are exist. D(p. D(q. s). s).A(p. s) (A) SF6 N 6+6 = =6 2 2 N 8+4 = =6 2 2 N 7+5 = =6 2 2 N 7+3 = =5 2 2 sp3d2 (B) XeF4 sp3d2 two lone pairs on central atoms (C) BrF5 sp3d2 one lone pair on central atom (D) ClF3 sp3d lone pair on central atom and T-shaped 21. one lone pair present on central atom (C) (D) NH+ 4 NH3 20. B(r. s). r. r. B(p. Answer . r). B(p. D(q. B(q. s).Regd. C(r. Aakash IIT-JEE . of valence electrons of centre atoms) + No. s). Plot No. s) (A) Na2B4O7·10H2O exists as OH O 2 Na + B O B OH O B O OH · 8H O 2 HO B O ∴ Therefore hybridisation of boron is sp3 and sp2. s). r. s). 4. r. s). q. Sector-11. r. Answer .Physical Chemistry Success Magnet (Solutions) 19.: 45543147/8 Fax : 25084124 (97) .A(q. s) N = (No.A(p. Answer . of atoms attached to this (A) (B) (C) (D) CaF2 is insoluble in water and ionic compound BeSO4 is soluble in water and has Be2+ + SO42– ionic bond as well as covalent bond Na2CO3 is soluble in water and ionic as well as covalent bond Ca2+ + CO32– has ionic and covalent bond insoluble in water 22. q. Dwarka. Office : Aakash Tower.A(p. s). s). C(r. C(p. C(r. r. s). New Delhi-75 Ph. B(r. C(s). r) (A) Br F5 + 3H2O → HBrO3 + 5HF sp3d2 (B) sp3 H3 BO3 + H2O → [B(OH)4]– + H+ sp2 sp3 (C) Cl F3 + Sb F5 → [Cl F2]+ [SbF6]– sp3d sp3 (D) P Cl5 + 4H2O → H3PO4 + 5HCl sp3d sp3 24. b depend on the nature of gas For ideal gas second viral coefficient is zero. q). (D) KMnO4 can be written as K+ + MnO 4 Hybridisation of Mn in MnO 4 is sp3 and both covalent and ionic bonds are present. 25. q. q. Dwarka. van der Waal force of attraction increases hence boiling point increases. Answer . D(q. D(p.A(p. Office : Aakash Tower. − − 23. Answer . van der Waal’s constant b depends on molecular size.A(q. D(r) (A) As molar mass increases. (C) a depends on polarizability of molecule. C(q. Plot No. r). Answer . r. s).A(p. q) Boyle temperature is the temperature at which real gases Behave like ideal gas TB = Ti = TC = a Rb 2a Rb 8a 27Rb a. (D) b depends on molecular size. (B) Second virial coefficient b− a RT as b depends on molecular size and a depends on polarizability of molecule hence second virial coefficient depends on these two factors. Sector-11. B(p. van der Waal’s constant a depends on polarizability of molecule. s). s). Aakash IIT-JEE . r). New Delhi-75 Ph. C(p. 4.: 45543147/8 Fax : 25084124 (98) .Success Magnet (Solutions) Physical Chemistry (C) K2 [Ni(CN)4] can be written as 2K+ + [Ni(CN)4]2– Hybridisation of [Ni(CN)4]2– is dsp2. B(p.Regd. Both ionic and covlaent bond are present. s). r). 4. B(s). Office : Aakash Tower.e. D(q) Vrms = VAV VMP PV = 3RT M 8RT πM 2RT M = = 1 mnc 2 3 Aakash IIT-JEE . s). Plot No.Regd. D(r. B(p. Answer . Dwarka. r. New Delhi-75 Ph.A(p. C(s).A(r). ⎜ KT ⎟ ⎝2 ⎠ 27. C(s). s). C(q). B(p).: 45543147/8 Fax : 25084124 (99) . r. D(p) (A) Z = PV = Compressib ility factor RT for ideal gas Z = 1 (B) When the pressure is low volume is high ~V (V – b) – a ⎞ ⎛ ⎜ P + 2 ⎟( V ) = RT V ⎠ ⎝ PV + a = RT V a V PV = RT − Z= PV ⎛ a ⎞ = ⎜1 − ⎟ RT ⎝ RTV ⎠ (C) a ⎞ ⎛ ~P At high pressure ⎜ P + 2 ⎟ – V ⎠ ⎝ P( V − b ) = RT PV = RT + Pb PV = 1 + Pb RT RT (D) Critical temperature TC = 8a 27Rb 28. Sector-11.Physical Chemistry Success Magnet (Solutions) 26. s) Rate of diffusion = Volume of gas diffused time Partial pressure = (mole fraction) × total pressure Force is rate change of momentum ΔP mv − mu = t t ⎞ ⎛3 Kinetic energy of ideal gas depends only on temperature i. Answer .A(r). q. Answer . The heat evolved in formation of H2O is the heat of combustion of H2. q. Partial pressure = (mole fraction) × total pressure Pressure ∝ temperature (C) Rate of diffusion ∝ 1 Molecular Mass Rate of diffusion also depends on temperature. s) (A) (B) Kinetic energy of gases depends only on temperature. Answer . 31. D(p) (A) (B) (C) (D) Heat of combustion. r). Plot No. D(r) (A) The critical temperature of CO2 is approximately 31. q. Answer . s). Answer . New Delhi-75 Ph. s). heat of formation of H2O and it is used in fuel cell. 4. s). Dwarka. q.A(q). p). Aakash IIT-JEE . The heat evolved of CO2 is the heat of combustion of carbon and heat of formation of CO2. D(p.2°. Office : Aakash Tower.A(s). C(p). C(q. C(p. Answer . Since temperature is below than this therefore CO2 exists as a liquid Critical temperature TC = (B) 89 27 Rb (C) 3 = PC VC 8 RTC When the compressibility factor is less than one then Vreal is less the Videal 32.: 45543147/8 Fax : 25084124 (100) . r.A(s). (D) Vapour pressure is directly proportional to the mole fraction and pressure also depends on temperature.A(q. B(q). D(r) (A) (B) At high the molecules come closer and intermolecular forces are dominated therefore gases follow the vander Waal’s equation Pressure is not too low but volume is high ~V ∴ (V – b) – a ⎞ ⎛ ⎜ P + 2 ⎟V = RT V ⎠ ⎝ a PV + = RT V a PV = RT − V (C) Force of attraction is negligible P (V – b) = RT PV – Pb = RT PV = RT + Pb a v2 ~0 – (D) At very high temperature and low pressure gas behaves as a ideal gas therefore follows the ideal gas equation PV = RT 30. B(q).Success Magnet (Solutions) Physical Chemistry 29. s). r. Sector-11. C(q.Regd. B(q. Heat of neutralisation. B(p. Physical Chemistry Success Magnet (Solutions) 33. Answer - A(r), B(s), C(p), D(r) Ag2CrO4 2Ag+ + 2s mol/L − CrO2 4 s mol/L − 2 Ksp = [Ag+ ]2 [CrO2 4 ] = (2s) · s 4s3 = Ksp ⎛ K sp ⎞ 3 ⎟ s= ⎜ ⎜ 4 ⎟ ⎝ ⎠ 1 AgCNS(s) Ag+ (aq) + CNS– (aq) s mol/L s mol/L Ksp = [Ag+][CNS–] = (s)(s) = (s)2 s= K sp 3Ca2+ (aq) 3s mol/L − + 2PO3 4 (aq) Ca3(PO4)(s) 2s mol/L − 2 Ksp = [Ca2+]3 [ PO3 4 ] = (3s)3 (2s)2 = 108 s5 ⎛ K sp ⎞ 5 ⎟ s= ⎜ ⎜ 108 ⎟ ⎝ ⎠ 1 Hg2Cl2 − Hg2 2 + 2Cl– 2s mol/L 2s mol/L − – 2 Ksp = [ Hg2 2 ] [Cl ] = (s) (2s)2 = 4 s3 ⎛ K sp s= ⎜ ⎜ 4 ⎝ ⎞3 ⎟ ⎟ ⎠ 1 33(a). Answer (4) (MX) S2 = 4.0 × 10–8 S = 2.0 × 10–4 (MX2) 4S3 = 3.2 × 10–14 S3 = 0.8 × 10–14 S3 = 8 × 10–15 S = 2 × 10–5 (IIT-JEE 2008) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (101) Success Magnet (Solutions) Physical Chemistry (M3X) 27S4 = 2.7 × 10–15 S = 1 × 10–4 ∴ Order is MX > M3X > MX2 34. Answer - A(s), B(r), C(p), D(q) (A) Variation of vapour pressure with temperature P = Ae ln − ΔHv RT P2 1⎤ ΔHv ⎡ 1 = ⎢ − ⎥ P1 R ⎣ T1 T2 ⎦ (B) Kirchhoff’s equation ∂ ( Δ G) = ΔCp ∂T (C) Gibb’s Helmholtz equation ⎡ ∂( ΔG) ⎤ ΔG = ΔH + T ⎢ ⎥ ⎣ ∂T ⎦ (D) Vant Hoff isochore d ln K p dT = ΔH0 RT 2 35. Answer - A(q), B(r), C(p), D(s) (A) We know that ∆G = ∆G° + RT lnK at equillibrium ∆G = 0 ⇒ ∆G° = –RT lnK ∆H = ∆E + P∆V + V∆P at constant pressure ∆P = 0 ⇒ ∆H = ∆E + P∆V (B) H = E + PV (C) Entropy change ΔS = QRe v T nRT ln = V2 V1 T V2 = nR ln V 1 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (102) Physical Chemistry Success Magnet (Solutions) (D) Negative of free energy change = workdone = QV – ∆G = (nF)E ∆G = – nFE 36. Answer - A(q), B(q, r, s), C(p, r, s), D(p) (A) NO (g) + O3 (g) NO2(g) + O2(g) ∆H = –200 kJ No. of moles in both sides is equal therefore no effect of pressure on decreasing the temperature equilibrium shift in forward direction. (B) 4NH3 (g) + 5O2 (g) ∆n = 1 So the reaction forwarded in forward direction by decreasing temperature, decreasing pressure and addition of inert gas at constant pressure (C) N2O4 (g) ∆n = 1 Reaction is endothermic therefore by increasing temperature, decreasing the pressure and addition of inert gas at constant pressure equilibrium shift in forward direction (D) N2 (g) + O2 (g) 2NO(g) ∆H = +180.5 kJ 2NO2(g) ∆H = 57.2 kJ 4NO(g) + 6H2O(g) ∆H = –905.6 kJ Reaction is endothermic reaction shift in forward direction by increasing temperature. No effect of pressure because there is no change in no. of gaseous moles 37. Answer - A(s), B(r), C(q), D(p) (A) Mixture of weak acids (HA + HB) [H+ ] = k 1C1 + k 2 C 2 (B) Mixture of strong acid and weak acid [H+ ] = C2 + C2 2 + 4k aC1 2 (C) Equivalent mixture of strong acid + weak base [H+ ] = (D) kw ⋅C kb Equivalent mixture of strong base + weak acid [H+ ] = k w ⋅ ka C 38. Answer - A(p, r), B(p), C(p, q), D(s) (A) Initial m. mol After reaction pH = = CH3COOH + NaOH 100 75 25 0 CH3COONa + H2O 0 25 0 25 pk a + log pk a + log [Salt ] [acid] 25 = pka − log 3 75 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (103) mol After reaction pk a + log 100 25 75 = pk a + log 3 25 CH3COOH + NaOH 100 0 100 0 CH3COONa + H2O 0 100 0 100 pH of salt of weak acid with strong base is pH = 1 1 1 pk w + pk a + log C 2 2 2 100 1 = M 200 2 Conc. B(q). D(p) (A) N2 (g) + 3H2 (g) ∆n = –2 Kp = Kc (RT)∆n = Kc (RT)–2 (B) 2SO2 (g) + O2 (g) ∆n = –1 Kp = Kc (RT)–1 (C) PCl5 (g) ∆n = 1 Kp = Kc (RT)1 (D) H2 (g) + I2 (g) ∆n = 0 Kp = Kc (RT)0 = Kc Aakash IIT-JEE .: 45543147/8 Fax : 25084124 2NH3 (g) 2SO3 (g) PCl3 (g) + Cl2 (g) 2HI (g) (104) . Dwarka.A(s). 4. Answer . of salt = ∴ pH = 1 1 1 pk w + pk a − log 2 2 2 2 39. mol After reaction pH = (D) Initial m.Success Magnet (Solutions) Physical Chemistry (B) Initial m. Sector-11. C(r). New Delhi-75 Ph.Regd. Plot No. mol After reaction pH = CH3COOH + NaOH 100 50 50 0 CH3COONa + H2O 0 50 0 50 pk a + log pk a + log [Salt ] [acid] 50 50 = = (C) pka CH3COOH + NaOH CH3COONa + H2O 75 0 0 75 0 75 Initial m. Office : Aakash Tower. Regd. of Cr = +6 (B) Peroxy linkage K2Cr2O7 ⇒ 2(+1) + 2x + 7 (–2) = 0 x=+6 n factor of K2Cr2O7 = 6 Equivalent mass = M 294 = = 49 6 6 Used in chromyl chloride test Aakash IIT-JEE . s). (C) (D) Addition of amount of reactants favours forward direction there is no effect of pressure and addition of inert gases of constant pressure because ∆ng = 0.Physical Chemistry Success Magnet (Solutions) 40. K1 = 10–7 (x – y) (x + y) H2S + OH– . New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (105) . 4.0999 M 42. r. C(p). C(q). Office : Aakash Tower.0001 = 0.A(q. K2 = 10–14 y (y + x) ∴ x >>> y K 1·C = 10 −7 × 0. Since (∆ng = –2) therefore high pressure favours forward direction and addition of inert gas at constant pressure favours backward reaction. Addition of reactant favours forward direction and ∆ng = 1 therefore addition of inert gas at constant pressure favours forward direction. No effect of addition of CaCO3.A(q. s) (A) Since CaCO3 is solid therefore its concentration remains constant. Since ∆ng= 1 therefor low pressure favour the forward direction reaction and addition of inert gas at constant pressure favours the forward direction because ∆ng > 0. Answer . (B) On addition of amount of reactant favours the forward direction and low temperature favours the forward direction because reaction is exothermic. B(p. B(r). q. Plot No. Answer .A(p. s). Dwarka.1 – 0. B(p. D(p) S2– + H2O C–x HS– + H2O x–y Since K1 >>> K2 [OH–] = pH = 10 [HS–] = [OH–] = 10–4 [H2S] = 10–14 [S2–] = C – x = 0. Since reaction is endothermic therefore high temperature favours forward direction reaction.1 = 10 − 4 M O –1 –1 (A) O +1 +2 +1 +1 O Cr +1 +1 O O –1 –1 Oxidation no. 41. s). Answer . C(p). D(p. q). D(p) –2 HS– + OH– . Sector-11. r). B(q).A(q). r). s) (A) +6 Reduction (Oxidising agent) S (B) –2 K2Cr2O7 → 2(1) + 2 x + 7 (–2) = 0 x = +6 +6 Reduction (Oxidising agent) 0 O (C) H2SO5 O= = S– O– O –H (two oxygen in form of peroxide) O–H Oxidation no. C(p.: 45543147/8 Fax : 25084124 (106) . of chromium = +6 Four oxygen in form of peroxide 44. 4. D(r) (A) Conductance = 1 ⎛ 1⎞ = ⎜ ⎟ = siemen resis tan ce ⎝ R ⎠ (B) ⎛a⎞ Resistivity = R⎜ ⎟ = ohm × m ⎝l⎠ 1⎛ l ⎞ ⎜ ⎟ = siemen m–1 R⎝a⎠ (C) Conductivity = (D) ⎛l⎞ Cell constant ⎜ ⎟ = m–1 ⎝a⎠ Aakash IIT-JEE . D(q. Answer . of chromium = + 6 H2SO4 → 2(+1) + x + 4(–2) = 0 x = +6 43. New Delhi-75 Ph.Regd. of sulphur = +6 –2 O –1 –1 +1 +2 +1 (D) CrO5 O O O O Cr +1 +1 –1 –1 Oxidation no.Success Magnet (Solutions) Physical Chemistry (C) K2CrO4 ⇒ 2(+1) + x + 4 (–2) = 0 x=+6 (D) CrO3 Oxidation no. Office : Aakash Tower. Answer . Dwarka. Sector-11.A(p). C(s). Plot No. B(p). s). D(p. D(p. s) (A) (B) (C) (D) Body diagnol will touch 2 corner and 1 body centre C4 axis diagnol will touch 2 face centre and 1 body centre Rectangular plane will contain 4 face centres.A(q).Physical Chemistry Success Magnet (Solutions) 45.: 45543147/8 Fax : 25084124 (107) .Regd. C(s). 4. Dwarka. s). s). Degree of dissociation increases with dilution and decreases with increase in conc. Answer . C(q). Answer . B(q. New Delhi-75 Ph. B(p.A(r). Sector-11. B(p. 47. Office : Aakash Tower. Resistance increases with increasing the distance between the plate resistance decreases with dilution. r) (A) (B) (C) (D) Specific conductance decreases with dilution Molar conductance increases with dilution and decreases with increase in conc. s). Plot No. r. Aakash IIT-JEE . C(q. s) (A) electrolysis ⎯→ H2 (g) ↑ + Cl2 ↑ H2O + KCl ⎯⎯ ⎯⎯ ⎯ cathode anode (K+ + OH– solution) (B) electrolysis ⎯→ Ag(s) + O 2 AgNO3 + H2O ⎯⎯ ⎯⎯ ⎯ cathode anode (H+ + NO3– solution) (C) ⎯→ Cu(s) + O2 CuSO4 + H2O ⎯⎯ ⎯⎯ ⎯ cathode electrolysis anode (H+ (D) + SO4– solution) ⎯→ H2SO4 ⎯⎯ ⎯⎯ ⎯ electrolysis H2 ↑ cathode + O2 ↑ anode H2O consumed so H2SO4 concentration increases and pH decreases. r). 2 edge centre and 1 body centre.A(q. D(q. Answer . D(p) ⎛ 1⎞ Conductance = ⎜ ⎟ = siemen ⎝R ⎠ (A) V = iR ⎛V⎞ R= ⎜ ⎟ ⎝ i ⎠ ⎛a⎞ Resistivity = R ⎜ ⎟ = ohm × m ⎝l⎠ ⎛l⎞ Cell constant ⎜ ⎟ = m–1 ⎝a⎠ (B) (C) (D) Resistance = volt amp 46. 2 face centre. C(p. s). B(s). Answer . 48. 4 edge centre and 1 body centre Body diagnol plane will contain 4 corner atom. s). q.A(p. α=β=γ 50.c. q). B(p. B(p. B(p).r. Office : Aakash Tower. like packing and Na+ is placed on all tetrahydral void. C(q). D(s) For Rock salt distance between nearest ions = For Fluorite distance between nearest ions = For CsCl distance between nearest ions = 54. S2 form hcp and Zn2+ are present in Half of tetrahedral voids. r. q). B(p). s). D(r) Rhombohedral a = b = c Cubic a = b = c Tetragonal a = b Hexagonal a = b ≠ 90° α = β = γ = 90° ≠c ≠c α = β = γ = 90° α = β = 90° γ = 120° 51. O–2 show c.c. D(r) (A) Rate = K[H2O2] first order reaction (B) Rate = K1[NH3 ] 1 + K 2 [NH3 ] a 2 3 a 4 3 a 2 High concentration of NH3.A(p.A(q). B(p. q) (A) (B) (C) (D) In Wurtzite structure.: 45543147/8 Fax : 25084124 (108) . r. C(q. s) (A) (B) (C) (D) Rock salt : Cl– constitutes fcc and Na+ is present in all oh voids Zinc blends : S2– constitute ccp and Zn2+ is present in alternate td voids Na2O : O2– constitute ccp and Na+ is present in all td voids CsCl ⇒ Cl– constitutes primitive cube and Cs+ is present at body centre.A(q. B(p). r).t. [NH3] K 1[NH3 ] K1 = =K K 2 [NH3 ] K 2 ⇒ Zero order Aakash IIT-JEE . Plot No.Success Magnet (Solutions) Physical Chemistry 49.A(r). C(q. Sector-11. Answer . 52.A(q. C(p.Regd. D(q. s). Answer . In Rock salt structure Cl– occupy f. C(s). r. Answer . D(r) (A) (B) (C) A 8× 1 8 B 12 × 1 4 = AB 3 A4 B4 + 4 = AB2 A B 1 1 8× +1 2× 8 2 4× 1 8 ⇒ A 2B ⇒ A 1 B C2 2 (D) A B 4× 1 4 C 1 2× + 1 2 53.A(p. q). r). q. 1 can be neglected w. and Na+ occupy octahedral voids. s). In Zinc Blend structure. s). q. s). 4. Answer . D(p. Answer .p. C(q. In antiflourite strucrure. Dwarka. Answer .p. r. New Delhi-75 Ph. S–2 form cubical closed packed structure and Zn+2 occupy half of tetrahedral voids. s). V∞ are volume of O2 obtained at time t and ∞ respectively K= V 2. K2[NH3] can be neglected w. New Delhi-75 Ph.A(p). D(q.t. Dwarka.Physical Chemistry Success Magnet (Solutions) (C) For very low concentration of NH3. C(q). Plot No. B(r). Vt and V∞ are volume of NaOH used at time 0.A(q). ⎯→ CH COOH + C H OH (C) CH3COOC2H5 + H2O ⎯⎯ 3 2 5 K= V − V0 2. Office : Aakash Tower. Vt are volume of KMnO4 used at time 0 and t respectively 56. t and ∞ respectively. r) (A) 4 2N2O 5 ⎯⎯⎯ → 4NO 2 + O 2 CCl K= 2.303 log 0 t Vt V0. Sector-11. 1. (D) 2H2O2 → 2H2O + O2 K= 2. t and ∞ respectively. Thus K1[NH3] first order. Answer . rt and r∞ are polarimeter readings after time 0.303 log ∞ t V∞ − Vt H+ Where V0.303 V∞ log t V∞ − Vt Vt = volume of O2 after time t V∞ = volume of O2 after ∞ time (B) H C12H22O11 + H2O ⎯⎯ ⎯→ C 6H12 O 6 + C 6HG12 O 6 d −sucrose d glu cos e l fructose + After the reaction is complete the equimolar mixture of glucose and fructose obtained is leavorotatory.303 log 0 t rt − r∞ r0. Answer .303 V∞ log t V∞ − Vt Vt.r.Regd.: 45543147/8 Fax : 25084124 . D(s) (A) (B) (C) For zero order reaction K = 2O3 x t 3O2 is first order reaction Hydrolysis of ester in basic medium is 2nd order reaction (109) Aakash IIT-JEE . (D) Rate = K[CH3CHO]2 Second order 55. C(s). B(p). 4. K= r − r∞ 2. Answer .4) 1 1 + 4 × 0.Success Magnet (Solutions) Physical Chemistry (D) K= 1 x(2a − x ) for third order reaction 2 t a 2 (a − x ) 2 57. Office : Aakash Tower.A(q). B(q. C(s).A(p). r). Plot No. Answer . C(r). New Delhi-75 Ph. B(r). C(p). D(q) (A) K4[Fe(CN)6] 1 (1 – α) i= i= 1 + 4α (Q α = 0. D(p) (A) (B) (C) 82Pb 207 → → → → 207 = 4n + 3 – actinium series 4 208 = 4n – thorium series 4 209 = 4n + 1 – neptunium series 4 208 82Pb 83Bi 209 (D) 82Pb 206 206 = 4n + 2 – uranium series 4 4K+ + [Fe(CN)6]4– 0 4α 0 before dissociation α after dissociation 60. Sector-11.A(p. D(p. Dwarka. 4. Answer . B(p.A(q). B(s). r). r).: 45543147/8 Fax : 25084124 (110) . Answer . D(q) (A) (B) (C) (D) Hydrolysis of ester in acidic medium is pseudounimolecular reaction Inversion of cane sugar is pseudounimolecular reaction Decomposition of H2O2 is first order reaction Hydrolysis of ester in alkaline medium is 2nd order reaction 59. s) α-emission increase β-emission decrease 1 0n n ratio p n ratio p → 1p1 + 0 –1 e (β-particle) Positron emission increases 1 1p n ratio p → 0n1 + 0 1e (positron ) Electron capture increases 1 1p n ratio p + –1e 0 → 0n 1 58.Regd.6 1 (B) NaCl Na+ + Cl– 1 (1 – α) 0 α 0 α before dissociation after dissociation Aakash IIT-JEE . C(p. s).4 = 2. s).9 1 Al3+ + 3Cl– 0 α 0 before dissociation i= (C) AlCl3 1 (1 – α) i= 3α after dissociation 1 + 3α (Q α = 0. s). not by true solutions. C(r.9 = 1.8) 1 1 + 2 × 0. C(r). D(p.6 1 i= 61. Answer . D(q) a and b : Brownian movement and Tyndall effect shown by colloidal solution. Sol : When solid is dispersed in liquid. Plot No. Sector-11. 63. B(p. C(p. s). r). B(q). q).8 = 2. r. (D) Depression in freezing point is a colligative property and the depression constant is also known as cryoscopic constant. (B) Osmotic pressure is a colligative property and it is measured by Berkley Hartley method. s). (C) Relative lowering in vapoure pressure is a colligative property. q) (A) Elevation in boiling point is a colligative property and the elevation constant is also known as ebullioscopic constant.8 1 Ca2+ + 2F– 0 α 0 before dissociation 2α after dissociation (D) CaF2 1 (1 – α) i= 1 + 2α (Q α = 0.6 = 2.: 45543147/8 Fax : 25084124 (111) . Answer . B(q. C(q). D(p) The ratio of effective molarity is equal to the ratio of osmotic pressure. q) Solid sol : When solid is dispersed in solid named solid sol.A(r). s). Answer . B(p. New Delhi-75 Ph.A(p.Physical Chemistry Success Magnet (Solutions) i= 1+ α (Q α = 0. Dwarka. s). D(r. Aakash IIT-JEE . Office : Aakash Tower.9) 1 1 + 0.A(q. Emulsion are colloidal solutions in which dispersed phase as well as dispersion medium are liquids.Regd.A(p. 4. For colloidal system particle size is 10–9 – 10–8 m. 62. Answer . r. True solutions are homogenous. Emulsion : When liquid is dispersed in liquid Gel : When liquid is dispersed in solid 64.6) 1 i= 1 + 3 × 0. suspension and emulsion. Answer . D(q) (A) Tyndall effect is due to Scattering of light by colloidal particles. Sector-11.Success Magnet (Solutions) Physical Chemistry 65. B(p.F : Subjective Type 1. s). (D) Activation of adsorbent is endothermic and removal of adsorbed material. (D) Electrophoresis is the movement of colloidal particles towards oppositely charged electrode. (C) Purple of cassius is the colloidal solution of gold. r) (A) Physical adsorption is exothermic. Section .A(p). (B) Brownian movement is the Zig-zag motion of colloidal particles. (C) Desorption is endothermic and removal of adsorbed material.55 gm 2 ⎢ 50 224 ⎦ ⎣ So mass percentage of Zn in the sample = 3. C(q.: 45543147/8 Fax : 25084124 (112) .Regd. r). D(r) (A) Argyrol is the colloidal solution of silver. Plot No. (B) Aquadag is the colloidal solution of graphite in water. B(p). Dwarka. (D) Colloidion is the colloidal solution of cellulose nitrate in ethanol.A(q). The reaction are 6Zn + 2KMnO4 + 9H2SO4 → 6ZnSO4 + K2SO4 + 2MnSO4 + H2 + 8H2O 1 O 2 → H2 O 2 x Let x mole of H2 is evolved and mole of O2 is required. 4. Answer . 2 x 3x Total moles of H2 and O2 = x + = 2 2 H2 + Molar contraction after spark = 22400 × x= 10 224 3x = 1500 2 3x 2 Equivalent of Zn = equivalent of KMnO4 + equivalent of H2SO4 = 20 + equivalent of H 2 1000 = 20 2 × 10 1 20 + = + 1000 224 50 224 Amount of Zn = 65 ⎡ 1 20 ⎤ + ⎥ = 3. 66. 10 Aakash IIT-JEE . (B) Chemisorption is exothermic and specific in nature. (C) Ultra filteration is the purification of colloids. Office : Aakash Tower. B(s). D(q.A(r). New Delhi-75 Ph. Answer .5% . C(p).55 × 100 = 35. 67. C(s). Dwarka.048 3 ⇒ …(ii) On solving. of FeSO4 = m.012 [Q I– → I+] 2 0.064 Zn and dil HCl will reduce all the Fe3+ ions to Fe2+ …(i) ions. Let the normality of FeC2O4 and FeSO4 solutions be x and y respectively milliequivalent of FeC2O4 = 75x milliequivalent of FeSO4 = 75y m.040 N.012 50 × = 0. 4. of K2Cr2O7 75x + 75y = 20 × 0.012 Equivalents of KI in 20 ml = 0.02 = 0. Sector-11. Aakash IIT-JEE . Plot No.015 – 0.005 moles of AgNO3 = 0.04 × 6 x + y = 0.04 Normality of FeC2O4 = 0. The equivalents of KIO3 reacting with 20 ml of KI solution in the second titration = 1 × 4 × 30 × 10 −3 10 [Q I5+ → I+] = 0.012 moles of KI in 20 ml = 0.85 % AgNO3 = 0.Physical Chemistry Success Magnet (Solutions) 2.024 N Normality of FeSO4 = 0.85 × 100 = 85%.: 45543147/8 Fax : 25084124 (113) . New Delhi-75 Ph. millimoles of Fe3+ ions from FeC2O4 = millimoles of Fe3+ ions from FeSO4 = 75 x (n factor of FeC2O4 = 3) 3 75 y (n factor of FeSO4 = 1) 1 75 x + 75 y final millimoles of Fe2+ = total millimoles of Fe3+ ions = 3 milliequivalent of Fe2+ ions = milliequivalent of KMnO4 ⇒ 75 x + 75 y = 36 × 0. of FeC2O4 + m.Regd.015 .eq. 3.005 mass of AgNO3 = 0.01 2 1 × 4 = 0.02 × 5 3 x + y = 0. Office : Aakash Tower.005 × 170 = 0.02 moles of KI in excess = 0.024 and y = 0.eq.eq.02 10 moles of KI consumed by AgNO3 = 0. 2 20 moles of KI in 50 ml = equivalents of KIO3 reacting with excess of KI = 50 × 10–3 × equivalents of KI in excess = 0. x = 0.01 = 0. 15 176 y = 1. Plot No.15 − 66 × 10 −3 × 0.7% 10 Aakash IIT-JEE .Regd.2793 × 100 = 2.2408 g % of H3PO4 in the sample = 0. At the methyl orange end point.408 % 10 2+ + 5.2793 gm ∴ y = 47 × 10 − 3 × 0.0714 = Equivalents of Fe2+ reacted Equivalents of Fe2+ produced = 0. Sector-11. Let the mass of H3PO4 and HIO3 in the sample are x and y gm. Mn H 2 O2 Fe S 2 ⎯ ⎯→ Fe 3+ + SO2 ⎯⎯ ⎯⎯→ Fe 2+ + Mn7 + ⎯⎯H ⎯ ⎯⎯ ⎯→ Fe 3+ (n =1) " 40 mL of 10 vol" Normality of H2O2 solution = 10 = 1. 4.6 Equivalents of H2O2 solution = 1.Success Magnet (Solutions) Physical Chemistry 4.0714 = Equivalents of Fe3+ reacted = moles of Fe3+ produced (n factor = 1) Moles of FeS2 = 0.0714 Mass of sulphur in FeS2 = 0. Dwarka.0714 × 2 × 32 = 4.15 = ⎜ × 1⎟ × 2⎟ + ⎜ ⎝ 98 ⎠ ⎝ 176 ⎠ …(2) Substracting equation (i) from (ii) ⎛ x ⎞ (85 × 10 −3 × 0.: 45543147/8 Fax : 25084124 (114) .57 Percentage of Sulphur in the sample = 4. New Delhi-75 Ph.15) = ⎜ ⎟ ⎝ 98 ⎠ 19 × 10 − 3 × 0. Office : Aakash Tower.785 5.793 % 10 % of HIO3 in the sample = +2 −1 +4 1.57 × 100 = 45.785 × 40 × 10–3 = 0.15 = x 98 x = 0.15 = ⎜ × 1⎟ × 1⎟ + ⎜ ⎝ 98 ⎠ ⎝ 176 ⎠ …(1) At the phenolphthalein end point Equivalents of NaOH = Equivalents of H3PO4 (n = 2) + equivalents of HIO3 ⎞ ⎛ x ⎞ ⎛ y 85 × 10 −3 × 0. Equivalents of NaOH = Equivalents of H3PO4 (n = 1) + Equivalents of HIO3 ⎞ ⎛ x ⎞ ⎛ y 66 × 10 −3 × 0.2408 × 100 = 12. 82 × 10 6 4 × 11.5 × 10 −9 ∴ n2 = 5. According to Bohr's calculation. 7. ⎡ 1 1⎤ = 1. 4.82 × 10 6 J / mole 9 4 × 11. New Delhi-75 Ph. Dwarka. ⎡ 1 1 1⎤ = R Hz 2 ⎢ 2 − 2 ⎥ λ ⎢ n1 n 2 ⎥ ⎣ ⎦ for n1 = 1.962 × 10 −17 = 1.Regd. n2 = ? 1 30.097 × 10 7 × 4 ⎢ 2 − 2 ⎥ n2 ⎥ ⎢ ⎣2 ⎦ According to de Broglie matter wavelength λ = We have.84 × 107 J/mole.82 × 10 6 9 9 = 1. Sector-11.023 × 1023 J mol–1 = 11.82 × 10 6 + + 11. second and third ionisation potential is 1 : 4 : 9. 8.82 × 10 6 J / mole 9 and second ionisation potential (IP2) = Therefore the energy of the reaction Li(g) → Li3+(g) + 3e– = IP1 + IP2 + IP3 = 11.962 × 0–17 × 6.18 × 10 −18 × (3)2 12 J atom −1 = 1. third ionisation energy of lithium (IP3) = 2.4 × 10 n2 = 2 −9 ⎡1 1⎤ = 1. p2 h2 1 2 = mv Kinetic energy E = = 2m λ2 (2 m) 2 h h = mv p ∴ λ= h 2 mE The rate of change of de Broglie wavelength λ with the kinetic energy E Aakash IIT-JEE .82 × 106 J/mole Since the ratio of first. Plot No.Physical Chemistry Success Magnet (Solutions) 6.: 45543147/8 Fax : 25084124 (115) . Office : Aakash Tower.097 × 10 7 × 4 ⎢ 2 − 2 ⎥ n2 ⎥ ⎢ ⎣1 ⎦ ∴ for n1 = 2 (first excited state) n2 = ? 1 108. Hence first ionisation potential (IP1) = 11. 2 × 36.208 ⎢134 .: 45543147/8 Fax : 25084124 (116) . Office : Aakash Tower. ⎜ ⎟ −34 2 dE (6.1 = 3. Sector-11.14 nm Ethreshold = hc 6.88.67 kJ 18 10.96 × 10 −21 J 23 NAV 6.03 × 1023 Average energy required/molecule = 6000 6000 = = 9. Aakash IIT-JEE . New Delhi-75 Ph. Plot No. 4. Energy of photon = E (required for melting) = Number of photons = 166.626 × 10 −34 × 3 × 10 8 = = 3.626 × 10 −34 × 3 × 10 8 105.78 and XH < XF XF = 1. Dwarka. = 6.023 × 10 11.6 − (104 .026 × 10 J.626 × 10 −34 × 3 × 10 8 = λ 230 × 10 −9 6.78 + 2.78 + XH = 1.6) 2 ⎥ ⎢ ⎥ ⎣ ⎦ XH ~ XF = 1. −9 10 ⎣ ⎦ hc 6.626 × 10 ) ⎝ ⎠at λ =1 Å 1 8 = RH λ 9 9.07 × 10 6 mJ −1 .626 × 10 −34 × 3 × 10 8 ⎡ 1 1 ⎤ –18 ⎢105. Let XF and XH are the electronegativity of F and H atm then XH ~ XF = 0. λ = 105.313 × 10 −19 = 5.208 EH − F − E H − H × EF − F ] 2 1 [ ] 1 1⎤2 ⎡ XH ~ XF = 0.Regd.67 × 10 3 3.313 × 10–19 J λ 600 × 10 −9 500 × 6 = 166 .Success Magnet (Solutions) 1 Physical Chemistry dλ = dE d(E) 2 · dE 2m h − 1 h λ3m =− 2 = −2× h 2 mE × E (1× 10 −10 )3 × 9.E.14 − 230 ⎥ = 1.14 × 10 −9 Eincident = K.1× 10 −31 ⎛ dλ ⎞ =− = −2. 6 × 10–19 × 10–3 kJ = 173. Moles of Mg = 1 24 These mole of Mg will be converted to Mg+ and Mg2+ assume a mole of Mg+ are formed then ⎛ 1 ⎞ a × 740 + ⎜ − a ⎟ × 2190 = 50 ⎝ 24 ⎠ a = 0. 13.Regd. New Delhi-75 Ph. Sector-11.41 – 3.80 × 6.023 × 1023 eV = 1. Office : Aakash Tower.: 45543147/8 Fax : 25084124 (117) . Plot No.28 1 24 % of Mg2+ = 31.80 eV ∴ ∆H of reaction per mole = 1. The molecule BrF5 has bromine atom in centre surrounded by six electron pair. Figure shows the structure of BrF5 molecule.02845 % of Mg+ = 0. F short F 90° F Br F 85° F long Aakash IIT-JEE . As we know that the non-bonded pair of electrons occupied more space than bonding electron pairs so geometry of BrF5 depart from that of a regular octahedron. F 15. It is better to imagine six electrons pair around the bromine atom directed towards the corner of octahedron with five of these corners occupied by fluorine atoms. (i) (ii) NO2 is paramagnetic.61 = 1.023 × 1023 × 1. V-shaped with sp hybridisation of I NO2+ is linear with bond angle 180° NO2– is bent with bond angle slightly less than 120°. The NO2– ion has the longest and the weakest bonds of the three. Dwarka.7 kJ.72.02845 × 100 = 68.Physical Chemistry Success Magnet (Solutions) 12. pπ – pπ back bonding in BF3 gives some double bond character which is absent in BF4–. (i) O (ii) O F Xe F F Square pyramidal 3 2 with sp d hybridisation of Xe F F F Trigonal bipyramidal 3 with sp d hybridisation of Cl O Cl (iii) ⎡Cl ⎢Cl ⎣ O I ⎤ ⎥ Cl ⎦ Cl – (iv) ⎡ ⎢ ⎣Cl 3 I ⎤ ⎥ Cl⎦ + Square pyramidal 3 2 with sp d hybridisation of I 16. (iii) The NO2+ ion has the shortest and strongest bond. ∆H/molecule of Li+ and Cl– = IP1 + EA = 5. 14. 17. 4.80 × 6. five of which are used to form bonds to fluorine atoms. 5 300 R moles of Cl2O = Cl2O → Cl2 + After reaction 1 O 2 2 total moles of Cl2 = moles of O2 = V 1.78 atm Partial pressure of helium in the mixture (1 – 0.: 45543147/8 Fax : 25084124 (118) . Let the formula of nitrogen hydride be NxHy and the initial volume of nitrogen hydride by aml NxHy ⇒ initial a 0 x y N2 (g) + H2 (g) 2 2 0 ax 2 0 ay 2 Thus ax ay + = 2a 2 2 x+y=4 when O2 is added …(i) H2 (g) + 1 O 2 (g) → H2O(l) 2 Since the gaseous mixture was needed to pass through alkaline pyrogallol solution.43 x = 0. Let the volume of each vessel be V L moles of Cl2 = V ×1 300 R V × 1 . After passing through pyrogallol solution volume left would be corresponding to nitrogen only Thus ax a = 2 2 (x = 1) y=4–1 Formula of nitrogen hydride is NH3. 20. Effective molar mass of the gas mixture = dRT 0.78 × 1 = 0.6 × 0.78) × 1 = 0.Regd.5 V 5V + = = 300 R 300 R 300 R 600 R 1.5 V = 2 × 300 R 600 R Aakash IIT-JEE . IV = II < III < I. Plot No. Office : Aakash Tower.5 V 2.22 atm. 4. The oxygen was left in excess and hydrogen would have been consumed completely.43 g / mol p 1 Let the mole fraction of methane in the gaseous mixture be x x(16) + (1 – x)4 = 13. Dwarka. 21. 19. Sector-11.Success Magnet (Solutions) Physical Chemistry 18.78 Therefor.5 V 1.082 × 273 = = 13. partial pressure of methane in the mixture = 0. New Delhi-75 Ph. ∴ At high temperature motion of molecules is so fast that P >>> a V2 Aakash IIT-JEE .: 45543147/8 Fax : 25084124 (119) .65 × 10 V R × 300 × = 1. Dwarka.695 ~ 1. Plot No.5 V =⎜ − x⎟ ⎜ ⎟ V V ⎝ 600 R ⎠ −3 x = 5. R V 22.Physical Chemistry Success Magnet (Solutions) 5V 1. Let the moles of gases left in one of vessel maintained at 27°C ⎛ 6 .Regd. New Delhi-75 Ph. For one mole of a real gas a ⎞ ⎛ ⎜ P + 2 ⎟ ( V − b) = RT V ⎠ ⎝ If volume correction is ignored a ⎞ ⎛ ⎜ P + 2 ⎟ v = RT V ⎠ ⎝ at STP V = 22. Thus moles of gases in another vessel maintained at 52°C would be ⎜ ⎜ 600 R − x ⎟ ⎟.4 L) At low pressure volume is so high that b <<< V ∴ a ⎞ ⎛ ⎜ P + 2 ⎟ V = RT V ⎠ ⎝ PV + a = RT V a V PV = RT − PV a = 1− RT RTV ∴ Z < 1.5 V Total moles of gases in two vessel after the reaction = 600 R + 600 R = 600 R When the two vessel are kept in water bath at different temperatures diffusion will take place till the pressure of the gases in two vessel becomes equal. Office : Aakash Tower. ⎝ ⎠ ⎞ R × 325 x × R × 300 ⎛ 6.65 × 10 V R −3 P = 5.4 L ∴ Pr < Pi (= 1 atm) If pressure correction is ignored P(V – b) = RT at STP Pr = Pi = 1 atm ∴ Vr > Vi (22. 4.7 atm . Sector-11.5 V ⎞ be x.5 V 6. 10 V2 25. New Delhi-75 Ph. (i) rv = rO 2 32 = 1. Office : Aakash Tower.61 × 104 J of course work done on the gas W is positive for compression.314 × 400 log⎜ ⎟ = 1.25 × 10 −3 × 1. Sector-11. Aakash IIT-JEE .Success Magnet (Solutions) Physical Chemistry P(V – b) = RT PV – Pb = RT PV = RT + Pb PV Pb = 1+ RT RT ∴ Z > 1.61 × 104 J P1 ⎝ 1⎠ also q = –W = –1.63 × 10–6 atm.4 J.44 cal/gm W = –P(V2 – V1) = –∆nRT = − ⎜ ⎠ ⎝ 18 Now q = 540 cal/gm ∆E = ∆q + W = 540 – 41. Since T is constant ∆U = 0 W = 2.33)2 = 18.314 × 400 log = –30635.303 × 3 × 8.303 RT 2.85 × 10 3 = −5.: 45543147/8 Fax : 25084124 (120) .224.303 × 4 × 8. The vapour pressure of mercury (in atm) is equal to Kp for the reaction Hg(l) Hg(g).303 nRT log 1 V1 = 2.56 cal/gm For isothermal process ∆G = 2. Because the standard state for elemental mercury is the pure liquid ∆Gf° = 0 for Hg(l) and ∆G° for the vapourisation reaction is simply equal to ∆Gf° for 1 mol of Hg(g) − ΔG° − 31.36 (iii) Compression factor (Z) = PV 50. 23.58 = log Kp = 2.303nRT log P2 ⎛5⎞ = 2.25 × 10 −3 m 3 0.303 × 8.013 × 10 5 = = 1.63 × 10–6 Since Kp is defined in units of atmosphere therefore the vapour pressure of mercury at 25°C is 2. Dwarka.44 = 498. Kp = PHg Note that Hg(l) is omitted from equilibrium constant expression because it is pure liquid.09 g / mol (ii) Molar volume V = 18 . 26.09 = 50 .Regd. Work done for irreversible process ⎛ 1 ⎞ − 0 ⎟ × 2 × 373 = –41.33 M 32 M= (1. 4.314 × 298 Kp = antilog(–5. Plot No.58) = 2. The heat q is negative because heat must flow from the gas to surrounding of constant temperature maintain the temperature of the gas at 400 K when it is compressed. 24.314 × 500 (iv) Repulsive forces dominate since the actual density is less than the density if it were ideal. RT 8. 35 kJ .Regd.17 kJ 1 1 Cl2O7(g) → ClO3 + O 2 2 1 1 1 O (g) + O(g) → O3(g) 2 2 2 2 O(g) → 1 O (g) 2 2 ClO2(g) + O(g) → ClO3(g) . ∆H3 = –53. Office : Aakash Tower.5 kJ . ∆H4 = –249. The enthalpy change for the given reaction is calculated as ClO2(g) + 1 1 O(3(g) → Cl2O7(g) .: 45543147/8 Fax : 25084124 (121) .5 4 = pKa + log[salt]1 + log2 …(i) …(ii) pH2 = 6 = pKa + log[salt]2 + log2 Equation (ii) – (i) gives log[salt]2 – log[salt]1 = 2 [salt ] 2 = 100 [salt ]1 After mixing equal volume of 2 buffers [ Acid] f = 0. 28.5 M [acid]1 2 [Salt ] f = [Salt ] f = [salt]1 + [salt] 2 2 101 [salt ]1 2 101 [salt ]1 2 [acid]1 (pH)f = pK a + log Aakash IIT-JEE .04 – 566.5 + 498. Dwarka.2 kJ. The given buffer being an acidic buffer its pH is given as (pH)1 = pKa + log 4 = pKa + log [salt ]1 [acid]1 [salt ]1 0.85 kJ 2 2 .Physical Chemistry Success Magnet (Solutions) 27. New Delhi-75 Ph.5 + 0.34 – 2 ECl – 0 ECl – O = 141. Sector-11.87 kJ Using the enthalpy of formation of ClO2(g) we calculate the bond enthalpy of Cl = O 1 Cl (g) + O2(g) → ClO2(g) 2 2 ∆H = 102.7 = 605. ∆H = –280. ∆H1 = –121.17 kJ 2ClO2(g) + O3(g) → Cl2O7(g) ∆H = –243. ∆H2 = 143. Plot No. 4.5 = 0.5 = 170.34 – 2 ECl = 0 ∴ ECl = 0 = 283. V.4 = 7.D.3 × 0.: 45543147/8 Fax : 25084124 .181 atm 2 (Kp)1 = pNH3 × PHI = (0.3 × 10 −2 mol / litre .61 ⎝ ⎠ = 2. 29. 0. Kp = Kc(RT)∆n. 4.6 1 X = 0. Sector-11. Dwarka.362 atm 760 Total pressure in atm = PNH3 = PHI = 0. Office : Aakash Tower. New Delhi-75 Ph.033 …(i) (122) Aakash IIT-JEE .61 Kp = PCF3COOH P( CF3COOH)2 We know. Here ∆n = 1 Kc = Kp RT = 2.6 2p 2×1 2CF3COOH(g) 2 2X (CF3COOH)2(g) 1 1–X initial vapour density equilibrium moles = equilibrium vapour density initial moles 114 1 + x = 70.61 ⎞ × 1⎟ ⎜ 1. Plot No.Success Magnet (Solutions) Physical Chemistry (pH)f = pK a + log [ salt]1 101 101 + log = (pH)1 + log [acid]1 2 2 (pH)f = 5.0821× 400 = = 70. = dRT 4.703.Regd.39 ×1 1.181)2 ~ 0.033 atm 2 But when HI starts dissociating the pressure will increase further when both the equilibria are established simultaneously NH4I(s) 2HI(g) x–y NH3(g) + HI(g) x x–y H2(g) + I2(g) y 2 y 2 (Kp)1 = x(x – y) = 0.362 = 0. Initially the constant pressure of 275 mm of Hg is obtained when gaseous NH3 and HI are in equilibrium with solid NH4I NH4I(s) NH3(g) + HI(g) 275 = 0.4 atm = 0.61 ⎛ 2 × 0.0821× 400 30. 35 g/litre = 0.8 = 0.05 × 0. Solubility of CH3COOAg = 8.6 × 10–49 H+ + HS– .04 ) + Ptotal = 0.05 mol/L Hence Ksp(CH3COOAg) = 0.404 atm = 307. New Delhi-75 Ph.04 0.: 45543147/8 Fax : 25084124 (123) . 32. Office : Aakash Tower. 4. Dwarka.015 (Kp)2 = ( x − y )2 y = 0.202 + (0.04 + 2 2 CH3COO– + Ag+ (x – y) x CH3COOH 0 y ∴ Ka = 1.202 − 0.37 M 167 0.1 Aakash IIT-JEE . Sector-11.37 – y)0. Plot No.3632 and 1 y = K a ( x − y )10 −3 61 .202 and y = 0.87 × 10–5 M.05 = 25 × 10–4 M2 Let the solubility of silver acetate in acidic buffer be x mol/L and y moles/litre be the amount of CH3COO– reacting with H+ in the buffer to form CH3COOH CH3COOAg CH3COO– + H+ x x–y 10–3 10–3 (buffer) Solubility of CH3COOAg in buffer x = Ksp = (x – y)x (0. 31. K2 = Neglecting second ionisation let x is the amount of HS– produced then x2 = 1× 10 −7 0. K1 = [H+ ] [HS − ] = 1× 10 −7 [H2 S] [H + ] [S 2− ] [HS − ] = 1× 10 −14 HS– H+ + S2– .37 = 25 × 10–4 ∴ y = 0. Ag2S H2S 2Ag+ + S2– .04 atm Ptotal = PNH3 + PHI + PH2 + PI2 = 0.04 mm of Hg.015 2 ( x − y) …(ii) On solving equation (i) and (ii) we get x = 0.Regd. Ksp = 1.Physical Chemistry Success Magnet (Solutions) ⎛ y ⎞⎛ y ⎞ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ = 0. 34. 33.6 × 10 −49 1× 10 −14 = 4 × 10–18 M. Plot No.99 × 10–1 atm (PO 2 )3 (PO 3 )2 = 1.Success Magnet (Solutions) Physical Chemistry x = 1 × 10–4 K2 = [H+ ] [S 2− ] [HS ] − = 1× 10 −14 Q ([H+) ~ [HS–]) 1.3 × 10 57 Kp = = (1.0591 log ⎢ + 2 − 2⎥ 2 ⎥ ⎢ [H ] [OH ] ⎦ ⎣ (124) Aakash IIT-JEE .46 × 10–30 atm. 4. Number of molecules of ozone present in 10 million cubic metre of air at 25°C = 2. New Delhi-75 Ph.6 × 10 −49 1× 10 −14 [S2–] = 1 × 10–14 K sp [S 2 − ] = [Ag2+]2 = [Ag+] = 1.99 × 10 −1 )3 (PO3 )2 = 1. Dwarka.3 × 10 57 PO 3 = 2.023 × 1023 = 6.46 × 10 −30 atm 1 m3 = 1000 litre Volume = 1010 litre.21× 720 mm of Hg = 1.Regd.21 So partial pressure of oxygen in air PO2 = 0. Sector-11.4) − ⎡ K sp × PH2 ⎤ 0. (i) The given cell representation can be reduced to ⎛ K sp ⎞ + −12 ⎟ Cd | Cd2 + ⎜ ⎜ (OH− )2 ⎟ H (10 M) H2 (1 atm) Pt ⎝ ⎠ The cell reaction is Cd + 2H+ → Cd2+ + H2 Ecell = E ° cell − [Cd2 + ] PH2 0.46 × 10 −30 × 1010 = 1× 10 −21 mole 0. T = 298 K. Office : Aakash Tower.0591 log 2 [H+ ] 2 0 = 0 − ( −0. Mole fraction of oxygen in air = 0.082 × 298 Number of ozone molecules = 1 × 10–21 × 6. P = 2.02 × 102.: 45543147/8 Fax : 25084124 . 095 g of ammonia = 0. Dwarka.55 × 2 [n-factor for H2 = 2] 22.386 g of NaNO2 = 0.373 Hydrogen gas will evolve through reaction 2H+ + 2e → H2 3. 0.373 ∴ Current efficiency for NaNO2 = Ammonia will be formed by the reduction of nitrate following the equation 9H+ + NO3– + 8e– → NH3 + 3H2O 0. 35.Regd.386 ×2F Amount of charge used to obtain 0.4 Equivalent of H2 evolved = Charge used to liberate hydrogen = 0.045 × 100 = 12.386 × 2 [n-factor of NO2– = 2] 69 0.44 × 1013 × 10–28 = 3.386 g of NaNO2 = 69 0.0591 ⎛ ⎟ log ⎜ ⎜ K2 ⎟ 2 w ⎠ ⎝ ⎛ K sp ⎞ 0. (ii) ∆G = –nFE = 0 ⎛ ∂E ⎞ ⎟ = 2 × 96500 × 0. New Delhi-75 Ph.095 × 8 F = 0.028 kJ.55 22.373 Aakash IIT-JEE .045 F 17 Equivalent of NH3 obtained = Charge used for this purpose = Current efficiency for NH3 = 0.: 45543147/8 Fax : 25084124 (125) . Office : Aakash Tower.0591 ⎝ w ⎠ Ksp = 3. 4.002 = 386 JK −1 ∆S = nF ⎜ ⎝ ∂T ⎠ P ∆H = ∆G + T∆S = 0 + 298 × 386 = 115028 J = 115. Sector-11.536 log ⎜ 2 ⎟ = ⎜ K ⎟ 0.095 × 8 [n-factor of NH3 is 8] 17 0.373 F 96500 ∴ Equivalent of NO2– reduced from NO3– = 0.3169 × 100 = 84 .3169 F Current efficiency for H2 = 0.95 % 0.386 mole NaNO2 = mole NO2– 69 69 5 × 2 × 60 × 60 = 0.011 × 100 = 2.Physical Chemistry Success Magnet (Solutions) 0 = 0.4 − K sp × 1 ⎞ 0. Total amount of charge passed = Sodium nitrite will be formed as 2H+ + NO3– + 2e → NO2– + H2O 0.386 0.4 × 2 = 13.095 mole NH3 17 0.55 litre of hydrogen = 3.06% 0. Plot No.44 × 10–15 M3.4 3.96% . C C(1 – α) CH3COO– + H+ 0 Cα 0 Cα Dissociation constant for acetic acid 37.1 log = 0.71 kJ Cell reaction H2O(l ) + 1 O (g) + H2(g) → 2H+ (aq) + 2OH–(aq) 2 2 Aakash IIT-JEE .059 0.701 = − ( −0.0316 M 0.(a).23 kJ 2 2 H2O(l ) → H+(aq) + OH–(aq). Sector-11.2 = = 0. The degree of dissociation of acetic acid is given by α= λc 5 .264 g.: 45543147/8 Fax : 25084124 (126) .Success Magnet (Solutions) Physical Chemistry 35. New Delhi-75 Ph. ∆G = 79. 2 (10 −7 )2 38. Office : Aakash Tower.01× 2 = E 96500 t = 19.013)2 = 1.1 log + 2 2 [H ] Since sufficient NaOH is to be added to cathode compartment to consume entire H+ the equivalent of HCl must be equal to the equivalent of NaOH. After the addition of sufficient NaOH the solution becomes neutral and we have [H+] = 10–7 M ∴ Ecell = 0.7 CH3COOH initially At equilibrium Ka = Cα2 = 0.69 × 10–5 M.3765 volt.059 log – 2 [H+ ] 2 0. Let the weight of NaOH added be x gm x × 1 = 0.76 − 0. Plot No. For calculating the minimum weight of NaOH which is supposed to be added to cathode compartment we are required to know the [H+] present in this compartment.3 × 104 s 36. 4.1 × (0.Regd. H2(g) + 1 O (g) → H2O(l ). The reaction occuring in the cell is Zn(s) + 2H+ → Zn2+ + H2(g) ∴ Ecell = E° + − E° H / H2 Zn 2 +/ Zn PH2 [ Zn 2 + ] 0.059 1× 0.013 λ ∞ 390. Answer (2) W it = E F IIT-JEE 2008 W 10 × 10 −3 × t = 0.0316 × 1 40 x = 1. Dwarka.76) − [H+] = 0. ∆G1 = –237. 34 − ( −0. Sector-11.71) = –77.9 kJ ∆H3 of cell reaction = –285. ∆H1 = –285.23 kJ 2 2 2[H2O(l ) → H+(aq) + OH+(aq)].81 ⎛ ∂E ⎞ = = −0.58 + 2(56. Plot No.403 V H2(g) + 1 O (g) → H2O(l ).059 [ Zn2 + ] log 2 [Cu2+ ] 1 = 0.763) − [ Zn2+ ] [Cu2+ ] = 3101 Let x be the amount of Cu2+ that is converted into Cu when the cell potential drops to 1. ∆G3 = –237.23 + (2 × 79. New Delhi-75 Ph. ∆G2 = 2 × 79. ∆H2 = 56.92 × 105 coulomb. Office : Aakash Tower.9) = –172.00164 VK −1 ⎜ ⎟ = ∂ T nFT 2 × 96500 × 298 ⎝ ⎠p 39.9993 × 96500 = 1.9993 mole 1− x Each half cell reaction requires 2 electrons. x = 0.303 × RT [ Zn2+ ] log 2F [Cu2+ ] 0.: 45543147/8 Fax : 25084124 (127) .85 kJ 2 2 H2O(l) → H+(aq) + OH–(aq).05 + 77.05 kJ.81 kJ 2 2 ∆G3 = –nFEcell = –2 ×96500 Ecell ∴ Ecell = 0. 40.71 kJ H2O(l ) + 1 O (g) + H2(g) → 2H+(aq) + 2OH–(aq). Dwarka. Ksp = [Hg22+][Br –]2 K sp [Br ] − 2 [Hg22+]anode = = K sp (0.Physical Chemistry Success Magnet (Solutions) The given cell reaction can be obtained from equation (i) are (ii) as H2(g) + 1 O (g) → H2O(l ). Cell reaction is Zn + Cu2+ → Zn2+ + Cu Ecell = E°cell – 2. Hg2Br2(s) Hg22+(aq) + 2Br–(aq). ∴ [ Zn2+ ] [Cu ] 2+ = 1+ x 1− x Hence 1+ x = 3101 .1) 2 reduction reaction of mercury half cell is Hg22+(aq) + 2e → 2Hg(l) Aakash IIT-JEE .0 V. ΔH3 − ΔG 3 − 172. ∆G1 = –237. 4. Therefore quantity of electricity delivered = 2 × 0.Regd. 1)8 E cell = EMnO − + 2+ 4 . Office : Aakash Tower. K sp = 41.75 4 4 4 = 0.1)(0.059 log K sp 2 reduction reaction of manganese half cell MnO4– + 8H+ + 5e → Mn2+ + 4H2O E MnO4 – . Plot No.1)2 0.Regd. 4. H / Mn − EHg2+ / Hg 2 ⎡ 0.1) log = 1. New Delhi-75 Ph.1)2 ⎤ 1.023 × 10 23 = = 8.1) 2 10 20 = 1× 10 −22 M3 .Success Magnet (Solutions) Physical Chemistry E + Hg2 2 / Hg = 0.5 8 (iv) Total fraction occupied = 42.112 × 10–24 cm³ = Z ×M a3 × NA 4 × 239 195.80 − log ⎥ 2 K sp ⎥ ⎢ ⎣ ⎦ log (0.21 = 1.51 − 0. Sector-11.42 V 5 (0.: 45543147/8 Fax : 25084124 (128) .059 (0.059 (0.1)2 = 20 K sp (0.112 × 10 −24 × 6.80 − (0.42 − ⎢0. H + / Mn2 + = 1.58 12 = 290 × 2 = 580 pm V of unit cell Density = 195. Edge length of unit cell 7 = 0.135 g/cm3 Aakash IIT-JEE . Dwarka. (i) A=8× 1 1 +6× =4 8 2 B = 12 × 1 +4=3+4 =7 4 Formula = A4B7 (ii) oh voids occupied Fraction occupied (iii) td void occupied Fraction occupied = = = = 3 (by edge centre) 3 = 0. Either oh or td voids are occupied. 4.414 (for maximum) r rA = 0.9799 log K = – 2. log K = log A – 2. of cationic vacancies = 6.693 0.303 × 1. Office : Aakash Tower.414 × 152.987 × 800 log K = log(1.35 pm 45. Maximum radius of atom (rA) is possible in case of oh & Minimum radius of atom (rA) is possible in case of td ∴ rA = 0. New Delhi-75 Ph.26 × 1013) – log K = 13.7 pm 4 ∴ Max.`21 pm Min.5 × 10 3 2.3194 × 10–3 sec–1 t1 / 2 = 0.3194 × 10 − 3 Aakash IIT-JEE . Plot No.8796 K = 1.: 45543147/8 Fax : 25084124 (129) . Dwarka.225 × 152.303 RT 58. radius possible = 0. K 1.02 × 10–5 × 1023 = 6. Doping of AlCl3–10 SrCl2 crystals bring replacement 3 Sr2+ ion by 2 Al3+ ions This produces 1 cation vacancy ∴ Cation vacancy produced by AlCl3 = 10–5 mol/mol of SrCl2 ∴ No.1003 – 15.225 (for minimum) r (r is radius of molecules constituting fcc) In fcc 4r = 2a r = 2 a = 4 2 × 432 = 152.02 × 1018 Ea 46.Regd. rsimple : cubic rbcc : rfcc a 2 : 3 a a : 4 2 2 2 : 3 : 2 44. radius possible = 0.693 = = 525 seconds.Physical Chemistry Success Magnet (Solutions) 43.7 ≈ 34. Sector-11.7 ≈ 63. 8 a= 5 2a + At t = 30 minutes a+ x= 3x = 284.7 gm N= 2 Disintegrated amount = 1 – 0. Office : Aakash Tower. For the (i) equation K eq = [NOBr2 ] [NO][Br2 ] or [NOBr2] = Keq[NO][Br2] Since. 210 ⎯ ⎯→ 2N2O5(g) a (a – x) 0 4NO2(g) + O2(g) 0 2x 2a 0 x 2 a 2 Number of moles ∝ pressure At t = ∞ a = 584 . 48. Total time = n × t1/2 69. At t = 0 At t = 30 min At t = ∞ 22400 × 0. Sector-11.5 2 284 .5 2 584.303 233.3 = 32 cm2 at STP.8 3 K= 2. 3r2 is reaction intermediate rate = K·[NOBr2][NO] rate = K·Keq[NO][Br2][NO] rate = K′[NO]2[Br2].Regd. Dwarka. New Delhi-75 Ph.8 log 30 200 K= K = 5.5 − 233.2 = n × 138.5 × 2 = 233. Plot No. Aakash IIT-JEE .7 = 0.303 a log t (a − x ) 2.206 × 10–3 minute–1.: 45543147/8 Fax : 25084124 (130) . 4. N0. volume of helium = 49.8 × 2 = 33.4 n= 1 2 n 1 ⎛ 1⎞ ⎛ 1⎞2 N = N0 ⎜ ⎟ = 1⎜ ⎟ ⎝2⎠ ⎝2⎠ 1 = 0.Success Magnet (Solutions) Physical Chemistry 47.3 gm −α 210 ⎯→ 206 84 Po ⎯⎯ 82 Pb So. m AB 4 = 1000 × 5.59 Molecular mass of B. x + 2y = 110.3 × 20 mAB2 = 110.2 i= 1 + α 1 + 0..0417 ∆Tf = 0. 2 × 10 −3 × 10 3 2 1000 60 m= = × 800 60 800 1000 m = 0 .: 45543147/8 Fax : 25084124 (131) . atomic weight of A is x and atomic weight of B is y So.093 Freezing point of solution = 0 – 0. Sector-11... For AB2. x = 25.1× 1 1.(1) .Regd. Office : Aakash Tower.Physical Chemistry Success Magnet (Solutions) 50. Plot No..64 . New Delhi-75 Ph.(2) 51. 4.093°C 52.2 1 1 ∆Tf = i × kf × m ∆Tf = 1.1× 1 2.2 = = 1.87 x + 4y = 196. α = 0.....2 × 1. For KCN.3 × 20 mAB4 = 196....704 = 1 × 1..87 For AB4.1892 i=2 Aakash IIT-JEE ..86 × 0. y = 42.0417 CH3COOH 1 (1− α ) CH3COO − + H 0 0 α α Given. Dwarka.. mAB2 = mAB2 = 1000 × k f × w ΔT × W 1000 × 5. ∆Tf = i × kf × m 0...15 By solving (1) & (2) Molecular mass of A.86 × 0.093 = – 0..15 Suppose.. 095 ΔTf m= k f 0.Regd.53 1. Whenever a mixture of gases is allowed to come in contact with a particular adsorbent under the same conditions.1892 – 0.095 n) + 0. 55.2849 n≈2 − So.1892 (0.095 – x) + nCN− 0. e.1892 + (0. 4.095 n = 0. Z and is given by Z= 4 πημ D Where η is called coefficient of viscosity.: 45543147/8 Fax : 25084124 (132) .1892 – nx) − Hg(CN)n n+ 2 0 x x may be taken as approx. D is dielectric constant of the medium and µ is the velocity of the colloidal particles when an electric field is applied. O2 etc. 54. − So.86 53.g. yet moisture is adsorbed more strongly on silica than the other gases of the air. 0. total molarity = K + + nCN− + Hg(CN)n n +2 = 0.025 × 1000 = 25 (mg) Aakash IIT-JEE .4734 – 0.095 n) + 0. [AgI] I– [AgI] Ag+ [As2S3] S2– + + solid + + K+ NO3– 2H+ – – – – Fixed layer Diffused layer Potential difference across this electrical double layer is called zeta potential or electrokinetic potential. New Delhi-75 Ph.095 due to the less amount of Hg(CN)2 & K+ is also present due to complete dissociation..1892 + (0. Plot No.1892 – 0. Office : Aakash Tower.095 = 0. Gold number is the weight of protective colloid in milligrams which prevents the coagulation of 10 cm3 of gold sol. Gold number of starch = 0.095 (0.Success Magnet (Solutions) Physical Chemistry This show that KCN is completely dissociated. the complex becomes Hg(CN)2 4 0. air besides moisture contains a number of gases such as N2. Dwarka. Sector-11. the more strongly adsorbable adsorbate is adsorbed to a greater extent irrespective of its amount present. Hg(CN)2 At t = 0 At t = t 0. There is formation of an electrical double layer of opposite charges on the surface of colloidal particles.