# Unit-I - Physical Chemistry - Solution(Final)

September 24, 2017 | Author: jassyj33 | Category: Chemical Equilibrium, Atomic Orbital, Chemical Bond, Cathode, Redox

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UNIT

Physical Chemistry

1

Section A : Straight Objective Type 1.

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.

x × 3y 3 xy = 2z 2z

Answer (2) Number of moles =

=

weight molecular weight 2.56 256

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 =

1.62 = moles of CaCl2 56

Mass of CaCl2 =

1.62 × 111 = 3.21 gm 56

%= 4.

3.21 × 100 = 32.1% 10

2O3

1000 – 3x

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

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x = 112 ml Volume of O3 at STP = 224 ml

224 = 0.01 22400

moles of O3 =

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.

x1y1 y2

⎞ ⎛y − x1 = x1 ⎜ 1 − 1⎟ ⎟ ⎜y ⎠ ⎝ 2

Answer (1) 2IClx ⎯→ I2 +

x Cl 2 2

moles of Cl2 =

112 = 5 × 10 −3 22400

moles of IClx =

1.625 127 + 35.5 x

∴ moles of Cl2 =

x ⎛⎜ 1.625 ⎞⎟ = 5 × 10 −3 2 ⎝ 127 + 35.5 x ⎠

1.625x = 10 × 127 × 10–3 + 10 (35.5 10–3 x) x (1.625 – 0.355) = 1.27 x= 7.

1.27 = 1 1.27

Answer (4) 2KClO3 ⎯→ KCl + KClO4 + O2 Molecular mass of KClO3

= (39 + 35.5 + 16 × 3) gm = 122.5

moles of KClO3 =

12.25 = 0.1 122.5

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moles of pure KClO3 = 0.1 ×

75 100

= 0.075 mass of residue = 9.1875 – 0.32 = 8.85 gm 8.

x F2 ⎯⎯→ XeF x 2

1 mole

1 mole

131 gm

(131 + 19x) gm

∴ 2 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 =

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 gm H2 gas. 2 x + 48

6( 0.1596 ) = 6 × 10 −3 2x + 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

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11. Answer (2) KMnO4 + H2O2 ⎯→ Product n=5

n=2

milli equivalents of KMnO4 = milli equivalents of H2O2 100 × 1 × 5 = milli eq. 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 =

500 ml 3

12. Answer (1) K2Cr2O7 + 14 HCl ⎯→ 2KCl + 2 CrCl3 + 7H2O + 3Cl2 Q 14 mole HCl produces = 3 moles Cl2 ∴ 1 mole HCl produces =

3 moles Cl2 14

MnO2 + 4 HCl ⎯→ MnCl2 + 2H2O + Cl2 1 mole

1 mole

∴ moles of MnO2 = Mass of MnO2 =

3 14 3 × 87 gm 14

= 18.642 gm 13. Answer (4) MnO2 + 4HCl ⎯→ MnCl2 + Cl2 + 2H2O equivalent mass of Cl2 =

71 = 35.5 2

6NaOH + Cl2 ⎯→ 5NaCl + NaClO3 + 3H2O equivalent mas of Cl2 =

6 × 71 = 42.6 10

14. Answer (3) n factor of HCl =

6 =3 14 7

equivalent mass of HCl = 36.5 ×

7 3

= 85.16 15. Answer (3) 2Mg + O2 ⎯→ 2MgO moles of O2 =

1120 × 20 = 0.01 22400 100

moles of Mg reacted = 0.02 mass of Mg = 0.02 × 24 = 0.48 gm Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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initial moles of Mg =

2·4 = 0.1 24

moles of Mg reacted with nitrogen is 0.1 – 0.02 3 Mg + N2 ⎯⎯→ Mg3N2 3 mole

1 mole

0.08

0.08 3

∴ mass of Mg3N2 =

0.08 8 × 100 = = 2.6 gm 3 3

16. Answer (1) Let the equivalents of Na2CO3 is X equivalents of NaHCO3 is Y Phenolphthalein indicator

X = 2.5 × 0.1 × 2 × 10–3 2 X = 1 × 10–3 (in 10 mL) ∴ In one litre = 1 × 10–1 mass of Na2CO3 = 5.3 gm methyl orange indicator

X +Y = 2.5 × 0.2 × 2 × 10–3 2 Y = 1 × 10–3 – 0.5 × 10–3 = 0.5 × 10–3 (in 10 mL) ∴ equivalents of NaHCO3 in 1 litre = 0.05 Mass of NaHCO3 = 0.05 × 84 = 4.2 gm 17. Answer (2) 1 4 × 10 1000

Number of equivalent of KMnO4 =

= 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 1.5

−3

× 100

= 47% Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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18. Answer (2) Molecular weight of lewsite

1.78 × 10 −22 + 2 × 12 + 1.25 × 10 −22 = 2 ×1+ 1.66 × 10 −24 1.66 × 10 −24 = 208.53 amu. 19. Answer (3) SO2 + H2O2 ⎯→ H2SO4 m. eq. of SO2 = m. eq. of H2SO4 = m. eq. of NaOH = 20 × 0.1 = 2 m. moles of SO2 =

2 =1 2

volume of SO2 at STP = 22400 × 10–3 = 22.4 ml. conc. of SO2 in air is 22.4 ppm 20. 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. 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. 8 4

3 4

equivalent mass of HNO3 =

63 3/ 4

63 = 4× 3 = 84 gm. 21. Answer (3) XZ and YZ planes are nodal planes. 22. Answer (3) ΔX = ΔP (ΔX)2 ≥

h 4π

ΔX =

h 4π

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ΔX · ΔV =

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h 4πm

h ΔV = h 4π 4πm ΔV = 1 h 2m π

23. Answer (2) Angular momentum (mvr) = =

nh 2π 3h = 1.5h 2π π

24. Answer (3) hν = hν0 + eVstop ⎛e⎞ ν = ν0 + ⎜ ⎟ Vstop ⎝h⎠

θ

So the given graph will be a straight line with slope equal to e 1.6 × 10 −19 = = 2.414 × 1014 h 6.626 × 10 −34

tan θ = ⎛⎜ e ⎞⎟ ⎝h⎠

ν ν 0 Stopping potential

25. Answer (1) 1 1 ⎞ ⎛ 1 = 109678 ⎜ 2 − 2 ⎟ λ ∞ ⎠ ⎝1

λ = 9.1176 × 10–6 cm = 911.76 Å 26. 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. 27. Answer (3) Light source is radiating energy at the rate 20 Js–1

hc 6.6 × 10 −34 × 3 × 10 8 = 3.3 × 10 −19 J Energy of single photon = λ = 600 × 10 −9 No. of photon ejected per second =

20 = 6.06 × 1019 3.3 × 10 −19

19 4 = 6.06 × 10 4 × 10 = 10 − 4 NAV 10

28. Answer (3) Angular momentum mvr = Angular momentum ∝ n

nh 2π

n∝ r

Angular momentum ∝

r

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29. 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. Answer (2) Let the electron be moving with momentum P its wavelength will be equal to Δx =

h P

h P

From Heisenberg’s uncertainty principle Δx · ΔP ≥

ΔP ≥

h 4π

h P ΔP 1 × ⇒ ≥ 4π h P 4π

Minimum percentage error in measuring velocity would be 100 ×

ΔV ΔP 100 = × 100 = = 7.96 ~ 8 . V P 4π

31. Answer (4) It has highest number of orbitals among all mentioned ones hence maximum orientation is possible for f-orbitals. 32. Answer (3) Cl(17) – 1s2, 2s2, 2p6, 3s2, 3p5 n = 3, l = 1, m = 1 33. Answer (2) KE = v2 =

1 mv2 = 4.55 × 10–25 2 2 × 4.55 × 10 −25 = 1 × 106 9.1× 10 −31

v = 103 m/s λ=

λ 6.626 × 10 −34 = = 7.28 × 10–7 m mv 9.1× 10 −31 × 10 3

34. Answer (2) Let the no. of photons required to be n

nhc = 10 −17 λ −17 −17 × 10 −9 n = 10 λ = 10 × 550 = 27.6 = 28 photons hc 6.626 × 10 −34 × 3 × 108

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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. 36. Answer (3) Ionisation energy of He = 13.6 Z2/n2 eV = 13.6 ×

22 eV 12

= 54.4 eV Energy required to remove both the electrons = binding energy + ionisation energy = 24.6 + 54.4 = 79 eV 37. Answer (4) Fe2+ –– 1s2, 2s2, 2p6, 3s2, 3p6, 3d 6 Cl– — 1s2, 2s2, 2p6, 3s2, 3p6 In Fe2+, d electrons are 6 while in Cl–, p electrons are 12 38. Answer (1) Ionisation energy = 13.6 Z2/n2 eV For excited state n = 2 Z = 1 I.E. = 13.6 ×

1 = 3.4 eV 4

39. Answer (4) KE = hν – hν0

3 hν = hν − hν 0 4 ν0 =

1 ν 4

= 1 × 3.2 × 1016 4 = 8 × 1015 Hz 40. Answer (2) hc =E λ

⎛h⎞ ∴ p=⎜ ⎟ ⎝λ⎠

⎛E ⎞ c = ⎜⎜ ⎟⎟ ⎝p⎠ Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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41. Answer (1) For M, n = 3 total no. of electrons = 18 ∴ total no. of orbitals = 9 42. Answer (2) For six energy level n = 6 No. of spectrum is UV region = 6 – 1 = 5 43. 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 =

9X 5

+13.6

= 3.4 eV 22 Difference in P.E. between n = 2 and n = 1 level Kinetic energy in first excited state =

U2 – U1 = 2 ×

13.6

13.6

= 20.4 eV 1 22 Potential energy in the first excited level 2

− 2×

…(i)

U2 = U1 + 20.4 eV If ground state is taken as zero potential level then U2 = 0 + 20.4 = 20.4 eV

…(ii)

Then equation (i) and (ii) Total energy = 20.4 + 3.4 eV = 23.8 eV. 45. Answer (2) 1 1⎤ ⎡1 = R × Z2 ⎢ 2 − 2 ⎥ λα 2 ⎦ ⎣1

1 1⎤ ⎡1 = RZ2 ⎢ 2 − 2 ⎥ λβ 3 ⎦ ⎣1 λβ λα

=

3 9 × 4 8

= 27 32 λβ =

27 × 0.32 Å = 0.27Å 32

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46. Answer (4) The lines in the Balmer series are emitted when the electrons jumps from n = 3, 4, 5...... orbits to the second allowed orbit. Since the difference in energy between the third allowed state and the ground state is 12.09 eV. The electrons will not be excited to the third allowed state and hence no line in the Balmer series will be emitted. 47. Answer (2) Absorption line in the spectra arise when energy is absorbed. i.e. electron shifts from lower to higher orbit out of (1) and (2), (2) will have lowest frequency as this falls in the Paschen series. 48. Answer (1) Ni(28) – 1s2, 2s2, 2p6, 3s2, 3p6, 3d8, 4s2 Total no. orbitals = 15 49. Answer (1) Radius in the third orbit = 9r (∴ rn ∝ n2) n3λ = 2πr3 3λ = 2π × 9r λ = 6πr 50. Answer (1) Orbital angular momentum =

l(l + 1)

h 2π

For s-orbital l = 0 ∴ Orbital angular momentum = 0. 51. Answer (1) Since it is feasible to remove only one electron from the element therefore element belong group 1. 52. Answer (4) Inert gases has most stable electronic configuration therefore has least electron affinity. 53. Answer (3) 1s2, 2s2, 2p6, 3s1, after removing the first electron it occupy the noble gas configuration therefore it is not feasible to remove 2nd electron. 54. Answer (3) BaO2 can exist in form of Ba2+ O22–. 55. 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. 56. Answer (3) 1 mole sodium = 23 gm sodium. Q 23 gm sodium requires 495 kJ energy for ionisation. ∴ 2.3 × 10–3 gm sodium requires = 49.5 kJ. 57. Answer (2) IE of Mg = 737 kJ/mol IE of Al = 577 kJ/mol IE of Na = 495.2 kJ/mol IE of Si = 786 kJ/mol Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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58. Answer (4) Alkali metal has low ionisation potential therefore can release an electron easily (oxidised) ∴ Good reducing agent. 59. Answer (3) CH3 CH3

R = =

μ12 + μ12 + 2μ1μ1 cos 60 º (0.36 )2 + (0.36 )2 + 2(0.36)2 × 1 2

= 0.36 × 3 = 0.36 × 1.732 = 0.62 D 60. Answer (4)

Cl Cl

μ = =

μ12 + μ12 + 2μ12 cos 60 º (1.5)2 + (1.5)2 + 2(1.5)2 1 2

= 1.5 3 = 1.5 × 1.732 = 2.6 D In fact it observed dipole moment is found to be much less due to bond angle diversion following ortho effect. 61. Answer (1) Bond angle

Molecules

180º

BeCl2

120º

BCl3

109º28′

CCl4

< 109º28′

PCl3

Therefore BeCl2 > BCl3 > CCl4 > PCl3. 62. Answer (3) XeF4 is square planar in shape BrF4– also have square planar in shape. 63. Answer (4) Nitrogen is chemically inert due to absence of bond polarity. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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N = 8+2 = 5 2 2 Attached atoms = 3 ∴ T-shaped. 65. Answer (1) Due to back bonding BF3 is weaker acid, among these given lewis acids back bonding is stronger in B – F. 66. Answer (4) In N2 there are pπ – pπ bonding itself and in CN– there is pπ – pπ bonding between C and N. 67. Answer (2)

P 60º

P

P P

68. Answer (2) NH3 hybridisation is sp3,

N = 5 + 3 = 4 sp3 2 2

PCl5

N = 5+5 = 5 → sp3d. 2 2

BCl3

N = 3+3 = 3 → sp2 2 2

In [PtCl4]2– hybridisation is dsp2. 69. Answer (1) Due to H-bonding H2O has higher boiling point than others. 70. Answer (4) SF4

N 6+4 = = 5, lone pairs = 5 – 4 = 1 2 2

CF4

N 4+4 = = 4, lone pairs = 4 – 4 = 0 2 2

XeF4

N 8+4 = = 6, lone pairs = 6 – 4 = 2 2 2

71. Answer (1) Greater electronegativity when bonding through axial position. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Bond order

Cl – O–

1

O = Cl – O–

1.5

O 1.66

Cl

O

O

O –

O = Cl – O

1.75

O Bond length is inversely proportional to bond order. 73. Answer (3) Dipole moment μ = charge (q) × distance 1.03 × 10–18 = charge × 1.275 × 10–8 Charge =

1.03 × 10 −18 1.275 × 10 −8

Percentage ionic character =

1.03 × 10 −18 × 100 1.275 × 4.8 × 10 −10 × 10 −8

74. Answer (4) sp3d and dsp3 have same geometry but d-orbitals that takes part in hybridisation are different. 75. Answer (2)

N = 6 + 3 + 1 = 5 , attached atoms = 3 2 2 ∴ T-shaped. 76. Answer (2) KO2 ⎯→ K+ + O2– In O2– unpaired electrons = 1 → paramagnetic Na2O2 ⎯→ 2Na+ + O22– In O22– – no unpaired electrons, hence diamagnetic. Na/NH3 conduct the electricity due to solvated ammonia electrons. 77. Answer (2) Diamond is sp3 hybridised Graphite is sp2 hybridised Acetylene is sp hybridised. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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n1 RT V

P2 =

n2 RT V

n2 P2 = n1 P1

⇒ n2 = n1 ×

P2 12 0.01 = × P1 120 100

Number of molecules left = n2 × N0 = 6 × 1018. 79. Answer (2) At low pressure the volume is high a ⎞ ⎛ ⎜ P + 2 ⎟( V − b) = RT V ⎠ ⎝

V −b ~ V

a ⎞ ⎛ ⎜ P + 2 ⎟( V ) = RT V ⎝ ⎠

PV +

a = RT V

80. Answer (2) a ⎞ ⎛ ⎜ P + 2 ⎟( V − b) = RT V ⎠ ⎝

Z < 1, V – b ~ V

RT a 1+ 2 V

Vr =

since P = 1

Vi = RT Vr < Vi ∴ Vr < 22.4 L 81. Answer (3) Volume is directly proportional to the number of molecules. 82. Answer (1) PV=

2 E 3

E=

3 PV 2

For 1 mole gas PV = RT E=

3 RT therefore E represent here translational kinetic energy. 2

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83. Answer (4) Because intermolecular force of attraction in NH3 is high. 84. 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. 85. Answer (2) Solubility of gases in liquids increases on increasing the pressure. 86. Answer (1) P=

dRT M

d=

PM RT

d ∝ P, d ∝

1 T

87. 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 = 1.51 atm. 88. 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 =

Mass of gas leaked out 1 –

32 2 = gm 48 3

2 1 gas = gm = 0.33 gm. 3 3

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89. Answer (4) 4A3O4 → 3A4 + 8O2 for the 3 moles of A4 8 moles of O2 required. Since 8 mole O2 produces 4 moles of gaseous product. Therefore pressure reduced to half. 90. Answer (2) N2O4(g)

2NO2(g)

1

0

1–α

α = 0.2 Total number of moles after equilibrium = 1.2 1 × V = 1 × R × 300

…(i)

P × V = 1.2 × R × 600

…(ii)

dividing equation (i) by (ii) then P = 2.4 atm 91. Answer (2) Since PV = K(constant) 92. Answer (1)

rA = rB

MB MA

rate of diffusion =

volume diffused 50 = time t

50 t = MB 40 MA t 5 = 4

MB 64

25 MB = 16 64

MB = 100 93. Answer (3) 4 HCOOH ⎯⎯2 ⎯ ⎯ → H2O(g) + CO(g)

H SO

X moles

X moles

H SO4 ⎯→ CO2 (g)+ CO(g)+ H2 O(g) COOH ⎯⎯2 ⎯ Y moles

Y moles

COOH y moles

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Y 1 = X + 2Y 6

6Y = X + 2Y X = 4Y X = 4 :1 Y

+ 2O2 → CO2 + 2H2O

Pressure (63 – X)mm C 2H2 + Pr essure X(mm )

volume of C2H2 is X mL

(63 – X)mm

5 O ⎯⎯→ 2CO 2 + H2 2 2 2 X( mm )

total pressure of CO2 = (63 –X) + 2X = 63 + X 63 + X = 69, X = 6 mm fraction of methane =

63 − X 57 = = 0.9 63 63

n H2 nH2 + nHe + nCH4

× 2.6

⇒ 1.6 atm 96. Answer (2)

rH2 rHe

=

nH2 nHe

1 mHe = 2 1 mH2 4

4 = 2 2 :1 2

97. Answer (3) The expression for standard heat of formation of gaseous carbon is C(graphite) → C(gas) ΔH = 725 kJ/mol As graphite is thermodynamically more stable than diamond so heat required to convert graphite to gaseous carbon should be more. 98. Answer (3) Change in enthalpy = Heat of evaporation × Number of moles = 9.72 × 5 = 48.6 kcal ΔH = ΔE + ΔnRT ΔE = 48.6 – (5 × 2 × 10–3 × 373) = 44.87 kcal. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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99. Answer (4) At constant temperature ΔT = 0 ΔE = 0, ΔH = 0 and at constant temperature, PV = K(constant) – Boyle’s law When temperature is constant PV is constant ΔH = ΔE + Δ(PV) = 0. 100. 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. 101. 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 . 3 102. 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 ΔH > 0 further more entropy increases ΔS > 0. 103. Answer (3) For a diatomic gas Cp =

Only

7 5 R , Cv = R 2 2

5 = 0.71 of energy supplied increases the temperature of gas. 7

The rest is used to do work against external pressure 0.71 × 60 = 42.6 kcal. 104. Answer (1) T1V1γ −1 = T2 V2γ −1

T2 ⎛ V1 ⎞ ⎟ =⎜ T1 ⎜⎝ V2 ⎟⎠

γ −1

= 2 γ −1

Since γ is more for the gas X. The temperature will also be more for it. 105. 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. The value of γ is maximum for helium monoatomic gas. Thus curve C should respond to helium. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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106. Answer (2) BaCl2·2H2O + aq → Ba2+(aq) + 2Cl–(aq) + 2H2O; ΔH = 200 kJ/mol BaCl2 + 2H2O → BaCl2·2H2O; ΔH = –150 kJ/mol On adding both equations we get BaCl2 + aq → Ba2+(aq) + 2Cl–(aq); ΔH = 50 kJ/mol. 107. Answer (4) By definition heat of neutralization we have,

1 1 H2C2O4 + NaOH → Na2C2O4 + H2O; ΔH = –53.35 kJ 2 2 1 1 H2C2O4 + OH– → C2O42– + H2O; ΔH = –53.35 kJ 2 2

…(1)

H+ + OH– → H2O; ΔH = –57.3 kJ

…(2)

Subtracting eq (1) from eq (2) we get 1 1 H2C2O4 → C2O42– + H+; ΔH = 3.95 2 2

H2C2O4 → C2O42– + 2H+; ΔH = 2 × 3.95 = 7.9 kJ. 108. Answer (2) Work done in expansion = P × V = 3(5 – 3) = 6 atm-litre We have, 1 atm-litre = 101.3 J Work done = 6 × 101.3 J = 607.8 J Let ΔT be the change in temperature PΔV = mSΔT 607.8 = 180 × 4.184 × ΔT ΔT = 0.81 K Tf = Ti + ΔT = 290.8. 109. Answer (1) N≡N+

+ – 1 (O = O) → N = N = O(g) 2

1 ⎛ ⎞ ΔHf = ⎜ 946 + × 498 ⎟ − ( 607 + 418 ) = 170 kJ mol–1 2 ⎝ ⎠

Resonance energy = observed heat of formation – calculated heat of formation = 82 – 170 = –88 kJ/mol. 110. Answer (3) S(g) + 6F(g) → SF6(g);

ΔH = –1100 kJ mol–1

S(s) → S(g);

ΔH = +275 kJ/mol

1 F (g) → F(g); 2 2

ΔH = 80 kJ/mol

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. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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111. Answer (1) The solution contains = 200 × 0.1 = 20 m.mol of NaOH 1 m.mol of CO2 reacts with 2 m.mol of NaOH 2NaOH + CO2 → Na2CO3 + H2O The resulting solution contains 18 m mol of NaOH and 1 m.mol of Na2CO3. On titration upto phenolphthalein end point, the NaOH will use 18 m. mol of acid and Na2CO3 will use 1 m.mol of acid, Hence Normality =

(18 + 1) = 0.095 N . 200

112. Answer (3) CH3COONa + HCl initially moles

0.1

after reaction moles 0

CH3COOH + NaCl

0.2

0

0

0.1

0.1

0.1

Since CH3COOH is weak acid therefore we assume that H+ ions from CH3COOH is so less than can be neglected ∴ HCl

H+

0.1

+

0.1

Cl– 0.1

[H+] = 0.1 M pH = – log[H+] = – log 0.1 =1 113. Answer (1) For CaCO3 (s)

CaO(s) + CO2(g)

Kp = PCO = 0.0095 atm 2

Since atmospheric pressure is 1 atm so percentage of CO2 in air =

PCO2 Ptotal

× 100 = 0.95% .

Thus to prevent the decomposition of CaCO3 at 100°C the % of CO2 in air must be greater than 0.95%. 114. Answer (1)

C Csalt ⇒ pH = pK + log salt pH = pK a + log C a C acid acid ∴

Csalt = Cacid.

pKa + pKb = 14 pKa = 14 – 4.7 = 9.3

pH = pK a

pH = 9.3.

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B+

0.2

0.2

Cl–

+

0.2 B+

0.1

0

0

0.1 – x

(x + 0.2)

x

initially moles after equilibrium

OH–

BOH

+

( x + 0. 2) x = 10 −5 0. 1 − x

Kb =

∴ x + 0.2 ~ 0.2 0.1 – x ~ 0.1 x

=

1 × 10 −5 = 5 × 10 −6 2

∴ degree of dissociation =

5 × 10 −6 = 5 × 10 −5 . 0.1

NH4+

C2(1 – α)

C2α

NaOH

–→

Na+

+

C1

+

OH– C2α

C2 = [NH4OH] = 0.15 M

OH– C1

+

Kb

[NH4 ][OH− ] C 2 α(C1 + C 2 α ) = = C1α = [NH4OH] C 2 (1 − α )

α

Kb −4 = C = 1.8 × 10 1

117. Answer (3) 2 Qp = PCO 2 .PNH3 = (20) (10)2

= 2000 atm3 Kp = 2020 atm3 Since Qp < Kp. So this pressure is not sufficient to maintain the system in equilibrium therefore total pressure in the chamber would be equal to 30 atm. 118. Answer (2) C

Kf Kb

D

conc. at t = 0

a

0

conc at equi.

a–x

x

when equilibrium is achieved kf (a – x) = kb.x kb =

k f (a − x ) x

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PCl3(g) + Cl2(g)

COCl2(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. As a result the equilibrium concentration of Cl2 will be less. So the equilibrium constant of the first reaction will also be disturbed and reaction quotient will be less than the equilibrium constant. 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. Answer (2) [S2–] =

5 × 10 −21 = 1.0 × 10 –19 0.05

[H+ ][S 2− ] = 1× 10 −7 × 1× 10 −14 [H2S]

[H+] =

1× 10 −21 × 0.1 1× 10 −19

pH = 1.50. 121. Answer (3)

Its equilibrium constant Keq =

=

K a × Kb Kw 3.24 × 10 −10 10 −14

= 1.8 × 1.8 × 104 122. Answer (2) Solubility of PbSO4 =

K sp = 1.44 × 10 –4 = 1.2 × 10 –4 M

Solubility of PbSO4 = 1.2 × 10–4 = 1.2 × 10–4 × 303 × 103. = 36.36 mg litre–1. Volume of water needed to dissolve 1 mg of PbSO4 =

1000 36.36

= 27.5 mL. 123. Answer (2) CO(g)

+

NO2(g)

CO2(g) +

NO(g)

1

1

1

1

t=0

1–x

1–x

1+x

1+x

at equilibrium

CO2 + Ba(OH)2 → BaCO3 ↓ +H2O white

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moles of BaCO3 =

Physical Chemistry

236.4 = 1.2 197

moles of CO2 at equilibrium ⇒ 1 + x = 1.2 x = 0.2 2

2

⎛ 1 + x ⎞ = ⎛ 1.2 ⎞ = 2.25 Kc = ⎜ ⎟ ⎜ ⎟ ⎝ 1− x ⎠ ⎝ 0 .8 ⎠

124. Answer (2) Kp = PCO2 = 2.25 Number of moles of CO2 =

2.25 × 1 0.0821× 600

Min. moles of CaCO3 required = 0.0457 Min. weight of CaCO3 required = 0.0457 × 100 = 4.57 gm. 125. Answer (2) NH4+ + OH–

NH3 + H2O +

Kb =

[NH4 ][OH − ] 1.5 × 10 −3 × 1.5 × 10 −3 = = 1.8 × 10 −5 [NH3 ] [NH3 ]

[NH3] = 0.125 M total [NH3] required = 0.125 + 1.5 × 10–3 = 0.1265 M 126. Answer (1)

X Cl2 = XCl = 0.50 2 PCl ( X Cl × PT )2 (0.5 × 1)2 Kp = P = X × P = (0.5 × 1) Cl2 Cl2 T

= 0.5 M 127. Answer (2) H2(g) + At t = 0 At eq. Kc =

S(s)

H2S(g)

0.2

1

0

0.2 – x

1–x

x

[H2S] x = 6.8 × 10 − 2 = [H2 ] (0.2 − x )

x = 1.27 × 10–2 moles/litre

PH2 S =

nRT 1.27 × 10 −2 × 0.082 × 363 = = 0.38 atm V 1

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128. Answer (3) pH = 11.27 – log[H+] = 11.27 [H+] = 5.37 × 10–12 [OH–] =

7.1× 10 –15 = 1.322 × 10 −3 5.37 × 10 –12

(Cα )2 (1.322 × 10 −3 )2 = = 1.75 × 10 – 5 C 0.1

Kb =

129. Answer (3) Since PCO < PCO2

Kp =

PCO2 PCO

=

1 500 × 10 −6

= 2000

Since 2.303 RTlogKp = − ΔGop 2.303 × 8.314 × T log2000 = 20700 + 12T T = 404.3 K 130. Answer (2) [Complex ] = 0.9 [H3BO3 ][Glycerine]

K=

[Complex ] 60 = = 1.5 [H3BO3 ] 40 1.5 = 0.9 [Glycerine]

K=

[Glycerine] =

1.5 = 1.7 M 0.9

% of

[In–]

=

[In − ] −

[In ] + [HIn]

× 100 =

10 × 100 = 91% 10 + 1

M2+ + 2OH– x

2x

pH = 10.6 pOH = 14 – 10.6 = 3.4 [OH–] = antilog(–3.4) = 3.98 × 10–4 M

x=

3.98 × 10 −4 = 1.99 × 10 − 4 M 2

Ksp = [M2+][OH–]2 = 4x3 = 4 × (1.99 × 10–4)3 = 3.15 × 10–11 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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133. Answer (4) Mg2+ (aq) + 2OH– (aq.)

Mg(OH)2(s) [Mg2+][OH–]2

= 1.2 × 10–11

[OH–]2 = 1.2 × 10–10 [OH–] = 1.1 × 10–5 pOH = – log 1.1 × 10–5 = 5 – log 1.1 = 5 – 0.04 = 4.96 pH = 14 – 4.96 = 9.04 134. Answer (4) Higher the reduction potential greater is tendency for reduction. 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

Fe ⎯⎯→ Fe 2+ + 2e

At cathode

Pb 2+ + 2e ⎯⎯→ Pb

Net reaction

Pb 2+ + Fe ⎯⎯→ Fe 2+ + Pb

º º = –0.13 V – (–0.44) = + 0.31V E ºcell = Ecathode − Eanode

Since the standard emf to the cell is positive the reaction is spontaneous. Hence more of Pb and Fe2+ are formed. 135. Answer (3) Na2S2O3

S

–2

+ — –1 +1

Na

–1 –

+

O — S+2 — O Na +2

+1

O

–2

Since total charge in the sulphurs are –2 and +6 each ∴

Oxidation no. of sulphur in hypo are –2 and +6

136. Answer (2) Equivalents of Cr deposited = equivalent of O2 evolved

2 .6 0.56 ×n = ×4 52 22.4 n = 2 i.e. Cr 2 + ⎯⎯→ Cr 137. Answer (2) E°cell = E°cathode – E°anode = E°H2O2 ( aq)| H2O( l) − E°Fe3+ ( aq)| Fe2+ ( aq ) = 1.763 – 0.769 = 0.994 V Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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138. Answer (3) As the Cu2+ ions lost from the solution are compensated by copper anode therefore concentration of the solution remain same At anode

Cu(s) ⎯⎯→ Cu2 + + 2e

At cathode

Cu2 + + 2e ⎯⎯→ Cu(s)

139. Answer (2) As the surface area of contact of an electrode with electrolyte increases. Conductance of electrolyte increase thereby time rate of electrolysis increases. 140. Answer (2) Half cell reactions of the given electrodes are First electrode

AmO 22+ + e ⎯⎯→ AmO 2+

Second electrode

AmO 22+ + 4H+ + 2e ⎯⎯→ Am 4 + + 2H2O

Third electrode

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. 141. Answer (1) H2 ⎯⎯→ 2H+ + 2e

Anode

10–6 M 2H+ + 2e ⎯⎯→ H2

Cathode

[H+ ] 2 cathode 0.059 log = 0.118 = 2 (10 − 6 )2

Ecell

On solving we get [H+]cathode = 10 –4 M 142. Answer (2) 2MnO2 + Zn2 + + 2e ⎯⎯→ ZnMn2O 4

EMnO2 =

t=

87 1

w × 96500 8 × 96500 = = 51.35 days −3 i ×E 2 × 10 × 87 × 3600 × 24

143. 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. O

HO HO

O

HO

O

HO

O

HO

S

S

S O

HO

O

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2H+ + 2e ⎯⎯→ H2 E = E°H+ / H −

PH 0.059 0.059 1 log + 2 2 = E°H+ / H − log + 2 2 2 2 [H ] [H ]

E′ = E°H+ / H −

0.059 100 log + 2 2 [H ]

2

2

EH+ / H − E′H+ / H = 2

2

0.059 100 log × [H+ ] 2 = 0.059 V + 2 2 [H ]

145. 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

1 Cl−A ⎯⎯→ Cl2 (g) + e − 2

Cathode

1 Cl2 (g) + e ⎯⎯→ Clc− 2

The Ecell is given by

Ecell = −

C C [Cl − ] 0.059 0.059 log 2 = 0.059 log 1 log − C = − 1 C1 C2 1 [Cl ] A

For Ecell to be positive C1 > C2 146. Answer (2) Let the formula of mercury ion is Hgnn + then the formula of mercury nitrate would be Hgn(NO3)n. The reactions occuring at two electrodes are Cathode

(Hg )

Anode

nHg ⎯⎯→ Hg nn+

Net reaction

(Hg )

n+ n C

n+ n C

+ ne ⎯⎯→ nHg

(

)

(

A

+ ne −

⎯⎯→ Hgnn+

)

A

Thus the given cell is electrolyte concentration cell E cell =

0.0591 [Hgnn+ ]C log n [Hgnn + ] A

⎛ 1⎞ ⎜ ⎟ 0.0591 2 0.0295 = log ⎝ ⎠ n ⎛ 1 ⎞ ⎜ ⎟ ⎝ 20 ⎠ n=2 It means mercury ion exists as Hg22 + Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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147. 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.e. AgCl is consumed. So during reaction quantity of AgCl decreases. 148. Answer (1) Q + 2H+ + 2e

QH2

Ecell = E°cell −

0.0591 1 log + 2 2 [H ]

Ecell = E°cell − 0.0591 pH intercept = Eocell = 0.699 V ⎛ 0.699 − 0.492 ⎞ ⎟ = 3. 5 pH = ⎜ 0.0591 ⎝ ⎠

149. Answer (1) H+ + OH–

H2O

H+ + e ⎯⎯→

ΔG1o = −RT ln K w

;

1 H ; 2 2

H2O + e– ⎯⎯→

ΔG 2o = − n F E o +

H / H2

1 H + OH– ; 2 2

ΔG3o = − n F EHo O / H 2

− 2 , OH

o o Since E H+ / H2 = 0 So ΔG 2 = 0

Thus ΔG1o = ΔGo3 o –RTln Kw = − n F EH2O / H2 , OH−

E oH O / H 2

2 , OH

=

RT ln K w sin ce (n = 1) F

150. Answer (4) o − EMnO− / Mn2 + = EMnO − / Mn2 + 4

4

[Mn 2+ ] 0.059 log 5 [MnO 4− ][H+ ]8

Let the initial conc. of H+ be x. When it is reduced to x/2 the electrode potential is given by E′MnO− / Mn2+ 4

o = E MnO−4 / Mn2 + −

0.059 [Mn 2 + ]8 log 5 [MnO −4 ]x 8

o = EMnO4− / Mn2 + −

[Mn2+ ] 0.059 0.059 − log 28 log − 8 5 5 [MnO 4 ]x

o = E MnO−4 / Mn2 + −

0.059 [Mn 2+ ] log − 0.02846 5 [MnO −4 ] x 8

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151. Answer (4) The metal which has high reduction potential reduce first ∴ Sequence of deposition of metal Mg < Cu < Hg, but Mg will not be deposited because H+ preferentially discharge. 152. Answer (3) o Cu2+ + 2e → Cu; ΔG1 = –2F(0.337) o Cu+ + e → Cu; ΔG2 = –1F(0.153) o Cu2+ + e → Cu+; ΔG3 = –1FE°

ΔGo3 = ΔG1o − ΔGo2 –FE° = –2F(0.337) + F(0.153) E° = 2 × 0.337 – 0.153 = 0.521 volt. 153. 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. 154. Answer (3)

Λ∞ = Λ∞NH4NO3 + Λ∞KOH − Λ∞KNO3 = 128 + 239 − 125 = 242 Degree of dissociation =

Λ

⎛ 24 ⎞ =⎜ ⎟ = 0.10 . Λ ⎝ 242 ⎠ ∞

1000 K 1000 × 19.5 × 10 −5 = = 19.5 N 0.01

⎛ Λ Degree of dissociation = ⎜ ∞ ⎝Λ

⎞ 19.5 = 0.05 ⎟= ⎠ 390

156. Answer (1) o o E ocell = E RP cathode − ERP anode = 0.13 – (–0.34) = 0.47

2Tl + Sn4+ → Sn2+ + 2Tl+ Ecell = 0.47 –

[ Tl+ ]2 [Sn 2+ ] 0.059 0.059 log log 2 = 0.47 − 0.059 = 0.411 V = 0.47 – 4+ 2 [Sn ] 2

Eocell = 1.23 − ( −0.44) = 1.23 + 0.44 = 1.67 ΔG° = – nFEocell = –2 × 96500 × 1.67 × 10–3 kJ = –322 kJ 158. Answer (4) Ni(s) + Cu2+(aq) → Cu(s) + Ni2+(aq) o o Eocell = ERP cathode − ERP anode = 0.34 − ( −0.25) = 0.59 V

Q

Eocell is positive therefore reaction will be spontaneous.

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159. Answer (1) Since oxidation potential of Zn is high therefore Zn will be oxidised ∴ Zn(s)|Zn2+(aq)||H+(aq)|H2(g), Pt 160. Answer (2) Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) E1 = E ocell −

0.059 ⎛ 0.01 ⎞ log⎜ ⎟ 2 ⎝ 1 ⎠

when the [Zn2+] = 1 M[Cu2+] = 0.01 M E2 = E ocell −

0.059 ⎛ 1 ⎞ log⎜ ⎟ 2 ⎝ 0.01 ⎠

It is clear from both equation E1 > E2 161. Answer (4) Fe2+ + 2e → Fe; ΔG1o = –2F(–0.44) Fe3+ + 3e → Fe; ΔGo2 = –3F(–0.036) Fe3+ + e → Fe2+; ΔGo3 = –1FE

ΔGo3 = ΔGo2 − ΔG1o –FE = –3F(–0.036) – 2F(0.44) E = –3 × 0.036 + 2 × 0.44 = 0.771 V 162. Answer (4) When the cell is completely discharged Ecell = 0 o = E cell

1.1 =

⎡ Zn2+ ⎤ 0.059 log ⎢ 2+ ⎥ 2 ⎢⎣ Cu ⎥⎦

0.059 ⎡ Zn2 + ⎤ log⎢ 2+ ⎥ 2 ⎣⎢ Cu ⎦⎥

⎡ Zn 2+ ⎤ log⎢ 2+ ⎥ = 37.3 ⎢⎣ Cu ⎥⎦

⎡ Zn2+ ⎤ ⎢ 2+ ⎥ = 1037.3 ⎣⎢ Cu ⎦⎥ 163. Answer (2) Since the cell reactions proceed in the standard condition and E ocell is negative therefore the electricity cannot be produced. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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164. Answer (2) ΔG1° = –2.303 RT log Ksp

Ag+(aq) + Cl–(aq) ;

AgCl(s)

Ag+(aq) + e → Ag(s) ;

ΔG2° = −1F E° Ag+ / Ag

AgCl(s) + e → Ag(s) + Cl–(aq) ;

ΔG3° = −1F E°Cl− / AgCl / Ag

From the given equations ΔG3° = ΔG1° + ΔG2° E° Cl− / AgCl / Ag = E° Ag+ / Ag + 2.303 RT log Ksp

–0.15 = 0.80 + 2.303 RT log Ksp −0.95 = −16.101 0.059

log Ksp = ∴

Ksp = 7.92 × 10–17.

165. Answer (2) Let the I amp current is passed for 2 hrs. Charge = 2 × 60 × 60 × I = 7200 I Moles of electrons passed =

7200 I 96500

1 O + H2O + 2e 2 2

At. anode

2OH– →

At. cathode

2H+ + 2e → H2

Moles of O2 released at anode =

7200 I 1 × 96500 4

Moles of H2 released at cathode =

7200 I 1 × 96500 2

Volume of H2 + volume of O2 = 672 (at S.T.P.) 22400 ×

7200 3 I × = 672 96500 4

I = 0.536 amp. 166. Answer (3) Copper is oxidised to Cu2+ 167. Answer (1) 2Cl– → Cl2 + 2e (anode) Cu2+ + 2e → Cu (cathode) ∴

1 mole of copper deposited at cathode

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Z×M N AV × a

3

=

4(78 ) 6.023 × 10 (5.46 × 10 23

−8 3

)

=

4 × 780 = 3.2 gm / Cm 3 6.023 × 5.46

169. Answer (2) 3 face centre + 1 corner atom forms tetrahedral void in fcc. 170. Answer (3) The centre atom is surrounded by six atom and one atom lies over this therefore C.N. = 7. 171. Answer (4) All have same number of formula unit (i.e. Z = 4). 172. Answer (3) Triclinic is most unsymmetrical crystal system a ≠ b ≠ c and α ≠ β ≠ γ = 90°. 173. Answer (1) Total volume of sphere = 4 ×

4 3 16 3 πr = πr 3 3

For fcc unit cell, a = 2 2r 3 3 3 Volume of cube = a = (2 2r ) = 16 2 r

16 3 πr π volume of sphere = 3 = Packing fraction = 3 volume of cube 3 2 16 2 r

d=

N AV × a 3

=

4 × 62 6.023 × 10 23 × (10 −8 )3

=

4 × 620 = 411.75 g / Cm 3 6.023

In Frenkel defect the density remains unaltered. 175. Answer (3) When one face plane is removed then four corners ions and one face centre ion of B are removed (i.e. effective one ion of B) and 4 ions of A are removed from edge centres (effective one ion) ∴

New formula

A+ B–

effective atoms

3

3

Since equal number of cations and anions are missing therefore defect is Schottky detect. 176. Answer (1) r+ + r– =

a 2

a = 2.3 × 2 = 4.6 Å Z×M

d=

N AV × a 3

=

4 × 78 6.023 × 10 23 × ( 4.6)3 × 10 − 24

4 × 780

=

6.023 × ( 4.6)3

= 5.32 gm/Cm3.

177. Answer (4) Total number of effective atoms in unit cell = 4 + 4 = 8. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

3a = 4r

In bcc,

r=

3a = 4

3 × 297 4

2a = 4r

In fcc,

4r

a=

2

= 2 2r

= 2 2×

=

3 × 297 4

3 × 297 2

= 363.79 pm. 179. Answer (2) Z ×M

d=

NAV × a 3

Z × 108

10.5 =

6.023 × 10

23

× (4.09 × 10 −8 )3

Z=4 ∴

Unit cell is fcc.

180. 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 – 1.5 = Effective number of ions of A = 4 − ∴

5 2

1 7 = 2 2

New formula of compound A7/2B5/2.

r– =

= 0.732 r+ 100 = = 136 .61 pm. 0.732 0.732

182. Answer (3) Orthorhombic exist in body centred, end centred, face centred as well as primitive unit cells. 183. Answer (1) Co-ordination number of each sphere is 6. 184. Answer (3) In hexagonal crystal system a = b ≠ c, α = β = 90°; γ = 120°. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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4 3 πr ⎛ π⎞ In simple cubic unit-cell the packing fraction = 3 = ⎜ ⎟. 3 ⎝6⎠ (2r ) 186. Answer (3) Number of alternate corners = 4 Number of alternate edges = 4 Number of alternate faces = 2 Hence,

Number fo A atoms =

1 1 ×4 = 2 8

Number of B atoms =

1 ×2 =1 2

Number of C atoms =

1 × 4 +1= 2 4

So, the formula of compound per unit cell is A1/2BC2, simplest formula of compound is AB2C4. 187. Answer (2) 1 d [H2 ] 1 d [NH3 ] =+ 3 dt 2 dt d [H2 ] 3 − = × 2 × 10 − 4 = 3 × 10 − 4 dt 2

So,

−d[H2 ] d[NH3 ] × = 3 × 10 − 4 × 2 × 10 − 4 = 6 × 10–8 dt dt

188. Answer (3) In the rate of reaction, reciprocal of coefficient is written not in the rate of appearance of product. 189. Answer (3) A (g) ⎯⎯→

B (g) +

2C(g) +

D (g)

At t = 0,

a

0

0

0

At t = ‘t’,

(a – x)

x

2x

x

At t = 0

a = P0

At t = ‘t’, a + 3x = Pt x=

Pt − P0 3

For first order reaction,

k=

a 2.303 log (a − x ) t

k=

2.303 log t

k=

2.303 3 P0 log t ( 4P0 − Pt )

P0 ⎛ Pt − P0 ⎞ P0 − ⎜ ⎟ ⎝ 3 ⎠

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190. Answer (3) 238 92 U is

disintegrated through (4n + 2) series.

191. Answer (1) Total time = n × t 1 2

69.2 = n × 138.4 n

=½ n

1

⎛ 1⎞ ⎛ 1 ⎞2 N = N0 ⎜ ⎟ = 1 ⎜ ⎟ 2 ⎝ ⎠ ⎝2⎠ 1

N=

2

=

1 = 0.7 1.414

Disintegrated amount = 1 – 0.7 = 0.3 g −24 He 206 210 84 Po ⎯⎯ ⎯→ 82 Po

Volume of helium accumulated

=

22400 × 0.3 g 210

= 32 ml. 192. Answer (2) A 2B5 ⎯⎯→ 2AB 2 + ½ B2

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 −

So,

2k1 = k2 = 4k3 193. Answer (3) λ=

2.303 N log 0 t Nt

log

N0 λ×t = Nt 2.303

Comparing with y = mx + C then Slope = +

λ 2.303

194. Answer (3) Lower is the activation energy, higher is the rate of reaction. 195. Answer (3) For exothermic reaction, Activation energy for reverse reaction = ΔH (only magnitude) + Activation energy for forward reaction = 20 + 30 = 50 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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196. Answer (2) 2A(g) + B(g) ⎯⎯→ 2C(g) +

1 d [C] 120 − 100 = =2 2 dt 10

+

d [C] = 4 mm / min dt

1 (a)n−1

(n = order of reaction)

198. Answer (2) Overall order of reaction is 2. 199. Answer (2)

238 ⎛ ⎞ × 0.2 ⎟ ⎜1 + 2.303 206 ⎝ ⎠ k= log t 1 0.693 2.303 = × 0.1 9 t 4.5 × 10

t = 1.5 × 109 years. 200. Answer (3) 2A + 3B ⎯⎯→ 4C + 5D

1 d [B] 1 d [C] =+ 3 dt 4 dt

4 d [B] d [C] =+ 3 dt dt

201. Answer (4) For ‘B M=

1000 × K b × 1 (M = molar mass of B) 1× 100

Kb =

M 10

For ‘A’ M1 =

1000 × M × 2 1× 10 × 100

(M1 = molar mass of A)

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202. Answer (2) From Roult’s law, Ps = PAo ·X A + PBo ·XB Ps = PAo ·X A + PBo (1 − X A ) (Q XA + XB = 1) Ps = PAo ·X A + PBo − PBo ·X A Ps = PBo − X A (PBo − PAo )

…(i)

Comparing the equation (i) with Ps = 210 – 120XA then we get

PBo = 210 and PAo = 90 203. Answer (1) ΔTf = Kf × m 1.86 = 1.86 × m m=1 It shows that 1 mole of urea dissolved in 1000 g of water. nurea = 1, nwater = xurea =

1000 = 55.5 18

nurea 1 1 = = nurea + n water 1 + 55.5 56.5

204. Answer (2) Freezing will start at –1.86°C not 0°C because ΔTf = 1.86 and as the value of ΔTf increases then the molality also increases due to the freezing of water. Glucose doesn’t freeze. 205. Answer (3) ΔTf = Kf × m

…(i)

ΔTb = Kb × m

…(ii)

By adding (i) and (ii) ΔTf + ΔTb = Kf × m + Kb × m = m(Kf + Kb) (Q ΔTf + ΔTb = 2.38) 2.38 = m(1.86 + 0.52) m = 1 for non electrolyte solute. Hence, answer is (3). 206. Answer (4) For NaCl, m = 0.1 For Ba(NO3)2, m = 0.1 ΔTf for NaCl = i × Kf × m = 2 × 1.86 × 0.1 = 0.372 ΔTf for Ba(NO3)2 = i × Kf × m = 3 × 1.86 × 0.1 = 0.558 Total depression in freezing point = ΔTf for NaCl + ΔTf for Ba(NO3)2 = 0.372 + 0.558 = 0.93 Hence, freezing point of solution = 0 – 0.93 = –0.93°C. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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207. Answer (2) M Ca(NO 3 )2 and 0.05 M Na SO , effective molarity are same. 2 4 20

In case of 208. Answer (1) M=

1000 × K f × w ΔT × W

62 =

1000 × 1.86 × 50 9.3 × W

W(unfreezed water) = 161.29 g Ice separated = 200 – 161.29 = 38.71 g 209. Answer (3)

9.5 m = 95 = 0.5 m 200 1000 (For MgCl2, i = 3) ΔTb = i × Kb × m = 3 × 0.52 × 0.5 = 0.78 Hence, B.P. of solution = 100 + 0.78 = 100.78°C ΔTf = i × Kf × m = 3 × 1.86 × 0.5 = 2.79 Hence, F.P. of solution = 0 – 2.79 = –2.79°C B.P. – F.P. = 100.78 – (–2.79) = 103.57°C. 210. Answer (1) H+ + X–

HX 1

0

0

(1 – 0.2)

0.2 0.2

before dissociation after dissociation

= 0.8 i=

1.2 = 1.2 1

ΔTf = i × Kf × m = 1.2 × 1.86 × 0.2 = 0.45 F.P. of solution = 0 – 0.45 = –0.45°C 211. 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. 212. Answer (1) Emulsifying agent stabilised the emulsion of oil in water by forming an interfacial film between suspended particles and the medium. 213. Answer (2) A negatively charged sol is formed due to adsorption of I–. Coagulation value ∝

1 . coagulating power

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Section - B : Multiple Choice Questions 1.

Answer (2, 4) The substances having same composition of atoms and similar crystal structure are isomorphous.

2.

Answer (2, 3) %C in C2H5OH =

24 × 100 = 52% 24 + 5 + 16 + 1

%C in C6H12O6 =

72 × 100 = 40% 72 + 12 + 96

%C in CH3COOH =

%C in C2H5NH2 = 3.

24 × 100 = 40% 12 + 3 + 12 + 32 + 1

24 × 100 = 53% 24 + 5 + 14 + 2

Answer (1, 4) HF + NaOH ⎯⎯→ NaF + H2O

moles of HF = 0.1×

100 = 0.01 1000

moles of NaOH = 0.1×

100 = 0.01 1000

number of moles of NaF formed = 0.01 [NaF] =

0.01 = 0.05 M 200 1000

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Physical Chemistry

4.

Success Magnet (Solutions)

Answer (2, 3) KCrO3Cl = 1 + x + 3 × –2 + (–1) = 0 x = +6

O O

O

CrO5

Cr O

O

Oxidation number of Cr = +6 5.

Answer (2, 3) Those substance can be oxidised and reduced, in which central element is neither in lowest nor in highest oxidation state. For Cl, range of oxidation number is from –1 to 7. In HCl, Cl is present in lowest oxidation state In HClO4, Cl is present in highest oxidation state.

6.

Answer (1, 2, 3) MnO2 + (NH4)2SO4 → MnSO4 + (NH4)2S2O8 n=2

n=1

equivalents of MnO2 = equivalents of (NH4)2SO4 1 × 2 = 1x x=2 7.

Answer (1, 2, 3) Resultant normality of solution N1V1 + N2V2 + N3V3 = N4V4 5 × 1 + 20 ×

1 1 + 30 × = N4(1000) 2 3

25 = N 4 1000 N4 = 1 N 40 Resultant [H+] =

Normality =

1 = 0.025 40

mass / equivalent mass vol (litre)

1 × x molecular mass of NaOH = 40 40 40 x = 1 gm 8.

Answer (1, 2, 3) SO2 + H2O2 ⎯→ H2SO4 m. equivalents of SO2 = m. equivalents of H2SO4

= m. equivalents of NaOH = 20 × 0.1 = 2

n factor of SO2 =

2 =1 2

Volume of SO2 at STP = 22400 × 10–3 = 22.4 ml. Concentration of SO2 in air is 22.4 pm Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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9.

Physical Chemistry

0

3Cu + 8HNO3 ⎯→ 3Cu(NO 3 )2 + 2NO + 9H2 O

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 =

6 3 or mole of electrons 8 4

3 4

Cu is oxidised of Cu2+ equivalents mass of HNO3 =

63 = 84 gm. 3/ 4

10. Answer (1, 3, 4) +4

−1

+2

0

MnO 2 + 4HCl → MnCl2 + Cl2 + 2H2 O n factor of HCl =

2 1 = 4 2

n factor of MnO2 = 2 equivalent mass of MnO2 =

molecular mass . 2

+7

Ba(MnO 4 )2 ⎯⎯→ Mn2+ ( n=10 )

milliequivalents of Ba(MnO4)2 = 100 ×

1 × 10 = 100 10

Fe2+ → Fe3+ + e n=1 milliequivalents of FeSO4 = 100 × 1 = 100 +2 +3

+3

+4

Fe C 2 O 4 → Fe 3 + + CO 2 (n = 3 )

milliequivalents of FeC2O4 =

100 × 3 = 100 3

equivalents of Ba(MnO4)2 = equivalents of FeSO4 = equivalents of FeC2O4. 12. Answer (2, 3) Ca(OH)2 + H2SO4 → CaSO4 + 2H2O n=2

n=2

equivalent mass of H2SO4 =

98 = 49. 2

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⇒ n = 1, l = 0

Radial node = (n – l – 1) = 1 – 0 – 1 = 0 For 3d, n = 3, l = 2 Radial nodes = 3 – 2 – 1 = 0 For 4f, or n = 4, l = 3 Radial nodes = 4 – 3 – 1 = 0 14. Answer (3, 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. 15. Answer (1, 2, 3, 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

⎧⎪ (n + 1)2 − 1 ⎫⎪ ⎨ ⎬ ⎪⎩ (m + 1)2 − 1⎪⎭

16. Answer (1, 2, 3) Number of scattered α-article in Rutherford experiment is inversely proportional to square of kinetic energy of incident α-particles. 17. Answer (1, 2, 3, 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.e. 18. Answer (3, 4)

∫ ψ dxdydz = 1 2

−∞

n = 4, l = 0, 1, 2, 3 for l = 2, m = –2, –1.0 + 1, +2 s= +

1 (correct) 2

n = 4, l = 0, 1, 2, 3 l = 2, m = –2, –10 + 1 + 2 s= −

1 (correct) 2

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19. Answer (2, 3, 4) There are three possible values of spin quantum number it means an orbital can accommodate 3 electrons. So 1s – 1s3, first period would have 3 vertical columns. 20. Answer (1, 3, 4) Kinetic energy of the ejected electron depends on the frequency of incident radiation not on the intensity. Intense and weak beam are having more or less number of photons. 21. Answer (1, 2) rn = n2r1 ⎛ πr 2 ⎛A ⎞ ln ⎜⎜ n ⎟⎟ = ln⎜ n ⎜ πr 2 ⎝ A1 ⎠ ⎝ 1

ln

⎞ ⎟ = ln n 4 = 4 ln(n) ⎟ ⎠

An A1 ln(n)

22. Answer (1, 2, 4) Since only six different wavelengths are emitted therefore highest excited state is n = 4 therefore (1) is correct. In the emitted radiation two wavelength are shorter than λ0 it means that initially atoms were in excited state therefore (2) is also correct. Transition corresponding 4 → 1, 3 → 1, 2 → 1 belongs to lyman series. 23. Answer (2, 3) Energy of orbital of hydrogen depend upon n and not on l. 24. Answer (2, 3, 4) N 6 + 3 −1 ⎫ = = 4 lone pair = 1 ; ∴ Hybridisation = sp 3 ⎪ 2 2 ⎪ 4 + 3 +1 − N 3⎪ CH3 = = 4 lone pair = 1 ; ∴ Hybridisation = sp ⎬ trigonal pyramidal 2 2 ⎪ N 5+3 ⎪ 3 NH3 = = 4 lone pair = 1 ; ∴ Hybridisation = sp ⎪ 2 2 ⎭

H3 O +

25. Answer (1, 2) Cu and Al, Si and Ge donot show inert pair effect. 26. Answer (1, 2, 3, 4) (1) KF combines with HF and forms KHF2 with exists as K+ + [F ------ H —— F]–. (2) Due to smaller size of Li+ ions it has high polarising power therefore predominantly covalent in nature. (3) CsBr3 exists as Cs+ + Br3–. (4) Sodium sulphate is soluble in water but BaSO4 is sparingly soluble because hydration energy of BaSO4 is less than its lattice energy. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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N 8+2 = = 5 attached atoms = 3; ∴ T-shaped 2 2

N 5 + 2+1 = = 4 attached atoms = 2; 2 2 36. Answer (2, 3) NH2

F

B–C≡C–B F sp2 sp sp sp2

∴ Angular

F F

∴ planar

SiH3 SiH3

N – SiH3

Lone pair of nitrogen is involve in pπ-dπ bonding there fore delocalised therefore nitrogen is sp2 hybridsed and planar. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

N = C = O – linear

S = C = S – linear 38. Answer (1, 2, 4) (a) On decreasing the electronegativity of central atom bond angle decreases (b) Bond angle of NH3 – 107° Bond angle of H2O – 104.5° Bond angle of F2O – 103° 39. Answer (1, 2, 3, 4) Electron affinity of anion is positive and non spontaneous due to electron –2 repulsion. 40. Answer (1, 2, 3)

O

CH3 O

O N

Br

CH3

H

C C

O CH3 (μ ≠ 0)

N

Cl

O O (μ = 0)

(μ ~ – 0)

C2H5

H

(μ ≠ 0)

41. Answer (2, 3, 4) Statement 2 and 3 are facts

Unit of P = Unit of

n2a V2

∴ Unit of a = atm L2mol–2 42. Answer (3, 4) Mutual attraction of molecules known as van der Waal intermolecular force. 43. Answer (3, 4) Z=

PV RT

Average KE =

3 RT 2

44. Answer (2, 3, 4) Above critical temperature gas cannot be liquified. 45. Answer (1, 3, 4) Z=

Vmolar ( Vm ) Videal ( Vid )

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Physical Chemistry

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When Vm > Vid

Z>1

Vm = Vid

Z=1

Vm < Vid

Z H3PO3 > H3PO4, hybridisation of phosphorus in all acids are not sp3. In H3PO3, H3PO2 there is P—H bond present, therefore these are reducing in nature. 51. Answer (2, 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. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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52. Answer (1, 2, 3, 4) van der Waal’s constant a measure the intermolecular force of attraction b is called excluded volume Vc = 3b. 53. Answer (1, 3)

Kinetic energy for 1 mole gas =

3 RT 2

1 mole of gas has molecules = Nav. 54. Answer (1, 2) On the expansion the volume increases therefore pressure decreases as the temperature is constant therefore kinetic energy of gas molecule remain same. 55. Answer (1, 2)

Pc =

a 27b 2

Vc = 3b. 56. Answer (1, 3, 4)

ump 2RT M 1

: uaverage : urms :

8RT πM

: 1.128

:

3RT M

: 1.224

57. Answer (2, 4) For spontaneous process ΔG < 0, ΔH < 0 and ΔE < 0 58. Answer (1, 4) Absolute values of entropy and internal energy cannot be calculated. 59. Answer (2, 4) Statement of IInd law of thermodynamics. 60. Answer (1, 2, 3, 4) All are well known relations. 61. Answer (2, 3, 4) If P, Q are arbitrarily chosen intensive variables then P/Q, PQ are intensive variables and

dP is intensive dQ

property. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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62. Answer (1, 2) ΔE is a state function ∴

It is zero is cyclic process internal energy of ideal gas depends only on temperature

In isothermal process ΔE = 0.

63. Answer (1, 2, 4) In isothermal process T = constant P1V1 = P2V2 (Boyle’s law at constant T) ΔU = 0 ΔH1 = ΔH2 64. Answer (2, 4) Standard heat of formation of all elements in their standard states is zero ΔHf(O) ≠ 0 and ΔHf (diamond) ≠ 0 because these are not standard state. 65. Answer (1, 3, 4) State function depends only on initial and final position therefore Enthalpy, Entropy, Gibbs free energy are state function. 66. Answer (2, 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. 67. Answer (1, 3, 4) Work done in reversible process is more than work done in irreversible process at equilibrium ΔG is zero. 68. Answer (3, 4) BOH +

HCl 3 x 4

x x−

3 x 4

0

pOH = pK b + log

BCl 0 ⎛ 3x ⎞ ⎜ ⎟ ⎝ 4 ⎠

+

H2O 0 ⎛ 3x ⎞ ⎜ ⎟ ⎝ 4 ⎠

[Salt ] 3x × 4 = 5 + log [BoH] 4× x

pH = 14 – 5 – log 3 = 8.523. 68.(a). Answer (3)

(IIT-JEE 2008)

At equivalence point

N1V1 = N2 V2 (base)

2.5 ×

(acid)

2 2 = ×V 5 15

V = 7.5 ml ∴ Milli equivalents of salt = 1 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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pH = 7 –

Physical Chemistry

1 1 pkb – log C 2 2

=7–6–

1 1 log 10 2

= 7 – 6 + 0.5 = 1.5 (H+) = 10–1.5 = 3.2 × 10–2 M 69. Answer (1, 3) At equilibrium ΔG = 0 ΔG° = – 2.303 RT Log K – nF E °cell = – 2.303 RT log K At 25°C E°cell =

0.0591 log K . n

70. Answer (1, 2, 3, 4) Pressure favours the forward reaction. The temperature at which atmospheric pressure is equal to vapour pressure is called boiling point if pressure boiling point will be increased . 71. Answer (1, 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. 72. Answer (1, 2, 4) NaCN

Na+ + CN–

CN– + H2O

HCN + OH– (basic)

CH3COONa

CH3COO– + Na+

CH3COO– + H2O Na2CO3

CH3COOH + OH– (basic)

2Na+ + CO32–

CO32– + 2H2O

H2CO3 + 2OH– (basic)

73. Answer (1, 2) N2(g) + 3H2(g)

2NH3(g)

ΔH = negative

Since the no. of moles of gases decreases in product sides therefore on increasing the pressure forward reaction favours. Catalyst increases the rate of reaction, therefore NH3 formation will be fast. 74. Answer (1, 2, 3) Electron deficient species are called lewis acid therefore BF3, Ag+ are electron deficient SnCl4 can expand its octet due to vacant d-orbital therefore behaves as a lewis acid. 75. Answer (2, 3) The aqueous solution of NH4Cl and CuSO4 are acidic. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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76. Answer (1, 3, 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. mol

2

6

0

0

after reaction

0

4

2

2

Since solution is basic therefore pH > 7. The aqueous solution of CH3COONH4 is generally neutral ∴

pH > 6.

77. Answer (3, 4) NO3– is a conjugate base of HNO3 which is strong acid HSO4– is a conjugate base of H2SO4 which is strong acid 78. Answer (1, 2, 4) Due to common ion effect solubility of AgCl will be less than water in NaCl, AgNO3 and CaCl2 solution. 79. Answer (1, 2) H+ + In–

Hln

[H+ ][In − ] [HIn]

Ka = [H+] =

K a [acid] [base]

for 75% red

⎡ 75 ⎤ [H+] = 10 −5 ⎢ ⎥ = 3 × 10 −5 ⎣ 25 ⎦ pH = 4.52

⎛ 25 ⎞ 1 ⎟ = × 10 −5 = 5.47 for 75% blue [H+] = 10 −5 ⎜ ⎝ 75 ⎠ 3

80. Answer (2, 3) On increasing the temperature ionic product of water increases so pH and pOH decreases but water will remain neutral. 81. Answer (1, 2, 4)

K ΔH ⎡ 1 1⎤ log 2 = ⎢ − ⎥ K1 2.303R ⎣ T1 T2 ⎦ Since in the option (3) ΔH = 0 because (Eaf = Eab) ∴

Therefore only this reaction is independent of temperature and (K2 = K1) on the other hand there is not ΔH = 0 therefore equilibrium constant depends on temperature.

82. Answer (2, 3, 4) N2(g) + 3H2(g)

2NH3(g)

initial moles

3

9

0

at equilibrium

3–x

9 – 3x

2x

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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, 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.e.,

t1 t1 and because initial molar ratio is 1 : 3. 3 2

W (N2 ) (3 − x )28 ⎛ 14 ⎞ = = ⎜ ⎟ remain same at t = t1 as well as t = t1 . W (H2 ) ( 9 − 3 x )2 ⎝ 3 ⎠ 2 3

83. Answer (1, 2, 4) Upto phenolphthalein NaOH is fully neutralised and Na2CO3 will be converted to NaHCO3. In next step NaHCO3 coming from Na2CO3 neutralised by HCl using methyl orange indicator. So y ml should be less than x ml that required for phenolphthalein end point. 84. Answer (2, 3) Due to smaller size of Li+ is more solvated than Na+ ion therefore conductivity is less than Na+ ion. 85. Answer (1, 3) (Pb2+)A(aq)

Pb(s) (Pb2+)c(aq)

Pb(s)

for this cell E°cell = 0 Ecell

= −

0.059 [Pb 2 + ] A log 2 [Pb 2+ ]C

Ksp = [Pb2+][SO42–] = s2 s=

K sp (PbSO4 )

Ksp = [Pb2+][I–]2 = 4s3 1

⎤3 ⎡ K sp (PbI2 )⎥ s= ⎢ ⎦ ⎣ 4

1

Ecell

⎡ K sp (PbI2 ) ⎤ 3 ⎥ ⎢ 4 0.059 ⎦ ⎣ log = 1 2 [K sp (PbSO 4 ] 2

Ecell = E°cell –

0.059 [Cd2+ ] log 2 [Cu2+ ]

on increasing the concentration of [Cu2+] and decreasing the concentration of [Cd2+] Ecell will be more positive. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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87. Answer (1, 2, 3) –2

O –1 +2

+1 –1

H — O — S — O — O — H total charge at s = +6 +1

+2

O

–2

–2 –1

O

–1

O

+1 +1

O

CrO5

total charge on Cr = +6.

Cr

O

+1 +2 +1

–1

O

–1

88. 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. 89. Answer (1, 2) +2

+2.5

0

−1

2Na2 S 2 O3 + I2 ⎯⎯→ Na2 S 4 O6 + 2N aI n =1

n=2

n=2

90. Answer (1, 3) Ecell = 0 −

(10 −5 )2 0.059 log = positive 2 (10 −3 )2

E°H+ / H = 0 2

Cr ⎯⎯→ Cr3+(aq) + 3e Cu2+(aq) + 2e ⎯⎯→ Cu

2Cr + 3Cu2+ ⎯⎯→ 2Cr3+ + 3Cu

Ecell = E°cell –

(Cr 3 + )2 0.059 log 6 (Cu2+ )3

= E°cell –

0.059 (0.1)2 6 (0.2)3

= E°cell +

0.059 3 log 2 6

91. Answer (2, 3) The value of the constant A for a given solvent and temperature depends on the type of electrolyte i.e., charge on cation and anion produced on the dissociation of the electrolyte in the solution. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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92. Answer (1, 2, 3) E A / A n+ = E° A / A n+ −

2.303 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 −

2.303 RT 1 log nF [ A n+ ]

93. Answer (1, 2, 3) If a given metal ion has negative reduction potential H+ will be reduced by metal. Similarly if reduction potential is positive metal will be reduced by H2. 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). In aqueous solution containing Zn2+, Na+ and Mg2+ the H+ will get preferentially reduced while in aqueous solution of Cu2+, Ag+, Au3+ these ions will be discharged ahead of H+. 94. Answer (1, 4) As cell proceeds Ecell tend to zero to attain equilibrium state reaction quotient also increases to reach the state of equilibrium. 95. Answer (1, 2, 3) Mole of Fe3+ = 0.1 × 1 = 0.1 Mole of electron =

3600 × 4 = 0.149 mole 96500

Fe3+ + e → Fe2+ 0.100 mol electron required to reduce all the Fe3+ to Fe2+ 0.049 mol electron to reduce the Fe2+ to Fe Fe2+ + 2e → Fe Mole of iron formed =

1 × 0.049 = 0.025 mole Fe 2

O

O

H–O–S–O–O–H O

H – O – S – O – O – S – OH O

O

O

97. Answer (2, 3) +2 −2

0

+4

Cu S ⎯⎯→ Cu+ SO 2 n= 4

0

M 4

equivalent mass =

M 4

−2

O 2 ⎯⎯→ H2 O

n= 4

equivalent mass =

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98. Answer (1, 2, 4) Factual type 99. Answer (3, 4) Original Formula

Number of A atoms =

1 1 ×4= 2 8

Number of B atoms =

1 × 2 =1 2

Number of C atoms =

1 × 12 + 1 = 4 4

Hence, the formula of compound is A1/2 BC4 or AB2 C8. On placing body diagonal plane, 2 corner atoms, 2 edge atoms, 1 body atoms are removed but 2 face atoms may or may not be removed. Possibility I : Suppose, 2 face atoms are removed.

Number of A atoms =

1 1 1 – 2× = 2 8 4

Number of B atoms = 1 – 1 = 0 ⎛1 ⎞ 1 Number of C atoms = 4 – ⎜ × 2 + 1⎟ = 2 ⎝4 ⎠ 2

Hence, formula of compound is A 1 C 5 or AC10 4

2

Possibility II : Suppose 2 face atoms are not removed.

Number of A atoms =

1 1 1 – ×2= 2 8 4

Number of B atoms = 1 – 0 = 1 1 ⎛1 ⎞ Number of C atoms = 4 – ⎜ × 2 + 1⎟ = 2 4 2 ⎝ ⎠

Hence, formula of compound is A 1 BC 4

2

1 2

or AB4 C10

100. Answer (1, 3, 4) In spinel structure (MgAl2O4), O2– ions occupy ccp lattice, two Al3+ occupy 50% octahedral voids and one Mg2+ occupies 12.5% tetrahedral voids. 101. Answer (1, 2, 4) Frenkel defect is shown by those ionic solids in which the difference in the size of cation and anion is very large. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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102. Answer (1, 2, 4) In F-centre, Cl– ions are removed. Ist Combination : Two corner Cl– ions are removed. Number of Na+ ions = 4 Number of Cl– ions = 4 – 2 ×

1 = 3.75 8

Hence, formula per unit cell is Na4Cl3.75. IInd Combination : Two face Cl– ions are removed Number of Na+ ions = 4 Number of Cl– ions = 4 –

1 ×2 = 3 2

Hence, formula per unit cell is Na4Cl3 IIIrd Combination : One corner and one face Cl– ions are removed. Number of Na+ ions = 4 Number of Cl– ions = 4 –

1 1 × 1 + × 1 = 3.375 8 2

Hence, formula per unit cell is Na4Cl3.375. 103. Answer (1, 3) r+ 1 . 69 = = 0 . 86 i.e. coordinate no. = 8 r− 1 . 95

∴ bcc structure ∴

2 (r+ + r− ) = 3a

2 (1.69 + 1.95) 1.732 104. Answer (1, 4) ∴

a=

Body diagonal touches corner & body centre. 105. Answer (1, 4) The given plane represents face plane in fcc. 106. Answer (3, 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. 107. Answer (1, 3) Octahedral voids form at the edge centre as well as the body centre at fcc. 108. Answer (1, 2, 4)

Square close packing coordination number = 4

Hexagonal close packing coordination number = 6

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109. Answer (1, 3) Zn2+ is present in alternate tetrahedral voids therefore its C.N is 4. In rock salt structure there is 4Na+ and 4Cl– ions present in a unit cell. 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. 111. Answer (1, 2, 3) Statement 4 is incorrect because K = A e–Ea/RT. 112. Answer (1, 3) Statement 4 is incorrect because of B decreases then C increases hence there must be a –ve sign. 113. Answer (3, 4) Maxwell and Ostwald theories are exclusively related to chemical kinetics. 114. Answer (2, 3) A has units of rate constant of reaction not rate of reaction. 115. Answer (1, 2, 3, 4) Rate of reaction depends on nature of reactants, temperature, nature of catalyst and surface area of reactants. 116. Answer (1, 2, 4)

log

[R0 ] k·t = [R] 2.303

Comparing with y = mx + C then we get (1) and (4) t½ does not depend on concentration for first order reaction. 117. Answer (1, 2, 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. Answer (1, 3, 4) From the rate expression, overall order of reaction is two & first order w.r.t. [I–] 119. Answer (1, 2, 4) The alkaline hydrolysis of ethyl acetate is second order while others are first order reaction. 120. Answer (1, 2) A is called pre exponential factor. 121. Answer (1, 2, 4) In α-decay, positron emission & k-electron capture, n/p ratio increases while in β-decay, n/p ratio decreases. 122. Answer (1, 2, 3) In (1), (2) and (3) options, effective molarity are same. 123. Answer (1, 2, 3, 4) In all options, effective molarity are same i.e. 2.2 M. So, boiling point of solutions is same. 124. Answer (2, 3, 4) Effective molarity of Ba3(PO4)2 is more than MgSO4. So, the osmotic pressure of Ba3(PO4)2 is more. SPM allow the movement of solvent only not the solute. So, no ppt. of BaSO4 is formed in right side. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Fe(s) Pt(s)

2NH3 (Haber’s process) 2SO3 (Contact process).

135. 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. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Section - C : Linked Comprehension C1. 1. Answer (2) Let the % of B-10 = X then % of B-11 = (100 – X) Average atomic mass =

10 X + (100 − X )11 = 10.2 100

10X + 1100 – 11X = 1020 X = 80 % of B–10 = 80 2. Answer (3) Average atomic mass =

3 × 35 + 1× 37 = 35.5 4

3. Answer (2) Since X–, Y2– and Z3– are isoelectonic therefore Number of electrons in increasing order will be X > Y > Z X–, Y2– and Z3– all have same no. of neutrons. Therefore atomic no. increasing order will be Z 0

2. Answer (2) Entropy change in isothermal process. = 2.303 nR log

ΔS

V2 V1

= 2.303 × 2 × 8.314 log

10 1

= 38.29 JK–1 mol–1 3. Answer (2) ΔH ΔS = T b Tb =

30 × 1000 = 400 K 75

(IIT-JEE 2008)

ΔG = ΔG° + RT ln Q at equilibrium ΔG = 0 Q = Keq ∴ ΔG° = –RT ln Keq C22. 1. Answer (4) Work done in AB process

= –PΔV W1 = 3 (30 – 10) = –60 atm×litre

Work done in B → C process = 0 Work done in C → A process V2 W2 = –2.303 nRT log V 1

= − 2.303(1× 30 ) log

10 30

= +2.303 × 30 × 0.5 = 34.54 atm × litre Net work done in the process = –60 + 34.54 = –25.45 atm × litre Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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2. Answer (1) At point A → PV = nRT T=

3 × 10 = 375 K 0.08 × 1

Temperature at point C = T =

pV 30 = = 375K nR 1× 0.08

3. Answer (2) Since process is cyclic therefore ΔH and ΔU will be zero because in cyclic process state functions will be zero. C23. 1. Answer (2) Since heat of ionisation for HCN is more than CH3COOH. It means HCN is weaker acid than acetic acid. Therefore Ka(HCN) will be less than Ka(CH3COOH) ∴

pK a (HCN) > pK a (CH3 COOH)

2. Answer (3) Na+(aq) + OH– (aq) → NaOH (aq) Heat of ionisation of strong base = 0

ΔH = ΔHf (NaOH) − ΔHf (Na + ) − ΔHf (OH− ) = 0

− 470.7 − ΔHf (Na+ ) − ( −228.8) = 0 ΔHf (Na+ ) = −241.9 kJ

3. Answer (2) ΔQ = nSΔT Molar heat capacity =

ΔQ nΔT

Since in ice water equilibrium there is change in temperature is zero. ∴

Molar heat capacity = ∞

C24. 1. Answer (1) For acidic buffer pH

[ salt ] = pKa + log [acid]

⎡ 2 1000 ⎤ ⎢ 59 × 500 ⎥ ⎣ ⎦ = 4.74 + log 0.1

= 4.74 + log

40 59

= 4.74 – 0.1739 = 4.57 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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2. Answer (3) pH = pKa + log

[ salt ] [acid]

[salt] = [acid] pH

= pKa = 4.74.

3. Answer (3) [ salt ] pH = 4.74 + log [acid]

6.74 = 4.74 + log

[ salt ] [acid]

[ salt ] 100 = [acid] 1 % of acid in the mixture =

1 × 100 101

=1% ∴ % dissociation of acid = 99% C25. 1. Answer (3) Keq

Kf = K r

Kf = Keq × Kr = 1.16 ×10–3 × 57 = 66.12 × 10–3 Kf = 6.612 × 10–2 2. Answer (1) Kf > Kr. 3. Answer (4) Kf ⎞ ⎛ Since reaction is endothermic on increasing the temperature equilibrium constant increases ⎜ K c = K ⎟ r ⎠ ⎝ but Kf increases more than Kr.

CH3COO–

1–α Ka =

α

+

H+ α + 0.01

α( α + 0.01) = 1.8 × 10 – 5 1− α

on solving we get pH = 1.937. 2. Answer (4) [OH–] = 2 × 10–7 pOH

= – log (2×10–7) = 7 – log 2 = 6.7

pH

= 14 – 6.7 = 7.3.

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= [H+] [H–] = 10–14

Ka

=

[H+ ][OH − ] 10 −14 = [H2O] 55.55

= 0.018 × 10–14 = 18 × 10–17. C27. 1. Answer (4) CH3COONa → CH3COO– + Na+ CH3COO– + H2O

CH3COOH + OH– (basic).

H+ + HS solution is acidic [H+] > 10–7 m

NaCl

solution is neutral [H+] = 10–7

NaNO2

solution is basic [H+] < 10–7

H2SO4

solution is strong acidic

0.01 MNaNO2 < 0.01 MNaCl < 0.01 MH2S < 0.01 MH2SO4 3. Answer (2)

CH3COOH + NaOH m.moles (initial)

10

After reaction

6

CH3COONa + H2O

4 0 buffer solution

0

0

4

4

[ salt ] pH = pKa + log [acid]

pH – pKa = log

4 6

⎛ 2⎞ = log ⎜ 3 ⎟ . ⎝ ⎠

C28. 1. Answer (3) There is no effect at equilibrium at constant volume. 2. Answer (4) NaNO3(s)

NaNO2(s) +

1 O (g) 2 2

∴ since no. of gaseous moles decrese in backward direction thererfore on increasing the presure reverse reaction favour. 3. Answer (2) C2H4(g) + H2(g)

C2H6(g), ΔH = – 136.8

Since reaction is exothermic therefore decrease in temperature favour forward reaction. Since no. of moles decreases in forward direction therefore on increasing the pressure forward reaction favours. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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C29. 1. Answer (1) H2(g) → 2H+ + 2e (anode) 2H+ + 2e → H2 (cathode) Ecell

= –

(PH2 )C 0.059 log 2 (PH2 ) A

0.059 log⎛ 1 ⎞ = – ⎜ ⎟ 2 ⎝2⎠ = +

=

0.059 log 2 2

0.059 × 0.3010 = 8.90 mV 2

2. Answer (1) H2 → (2H+)A + 2e (2H+)C + 2e → H2

Ecell

[ ] [ ]

2 A 2 C

H+ = 0 − 0.059 log 2 H+

⎛ C1 ⎞ = – 0.059 log ⎜ C ⎟ ⎝ 2⎠

for non spontaneous process C1 > C2. 3. Answer (1) Zn → Zn2+ + 2e 2H+ + 2e → H2 Ecell

[ ] [ ]

Zn2+ = E°cell − 0.059 log 2 2 H+

0.28 = – 0.78 –

0.059 log 2

C1

(KC )

2

a

2

2.12 = log⎛ C1 ⎞ ⇒ log⎛ C1 ⎞ = −36 ⎜ ⎟ ⎜ ⎟ . 0.059 ⎝ K aC2 ⎠ ⎝ K aC2 ⎠

C1 = 10−36 K aC2 Ka =

C1 × 1036 C2

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C30. 1. Answer (2) Specific conductivity is directly proportional to the concentration. 2. Answer (3)

1 ⎛l⎞ K = R ×⎜a⎟ ⎝ ⎠ 1 ⎛l⎞ 1.2 = 55 × ⎜ a ⎟ ⎝ ⎠ ⎛l⎞ ⎜ ⎟ = 1.2 × 55 = 66 m −1 ⎝a⎠

or 66 × 10–2 cm–1. 3. Answer (4) On doubling the edge length, volume of cube becomes 8 times and the concentration of solution decreases 8 times. Specific conductance is directly proportional to concentration. Hence, it decreases 8 times. C31. 1. Answer (3) Λ=

1000 × K = 1000 × 4.95 × 10 –5 N 0.001028 = 48.15

⎛ Λ ⎞ ⎛ 48.15 ⎞ degree of dissociation = ⎜ ∞ ⎟ = ⎜ 390.5 ⎟ ⎠ ⎝Λ ⎠ ⎝ = 0.123. 2. Answer (3)

Λ∞m = 119 + 160 = 279

degree of dissociation

α

Λ = Λ∞ m

0.1 = Λ

Λ 279

= 27.9.

1000 × K M

K

27.9 × 0.1 1000

=

= 2.79 × 10–3. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Anion is bigger Q it will constitute lattice

r + 0.65 = = 0.294 (i.e. range of coordination number = 4) r − 2.21 2. Answer (1)

r+ = 0.414 (for maximum packing efficiency) r− r− =

80 = 193.23 pm 0.414

3. Answer (2) oh voids in fcc is situated at all edge centres and body centre and td voids are present at each body diagonal. Nearest oh and td void pair is oh void at body centre and td void at edge centre.

distance =

3 a 4 td void

√3 a 4

body √3a diagonal

oh void

td void

1 8

B

1 1 2 × + 12 × 2 4

C

1 +4 2

⇒ AB4C6 2. 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. If B is removed, formula of left compound = AB3C6 If C is removed, formula of left compound = AB4C5 3. Answer (1) A at corner B at 2 face centre and C at rest 4 face centres i.e. the basic lattice is ccp ∴

No. 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 = 0.58 12

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C34. 1. Answer (3) Density of CsCl Z = 1 (1 formula unit is present)

3a = 2(r+ + r− )

For bcc structure

a=

Volume =

ρ=

[2(r+ + r− )]3 ( 3)

3

2(r+ + r− ) 3

pm

× 10 −30 cm3

MCsCl × ( 3 )3

[2(r+ + r− )]3 × NA × 10 −30

ρfcc 4 (2.8)3 = × = 5.48 3 ρbcc (2) 2 3. 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

body diagonal plane of fcc.

Rectangular plane (fcc)

body diagonal plane (Bcc)

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C36. 1. 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. Answer (2) % of voids occupied =

2 (oh) + 1( td) × 100 = 25% 12

C37. 1. Answer (1) On increasing temperature, K increases but Ea and A remains same. 2. Answer (3) log

K2 E a ⎡ T1 − T2 ⎤ =− ⎢ ⎥ K1 2.303R ⎣ T1 ⋅ T2 ⎦

log 4 = −

Ea ⎡ 700 − 800 ⎤ 2.303 × 8.314 ⎢⎣ 700 × 800 ⎥⎦

By solving, Ea = 64 kJ 3. Answer (4) According to Arrhenious equation, log K = log A −

When

Ea 2.303RT

Ea becomes zero then K = A. 2.303RT

C38. 1. Answer (3) On β-emission,

n n ratio decreases while in K-electron capture, α-emission and positron emission, p p

ratio increases. 2. Answer (4) 0 13 13 +1 e 7 N ⎯⎯⎯→ 6 C unstable stable

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(IIT-JEE 2008)

Z (At no.)

38(a).

n ratio i.e. 2. p

45o no. 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. C39. 1. Answer (3) Rate ∝ [A]x 2 ∝ (4)x (4)1/2 ∝ (4)x

1 2

x=

2. Answer (1) In the 1st experiment, Rate ∝ [A]x (3)3 = 27 ∝ (3)x x=3 In the IInd experiment, Rate ∝ [A]x [B]y 8 ∝ (2)3 (2)y y=0 3. Answer (4) The order w.r.t. B is zero. C40. 1. Answer (3) Lesser is the half life, more is the radioactivity. 2. Answer (4)

K=

2.303 a log t (a − x )

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Physical Chemistry

100 0.693 2.303 = log 69.3 t 20

t = 161 minutes 3. Answer (1) K1 =

100 2.303 log 60 10

…(1)

K2 =

100 2.303 log t 10

…(2)

K1 = K2 100 2.303 100 2.303 log = log 10 60 t 10

t = 45 minutes C41. 1. Answer (3) 1 4 = 0.2. xtoluene = = 0.8 5 5

xbenzene =

According to Roult’s law o o PT = Pbenzene × x benzene + Ptoluene × x toluene

PT = 700 × 0.2 + 600 × 0.8 PT = 140 + 480 = 620 mm 2. Answer (2) Suppose, x g benzene and x g toluene are mixed nbenzene =

x x , 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 + 275.29 PT = 654.11 mm Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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3. Answer (2) Mole fraction of benzene in vapour phase =

Pbenzene 140 = = 0.225 PT 620

C42. 1. Answer (2) 3 3 1000 × = 0.1 m = 60 = 500 60 500 1000

In beaker ‘A’ ΔTb = i × Kb × m 0.17 = i × 1.7 × 0.1 i=1 It shows that acetic acid remains normal molecule in acetone. In beaker B ΔTb = i × Kb × m 0.13 = i × 2.6 × 0.1 i = 0.5 It shows that acetic acid is 100% dimerised in benzene. 2. Answer (1) In acetone, acetic acid remains the normal molecule. Hence, molecular weight of acetic acid will be 60. 3. Answer (2) In benzene, acetic acid dimerises 100%. Hence molecular weight of acetic acid will be 120. C43. 1. Answer (3) K3[Fe(CN)6]

3K+ + [Fe(CN)6]3–

1

0

0

before dissociation

(1 – α)

α

after dissociation

i=

1 + 3α (Q α = 0.6) 1

i=

1 + 3 × 0.6 = 2.8 1

2. Answer (4) The ratio of effective molarity of 0.5 M AlCl3, 2 M urea and 0.2 M K4[Fe(CN)6] is 2 : 2 : 1. Hence, the ratio of osmotic pressure is also 2 : 2 : 1. 3. Answer (2) 2CH3COOH

(CH3COOH)2

1

0

before association

(1 – α)

α 2

after association

Here, α = degree of dimerisation

i=

1 − 0.5α = 0 .8 1

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α=

Physical Chemistry

0.2 = 0.4 0.5

Percentage of dimerisation = 40%. C44. 1. Answer (3) Fe(OH)3 sol is positively charged. 2. 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. 3. Answer (3) As2S3 is negatively charged sol. C45. 1. Answer (3) According to Freundlich adsorption isotherm, log

1 x = log K + log P m n

Comparing the above equation with y = mx + c then we get, intercept = log K and slope =

1 . n

2. Answer (2) According to Langmuir adsorption isotherm, m b (1) = + x a aP

Comparing the above equation with y = mx + c then we get, slope =

1 b , intercept = . a a

3. Answer (1) According to Langmuir adsorption isotherm, x aP = m 1 + bP

At very low pressure, bP is negligible in comparison to 1. Then we get x = aP . m

Section - D : Assertion - Reason Type 1.

Answer (2) In this reaction Br2 is oxidised as well as reduced.

2.

Answer (2) Volume strength = 5.6 × normality

3.

Answer (3) Mg2+, Na+, O2–, F– all have 10 electrons therefore isoelectronic.

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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.

5.

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.

6.

Answer (2) Oxidation number of sulphur in H2SO4 is +6. It is in highest oxidation state therefore can acts as a oxidising agent.

7.

Na 2 S2 O 3 + I2 ⎯⎯→ Na 2S 4O 6 + 2NaI−1 n=2

8.

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 = 1.2 × 5 = 6 9.

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 = 6.023 × 1024.

10. Answer (3) Fe(26) → 1s2, 2s2, 2p6, 3s2, 3p6, 3d 6, 4s2 Fe2+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d 6 – unpaired electrons = 4 Fe3+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d 5 – (Half filled d-orbital more stable) unpaired electrons = 5 11. Answer (2) Cu(30) → 1s2, 2s2, 2p6, 3s2, 3p6, 3d 10, 4s1 Cu+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d 10 – No unpaired electrons

∴ diamagnetic

Cu2+ → No. of unpaired electrons = 1 (paramagnetic) 12. Answer (1) Heisenberg’s uncertainty principle is applicable for microscopic particle. 13. Answer (2) He(2) – 1s2

1s Maximum number of electrons in an orbital is two. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

14. 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. 15. Answer (4) Fixed circular path around nucleus is not possible because it violates the Heisenberg’s uncertainty principle. 16. 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. It is clear from this expression λ =

h mV

17. Answer (4) de Broglie wavelength λ =

h . It is applicable to moving particle. mV

18. Answer (2) Zn(30) → 1s2, 2s2, 2p6, 3s2, 3p6, 3d 10, 4s2 Zn2+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d 10 Since Zn2+ has no unpaired electrons therefore it is diamagnetic. 19. Answer (4)

y

x

dx2 – y2 In d x 2 − y 2 the electron density lies along the x and y axis. 20. Answer (1) The value of l depends on the n. For n = 3 l = 0, 1, 2

(because value of l ranging from 0 to (n – 1)

Value of m depends on l and ranging from –l to +l. 21. Answer (3) Energy of entron in nth orbit for H-atom or H-like ions = −13.6

Z2 n2

It is clear the energy of electrons depends on principal quantum number. In other atom the energy depends on n as well as value of l. 22. Answer (3) Orbital angular momentum = l (l + 1)

h 2π

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23. Answer (3) Be(4) → 1s2, 2s2 (full filled s-orbital) - more stable B(5) → 1s2, 2s2, 2p1 - less stable 2p orbital has higher energy than 2s-orbital. 24. Answer (3) Due to smaller size of fluorine atom there is great electron-2 repulsion therefore electron affinity of fluorine is less than chlorine. 25. Answer (3) Atomic volume is a guide to the size of atoms. If atomic radius increases atomic volume also increases. 26. Answer (4) Be and Al shows diagonal relationship. 27. Answer (2) BaSO4 is insoluble in water. Ionic radius of Mg2+ is smaller than Ba2+. 28. Answer (2) Oxidising power of fluorine is more than oxygen, due to greater reduction potential. 29. Answer (1) OF2 oxidation number of oxygen x + 2(–1) = 0 x = +2 The electronegativity of fluorine is highest in periodic table.

:

P

:P

P:

Hybridisation of p is sp3 and each p atom is liked with three other p atoms.

:

P

:

:

N

N

F

F F Subtraction

H

therefore dipole moment of NH3 is more than NF3

I

:

⎡ ⎢ ⎣

I :

I Linear

⎤ :⎥ ⎦

Hybridisation of central atom is sp3d

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Physical Chemistry

:

O

:

O:

O:–

:

O:

:

:

N :

N –

:

:

:

N 5 +1 = = 3 attached atoms = 2 2 2

shape is angular

34. Answer (3) Bond order of O2 ⎡π 2 py2 ⎤ ⎥ O2(16) — σ1s 2 σ * 1s 2 σ2s 2 σ * 2s 2 σ2 px 2 ⎢ ⎢⎣π 2 pz2 ⎥⎦

B.O. =

10 − 6 =2 2

⎡π * 2 p1y ⎤ ⎢ ⎥ ⎢⎣π * 2 p1z ⎥⎦

Number of unpaired electrons = 2

He2 ( 4) − σ1s 2 σ * 1s 2 B.O. =

2−2 =0 2

He +2 (3) bond order =

2 −1 1 = 2 2

Since B.O. of He2 is zero therefore it does not exist. 36. Answer (2) B.O. of N2− = 2.5 = B.O. of N+2 But B.O. of N22− is 2. Hence it is least stable. ⎡π2p 2y ⎤ 2 2 2 2 ⎥ σ2p 2x N2(14) – σ1s σ * 1s σ2s σ * 2s ⎢ ⎢⎣π2p 2z ⎥⎦

37. 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. 38. Answer (3) Due to available vacant d-orbital only SiCl4 reacts with water both are covalent compound. 39. Answer (3) LiCl is covalent in nature due to smaller size of Li+ ion. The electronegativity of Li and Cl is not small. 40. Answer (2) Hydrogen in non polar molecule due to absence of electronegativity difference between the two H-atoms. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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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.

3 RT) 2

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.

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 ⎜ V 2 ⎟⎠ ⎝

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

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Physical Chemistry

53. Answer (4) V1 V2 = T1 V2 V2 =

V V = 2 300 400

4 V 3

At constant pressure gases follows the Charle’s law V ∝ T (at constant P) 54. Answer (1) ΔE is a state function because it depends only on initial and final position only. 55. Answer (1) In cyclic process the change in enthalpy (ΔH), change in entropy (ΔS) and change in free energy (ΔG) 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 ΔG must be negative. ΔG = ΔH – TΔS 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) ΔG = ΔH – TΔS If ΔS is negative then at high temperature TΔS will be more than ΔH therefore ΔG will be positive and reaction will be non spontaneous at low temperature. ΔH will be more than TΔS therefore ΔG will be negative and reaction will be spontaneous. 61. Answer (3) ΔG = ΔH – TΔS Since ΔS = negative ∴

TΔS is positive.

If ΔG = (–) then ΔH 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

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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

pH + pOH = pKw ∴

pH =

pKw 2

66. Answer (1) CuO + H2

Cu + H2O at 75ºC the water in form of liquid

Kp = Kc(RT)–1

ΔH = –1

Kc > kp ∴

But at 175º water in forms of gaseous

ΔH = 0

Kp = Kc 67. Answer (4) The aqueous solution of salt of strong base with weak acid is basic

∴ pH > 7

due to cationic hydrolysis the solution becomes acidic. ∴

pH < 7

68. Answer (3) Solubility of salt increases on dilution the solubility product (Ksp) depends only on temperature. 69. Answer (3)

H3PO 4

H2PO−4 Ka1 > Ka2

H+ + H2PO −4

Ka1

H+ + HPO24−

Ka2

∴ H3PO4 is stronger acid then H2PO−4

pKa1 < pKa2 70. Answer (1) 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)

B(g) + C(g)

Initial moles

1

0

0

at equilibrium

1–x

x

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

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Physical Chemistry

CH3COONa + HCl 2M 1M 1M

CH3COOH + NaCl 0 0

0

1M

1M

Acidic buffer solution

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. NH4OH

75. Answer (1) Blood is a basic buffer solution of (H2CO3 + HCO3− ) . 76. Answer (2)

NH+4 H + H—N—H

form of charge on nitrogen is +1.

–1

O

+1

–1

O +2

O

Cr O

+1

+1

+1

oxidation number of chromium is +6.

O

–1

–1

78. 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 = 0.6 × 5 79. Answer (2) SO 2 + 2H2S ⎯⎯→ 3S + 2H2O

Bleaching action of SO2 is due to reduction. 80. Answer (2) In electrochemical cell the chemical energy is converted to electrical energy for this process the cell reaction should be spontaneous. For spontaneous process ΔG must be negavite. 81. Answer (3) Reduction potential of Na is very less than water therefore water reduced first and H2 gas evolved at cathode. Liberation or deposition depends on the reduction potential. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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82. Answer (2) At in finite dilution, molar conductivity of any electrolyte (strong or weak) is equal to the sum of molar conductivities of the ions produced by it. 83. Answer (1) KCl is used in salt bridge because K+ and Cl– ions have nearly same ionic mobility. 84. Answer (2) º For concentration cell E cell =0

º E cell =−

0.0591 ⎛ C1 ⎞ ⎟⎟ log⎜⎜ 2 ⎝ C2 ⎠

For spontaneous process C1 < C2 85. Answer (3) Equivalent conductance is directly proportional to the dilution because mobility of ions increase. 86. Answer (3) When CuSO4 is electrolysed then Cu deposited at cathode and O2 gas is evolved at anode. 87. Answer (1) When Zn electrodes are used in ZnSO4 solution then it acts as attacking electrodes. 88. Answer (3) 1 Faraday electricity is required to deposit 1 equivalent of substance not one mole 89. Answer (2) NaCl ⇒ coordination no. ratio = 6 : 6 Inverse simplified ratio = 1 : 1 Molar ratio = 1 : 1 For CaF2 ⇒ coordination no. ratio = 8 : 4 Inverse simplified ratio = 1 : 2 Molar ratio = 1 : 2 ∴ Statement (1) and (2) both are correct 90. Answer (1) r+ = 0.414 to 0.732 (for coordination no. = 6) r−

When

r+ = 0.414 size of r+ is least to be fitted in oh void r−

but when

r+ = 0.732 size of r+ is maximum to be fitted in oh void which would cause expansion of lattice. r−

91. Answer (2) Cubic unit cell has no. of atoms = 1 bcc unit cell has no. of atoms = 2 fcc unit cell has no. of atoms = 4 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

92. Answer (4) Inverse and normal spinel structure have same packing efficiency. 93. Answer (1) Ions come closer in crystals suffering from frenkel defects. 94. Answer (2) ccp has ABC ABC..... pattern of stacking. 95. Answer (1) Distance between 2 nearest spheres in fcc = Distance between 2 nearest spheres in bcc =

2 a 2

3 a 2

96. Answer (2) Ferromagnetic substances are those paramagnetic substance which persists their magnetic moment (i.e. spin alignment) even in absence of magnetic field. They turn to paramagnetic substance when heated above curies temperature. 97. Answer (3) Common salt has F electrons, responsible for colour. 98. Answer (4) 10–3 mole of SrCl2 will develop 10–3 NA cationic vacancies. 99. Answer (2) Each tetrahedral void is at ¼th distance from corner at body diagnol. 100. Answer (1) Co-ordination number in bcc is 8. 101. Answer (1) For LiCl, Cl– will constitute lattice. 102. Answer (1) ZnS :- 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. Answer (4) The value of Arrhenious constant is not affected by temperature. 104. Answer (1) The end product of (4n + 2) series is

206 82 Pb .

105. Answer (1) For second order, t1/2 is inversely proportional to concentration. 106. Answer (2) The rate law is −

1 d[ A ] d[B] d[C] =− =+ = Rate of reaction 2 dt dt dt

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Physical Chemistry

Section - E : Matrix-Match Type 1. Answer - A(p, q, r), B(q, s), C(p, q), D(q) (A)

1.7 gm NH3 moles =

1.7 = 0.1 17

molecules = 0.1 N0 atoms = 0.4 N0 volume = 2.24 litre no. of electrons = 0.1 N0 × 10 = N0 (B)

3.2 gm oxygen moles =

3.2 = 0.1 32

molecules = 0.1 N0 atoms = 0.2 N0 volume = 2.24 L (C)

2.6 gm C2H2 moles =

2 .6 = 0 .1 26

molecules = 0.1 N0 atoms = 0.4 N0 volume = 2.24 L (D)

6.4 gm SO2 moles =

6 .4 = 0 .1 64

molecules = 0.1 N0 atoms = 0.3 N0 volume = 2.24 L 2. Answer - A(p, q), B(q, r), C(r, s), 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)

64.4% metal in metal oxide

Q 35.6 gm oxygen reacts with 64.4 gm metal Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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∴ 8 gm oxygen reacts with =

8 × 64.4 = 14.5 35.6

⇒ Equivalent weight of metal = 14.5 Equivalent weight of oxygen = 8 (C)

29% metal in metal chloride

Q 71 gm chlorine reacts with 29 gm metal ∴ 35.5 gm chlorine reacts with =

35.5 × 29 = 14.5 71

⇒ Equivalent weight of metal = 14.5 (D)

Equivalent weight of Mg =

24 = 12 2

3. Answer - A(r, s), B(q), C(p, r) D(r) Molarity = No. of moles of solute (Temperatur e dependent ) Volume (litre)

Molality =

No. of moles of solute (Temperatur e independen t ) Mass of solvent (kg)

Formality =

No. of gram formula mass of solute (Temperatu re dependent) Volume (litre)

Strength of solution – It is defined as the amount of solute in grams present in one litre of solution. (temperature dependent) ∴

Volume of solution depends on the temperature.

4. Answer - A(p, r), B(p, q, r), C(s), D(r) HNO3 ⇒ Oxidation number of nitrogen is +5. Since it is in highest oxidation state therefore reduction is possible hence acts as a oxidising agent +5

HNO2 ⇒ Oxidation No. of Nitrogen = +3 ∴

Oxidation

acts as a reducing as well as oxidising agent

+3 Reduction

N2O – Oxidation No. of Nitrogen = +1 –3

N3H is called hydrazoic acid 5. Answer - A(r), B(p), C(q), D(s) (A)

+3

+2 +3

+4

Fe C2 O 4 ⎯⎯→ Fe 3+ + CO 2

n factor of FeC2O4 = 1 + 1 × 2 = 3 +1 −2

(B)

+2

+4

Cu2 S ⎯⎯→ Cu2+ + SO 2

n factor of Cu2S = 1 × 2 + 6 = 8 (C)

+2 −1

+3

+4

Fe S 2 ⎯⎯→ Fe3+ + SO2 n factor of FeS2 = 1 + 5 × 2 = 11

(D)

+3

+5

+2

+2

Fe(NO3 )3 ⎯⎯→ Fe + 2 + NO

n factor of Fe(NO3)3 = 1 + 3 × 3 = 10 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

6. Answer - A(s), B(r), C(q), D(p) (A)

N2 = No. of electron = 2 × 7 = 14 e– CO = No. of electron = 6 + 8 = 14

e–

CN– = No. of electron = 6 + 7 + 1 = 14 e– (B)

238 92U

⎫ ⎪ ⎬ (isoelectronic ) ⎪ ⎭

no. of neutrons = 238 – 92 = 146

(n – p) = 146 – 92 = 54 234 90U

no. of neutrons = 234 – 90 = 144

(n – p) = 144 – 90 = 54 ∴

(n – p) is same in both, hence isodiaphers

(C)

15 14 16 7N 6N 8O

all have 8 no. of neutrons therefore isotones

(D)

14 14 7N 6N

since mass no. is same therefore isobars

7. Answer - A(p, r) B(p, s), C(q, r), D(s) No. of spectrum =

(n2 − n1 )(n 2 − n1 + 1) 2

(A)

No. of spectrum =

(5 − 2)(5 − 2 + 1) = 6 belongs to visible region 2

(B)

No. of spectrum =

( 6 − 3)( 6 − 3 + 1) = 6 belongs to infrared region 2

(C)

No. of spectrum =

( 4 − 2)( 4 − 2 + 1) = 3 belongs to visible region 2

(D)

No. of spectrum =

(8 − 4)(8 − 4 + 1) = 10 belongs to infrared region 2

8. Answer - A(p, r), B(q, s), C(q, s), D(s) Zn2+ —

1s2, 2s2, 2p6, 3s2, 3p6, 3d10 4s0 No. of unpaired electrons = 0

Cr3+ —

1s2, 2s2, 2p6, 3s2, 3p6, 3d3 No. of unpaired electrons = 3

Co2+ —

∴ Paramagnetic

1s2, 2s2, 2p6, 3s2, 3p6, 3d 7 No. of unpaired electrons = 3

Cu2+ —

∴ Diamagnetic

∴ Paramagnetic

1s2, 2s2, 2p6, 3s2, 3p6, 3d 9 No. of unpaired electrons = 1

∴ Paramagnetic

9. Answer - A(r, s), B(p, s), C(r, q), D(p, q) (A)

No. of spectrum = =

(n2 − n1)(n2 − n1 + 1) 2 (6 − 3)( 6 − 3 + 1) = 6 Infrared region 2

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Physical Chemistry

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(B)

No. of spectrum =

(7 − 3 )(7 − 3 + 1) = 10 – Infrared region 2

(C)

No. of spectrum =

(5 − 2)(5 − 2 + 1) = 6 – Visible region 2

(D)

No. of spectrum =

( 6 − 2)( 6 − 2 + 1) = 10 – Visible region 2

10. Answer - A(p), B(p, q, r, s), C(p), D(p, q) Hydrogne atom Z = 1, –1s1 Hydrogen and He+ are monovalent therefore energy deciding factor is only principal quantum number. But in case of nitrogen multielectronic species energy depends on both principal and azimuthal quantum number N(7) – 1s2, 2s2, 2p3 Here energy of valence electrons depends on the exchange energy and symmetry also. 11. Answer - A(r), B(q), C(p, r), D(p, r, s) (A)

Orbital angular momentum =

l(l + 1)

h 2π

for p-orbital l = 1 ∴ Orbital angular momentum =

2

h 2π

nh where n is principal quantum number 2π

(B)

Angular momentum mvr =

(C)

p, d-subshell has 3 and 5 degenarate orbitals respectively

(D)

For N-shell n = 4 ∴ l = 0, 1, 2, 3 Therefore in N-shell d and p-subshell will be present and number of waves = principal quantum number = 4

12. Answer - A(q), B(r), C(s), D(p) (A)

Fluorine has maximum electronegativity

(B)

Chlorine has maximum electron affinity

(C)

Fe is transition element therefore has variable valency

(D)

He is inert gas (most stable configuration) therefore has maximum ionisation energy

13. Answer - A(q), B(p), C(s), D(r) (A)

Alkali metals are group 1 elements K belongs to group one

(B)

Alkaline earth metals are group 2 elements Ba belongs to group 2

(C)

(D)

As is metalloid.

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Physical Chemistry

14. Answer - A(q, s), B(r), C(p, q, s), D(p, s) (A)

Chlorine has highest electron affinity and forms oxy acids like HClO, HClO2

(B)

Fluorine is most electronegative elements

(C)

Sulphur exists in allotropic forms oxyacids like H2SO3, H2SO4

(D)

Phosphorus exist in allotropic form and forms oxyacids like H3PO4, H3PO3

15. Answer - A(q, s), B(p, r), C(s), D(p) (A)

van der Waal’s radius is bigger the covalent radius and used for gaseous molecules

(B)

Covalent radius is used for covalent molecules HCl, Cl2

(C)

(D)

Stevenson equation is used to calculate actual bond length of polar covalent molecules.

16. Answer - A(p, r), B(q, s), C(p, s), D(p, r)

O=C=O + F H

Subtractive, μR = 0 H

C=C

CH 3

F Substractive, μR ≠ 0 (both have different bond moment)

I F B

F Substractive, μR = 0.

F

17. Answer - A(p), B(q), C(p, r), D(q, s) (A)

Acid has pH < 7

(B)

Base has pH > 7

(C)

Acidic buffer is a mixture of weak acid + salt of this acid with strong base

(D)

Basic buffer is a mixture of weak base + salt of this base with strong acid

18. Answer - A(r, s), B(s), C(p), D(q, 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

=

Nb − Na 10 − 6 = =2 2 2

2 unpaired electrons

∴ Paramagnetic

Bond order of O−2

=

10 − 7 = 1.5 2

1 unpaired electrons

∴ Paramagnetic

Bond order of O22−

=

10 − 8 =1 2

zero unpaired electron ∴ Diamagnetic

Bond order of O+2

=

10 − 5 = 2.5 2

one unpaired electron ∴ Paramagnetic

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19. Answer - A(p, s), B(q, s), C(r, s), D(r) (A)

NO2+

N 5 −1 = =2 2 2

∴ hybridisation sp

(B)

NO 3−

N 5 +1 = =3 2 2

∴ hybridisation sp2

(C)

NH+4

N 5 + 4 −1 = =4 2 2

∴ hybridisation sp3

(D)

NH3

N 5+3 = =4 2 2

∴ hybridisation sp3, one lone pair present on central atom

20. Answer - A(p, r), B(r, s), C(r, s), D(q, s) (A)

SF6

N 6+6 = =6 2 2

sp3d2

(B)

XeF4

N 8+4 = =6 2 2

sp3d2

two lone pairs on central atoms

(C)

BrF5

N 7+5 = =6 2 2

sp3d2

one lone pair on central atom

(D)

ClF3

N 7+3 = =5 2 2

sp3d

lone pair on central atom and T-shaped

21. Answer - A(q, s), B(p, r, s), C(p, r, s), D(q, r, s) N = (No. of valence electrons of centre atoms) + No. of atoms attached to this (A)

CaF2 is insoluble in water and ionic compound

(B)

BeSO4 is soluble in water and has Be2+ + SO42– ionic bond as well as covalent bond

(C)

Na2CO3 is soluble in water and ionic as well as covalent bond

(D)

Ca2+ + CO32– has ionic and covalent bond insoluble in water

22. Answer - A(p, q, r, s), B(p, q, r, s), C(r, s), D(p, r, s) (A)

Na2B4O7·10H2O exists as

OH O +

2 Na

HO

B

O O

B O

B

B

O

OH · 8H O 2

OH ∴ Therefore hybridisation of boron is sp3 and sp2. (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. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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(C)

Physical Chemistry

K2 [Ni(CN)4] can be written as 2K+ + [Ni(CN)4]2– Hybridisation of [Ni(CN)4]2– is dsp2. Both ionic and covlaent bond are present.

(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. 23. Answer - A(p, r), B(r, s), C(q, r), D(q, r) (A)

Br F5 + 3H2O ⎯⎯→ HBrO3 + 5HF sp3d2

(B)

H3 BO3 + H2O ⎯⎯→ [B(OH)4]– + H+ sp2

(C)

sp3

Cl F3 + Sb F5 ⎯⎯→ [Cl F2]+ [SbF6]– sp3d

(D)

sp3

sp3

P Cl5 + 4H2O ⎯⎯→ H3PO4 + 5HCl sp3d

sp3

24. Answer - A(p, q, s), B(p, q), C(p, q, r), D(p, q) Boyle temperature is the temperature at which real gases Behave like ideal gas TB =

a Rb

Ti =

2a Rb

TC =

8a 27Rb

a, b depend on the nature of gas For ideal gas second viral coefficient is zero. 25. Answer - A(q, s), B(p, r, s), C(s), D(r) (A) As molar mass increases, van der Waal force of attraction increases hence boiling point increases. van der Waal’s constant a depends on polarizability of molecule. van der Waal’s constant 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. (C) a depends on polarizability of molecule. (D) b depends on molecular size. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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26. Answer - A(p, q, r, s), B(p, r, s), C(s), D(r, 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.e. ⎜ KT ⎟ ⎝2 ⎠ 27. Answer - A(r), B(s), C(q), 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

PV = RT − Z=

(C)

a V

PV ⎛ a ⎞ = ⎜1 − ⎟ RT ⎝ RTV ⎠

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. Answer - A(r), B(p), C(s), D(q) Vrms =

3RT M

VAV

=

8RT πM

VMP

=

2RT M

PV

=

1 mnc 2 3

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Physical Chemistry

29. Answer - A(s), B(q), C(p), D(r) (A)

At high the molecules come closer and intermolecular forces are dominated therefore gases follow the vander Waal’s equation

(B)

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

a v2

~0 –

P (V – b) = RT PV – Pb = RT PV = RT + Pb (D)

At very high temperature and low pressure gas behaves as a ideal gas therefore follows the ideal gas equation PV = RT

30. Answer - A(s), B(p, q, s), C(p, q, r, s), D(p, q, s) (A)

Kinetic energy of gases depends only on temperature.

(B)

Partial pressure = (mole fraction) × total pressure Pressure ∝ temperature

(C)

Rate of diffusion ∝

1 Molecular Mass

Rate of diffusion also depends on temperature. (D)

Vapour pressure is directly proportional to the mole fraction and pressure also depends on temperature.

31. Answer - A(q, s), B(q), C(q, p), D(r) (A)

The critical temperature of CO2 is approximately 31.2°. Since temperature is below than this therefore CO2 exists as a liquid

(B)

Critical temperature TC =

(C)

3 = PC VC 8 RTC

89 27 Rb

When the compressibility factor is less than one then Vreal is less the Videal 32. Answer - A(q), B(q, r, s), C(q, r), D(p) (A)

Heat of combustion.

(B)

The heat evolved in formation of H2O is the heat of combustion of H2, heat of formation of H2O and it is used in fuel cell.

(C)

The heat evolved of CO2 is the heat of combustion of carbon and heat of formation of CO2.

(D)

Heat of neutralisation.

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33. Answer - A(r), B(s), C(p), D(r) Ag2CrO4

2Ag+ +

CrO24−

2s mol/L

s mol/L

Ksp = [Ag+ ]2 [CrO24− ] = (2s)2 · s 4s3 = Ksp 1

⎛ K sp ⎞ 3 ⎟ s = ⎜⎜ ⎟ ⎝ 4 ⎠

Ag+ (aq) + CNS– (aq)

AgCNS(s)

s mol/L

s mol/L

Ksp = [Ag+][CNS–] = (s)(s) = (s)2 s=

K sp 3Ca2+ (aq)

Ca3(PO4)(s)

3s mol/L

+ 2PO34− (aq) 2s mol/L

Ksp = [Ca2+]3 [ PO34− ]2 = (3s)3 (2s)2 = 108 s5 1

⎛ K sp ⎞ 5 ⎟ s = ⎜⎜ ⎟ ⎝ 108 ⎠

Hg22 −

Hg2Cl2

2s mol/L

+

2Cl– 2s mol/L

Ksp = [ Hg22 − ] [Cl–]2 = (s) (2s)2 = 4 s3 ⎛ K sp s = ⎜⎜ ⎝ 4

1

⎞3 ⎟ ⎟ ⎠

(IIT-JEE 2008)

(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 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

27S4 = 2.7 × 10–15

(M3X)

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

ΔH0

=

dT

RT 2

35. Answer - A(q), B(r), C(p), D(s) (A) We know that ΔG = ΔG° + RT lnK at equillibrium ΔG = 0 ⇒

ΔG° = –RT lnK

(B) H = E + PV ΔH = ΔE + PΔV + VΔP at constant pressure ΔP = 0 ⇒

ΔH = ΔE + PΔV

(C) Entropy change ΔS =

nRT ln =

QRe v T

V2 V1

T

V2 = nR ln V

1

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(D) Negative of free energy change = workdone = QV – ΔG = (nF)E ΔG = – nFE 36. Answer - A(q), B(q, r, s), C(p, r, s), D(p) (A)

NO2(g) + O2(g) ΔH = –200 kJ

NO (g) + O3 (g)

No. of moles in both sides is equal therefore no effect of pressure on decreasing the temperature equilibrium shift in forward direction. (B)

4NO(g) + 6H2O(g) ΔH = –905.6 kJ

4NH3 (g) + 5O2 (g) Δn = 1

So the reaction forwarded in forward direction by decreasing temperature, decreasing pressure and addition of inert gas at constant pressure (C)

2NO2(g) ΔH = 57.2 kJ

N2O4 (g) Δn = 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)

2NO(g) ΔH = +180.5 kJ

N2 (g) + O2 (g)

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+ ] =

(C)

2

Equivalent mixture of strong acid + weak base

[H+ ] = (D)

C 2 + C 22 + 4k aC1

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)

CH3COOH + NaOH

CH3COONa + H2O

Initial m. mol

100

25

0

0

After reaction

75

0

25

25

pH =

pk a + log

[Salt ] [acid]

=

pk a + log

25 = pka − log 3 75

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(B)

Physical Chemistry

CH3COOH + NaOH

CH3COONa + H2O

Initial m. mol

100

50

0

0

After reaction

50

0

50

50

pH =

pk a + log

[Salt ] [acid]

=

pk a + log

50 50

=

pka

(C)

CH3COOH + NaOH

CH3COONa + H2O

Initial m. mol

100

75

0

0

After reaction

25

0

75

75

pH =

pk a + log

(D)

75 = pk a + log 3 25

CH3COOH + NaOH

CH3COONa + H2O

Initial m. mol

100

100

0

0

After reaction

0

0

100

100

pH of salt of weak acid with strong base is pH =

1 1 1 pk w + pk a + log C 2 2 2

Conc. of salt = ∴ pH =

100 1 = M 200 2

1 1 1 pk w + pk a − log 2 2 2 2

39. Answer - A(s), B(q), C(r), D(p) (A)

N2 (g) + 3H2 (g)

2NH3 (g)

Δn = –2 Kp = Kc (RT)Δn = Kc (RT)–2 (B)

2SO2 (g) + O2 (g)

2SO3 (g)

Δn = –1 Kp = Kc (RT)–1 (C)

PCl5 (g)

PCl3 (g) + Cl2 (g)

Δn = 1 Kp = Kc (RT)1 (D)

H2 (g) + I2 (g)

2HI (g)

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40. Answer - A(q, s), B(p, r), C(p), D(p, q, s) (A)

Since CaCO3 is solid therefore its concentration remains constant. No effect of addition of CaCO3. Since reaction is endothermic therefore high temperature favours forward direction reaction. Since Δng= 1 therefor low pressure favour the forward direction reaction and addition of inert gas at constant pressure favours the forward direction because Δng > 0.

(B)

On addition of amount of reactant favours the forward direction and low temperature favours the forward direction because reaction is exothermic. Since (Δng = –2) therefore high pressure favours forward direction and addition of inert gas at constant pressure favours backward reaction.

(C)

Addition of amount of reactants favours forward direction there is no effect of pressure and addition of inert gases of constant pressure because Δng = 0.

(D)

Addition of reactant favours forward direction and Δng = 1 therefore addition of inert gas at constant pressure favours forward direction.

41. Answer - A(q, s), B(r), C(q), D(p) S2– + H2O

HS– + OH– ; K1 = 10–7

C–x

(x – y) (x + y)

HS– + H2O

H2S + OH– ; K2 = 10–14

x–y

y

∴ x >>> y

Since K1 >>> K2 [OH–] =

(y + x)

K 1·C = 10 −7 × 0.1 = 10 − 4 M

pH = 10 [HS–] = [OH–] = 10–4 [H2S] = 10–14 [S2–] = C – x = 0.1 – 0.0001 = 0.0999 M 42. Answer - A(p, q), B(p, r, s), C(p), D(p) –2

O –1

(A)

–1

O +1 +2 +1 +1 O Cr +1 +1 O O –1

–1

Oxidation no. of Cr = +6 (B)

K2Cr2O7 ⇒ 2(+1) + 2x + 7 (–2) = 0 x=+6 n factor of K2Cr2O7 = 6 Equivalent mass =

M 294 = = 49 6 6

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(C)

Physical Chemistry

K2CrO4 ⇒ 2(+1) + x + 4 (–2) = 0 x=+6

(D)

CrO3 Oxidation no. of chromium = + 6

43. Answer - A(p), B(q), C(p, r), D(q, s) (A)

H2SO4 → 2(+1) + x + 4(–2) = 0 x = +6

+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. of sulphur = +6 –2

O –1

(D)

CrO5

O

–1 +1

+2

+1

O

Cr O

+1 +1

–1

O

–1

Oxidation no. of chromium = +6 Four oxygen in form of peroxide 44. Answer - A(q), B(p), C(s), D(r) 1 ⎛ 1⎞ = ⎜ ⎟ = siemen resis tan ce ⎝ R ⎠

(A)

Conductance =

(B)

⎛a⎞ Resistivity = R⎜ ⎟ = ohm × m ⎝l⎠

(C)

Conductivity =

(D)

⎛l⎞ Cell constant ⎜ ⎟ = m–1 ⎝a⎠

1⎛ l ⎞ ⎜ ⎟ = siemen m–1 R⎝a⎠

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45. Answer - A(r), B(s), C(q), D(p)

(A)

⎛ 1⎞ Conductance = ⎜ ⎟ = siemen ⎝R ⎠

V = iR ⎛V⎞ R= ⎜ ⎟ ⎝ i ⎠

(B)

⎛a⎞ Resistivity = R ⎜ ⎟ = ohm × m ⎝l⎠

(C)

⎛l⎞ Cell constant ⎜ ⎟ = m–1 ⎝a⎠

(D)

Resistance =

volt amp

46. Answer - A(p, r), B(q, s), C(q, s), D(p, 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)

electrolysis

⎯→ Cu(s) + O2 CuSO4 + H2O ⎯⎯ ⎯⎯ ⎯ cathode

(H+ (D)

electrolysis

⎯→ H2SO4 ⎯⎯ ⎯⎯ ⎯

H2 ↑ cathode

+

SO4–

anode

solution)

+ O2 ↑ anode

H2O consumed so H2SO4 concentration increases and pH decreases. 47. Answer - A(q), B(p, s), C(p, s), D(q, r) (A)

Specific conductance decreases with dilution

(B)

Molar conductance increases with dilution and decreases with increase in conc.

(C)

Degree of dissociation increases with dilution and decreases with increase in conc.

(D)

Resistance increases with increasing the distance between the plate resistance decreases with dilution.

48. Answer - A(q, s), B(p, s), C(s), D(p, r, s) (A)

Body diagnol will touch 2 corner and 1 body centre

(B)

C4 axis diagnol will touch 2 face centre and 1 body centre

(C)

Rectangular plane will contain 4 face centres, 4 edge centre and 1 body centre

(D)

Body diagnol plane will contain 4 corner atom, 2 face centre, 2 edge centre and 1 body centre.

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Physical Chemistry

49. Answer - A(q, r, s), B(p, q, s), C(p, q), D(p, q) (A)

In Wurtzite structure, S2 form hcp and Zn2+ are present in Half of tetrahedral voids.

(B)

In Zinc Blend structure, S–2 form cubical closed packed structure and Zn+2 occupy half of tetrahedral voids.

(C)

In antiflourite strucrure, O–2 show c.c.p. like packing and Na+ is placed on all tetrahydral void.

(D)

In Rock salt structure Cl– occupy f.c.p. and Na+ occupy octahedral voids.

50. Answer - A(p, s), B(p, q), C(q, r), D(r)

≠ 90°

Rhombohedral a = b = c

α=β=γ

Cubic a = b = c

α = β = γ = 90°

Tetragonal a = b

≠c

α = β = γ = 90°

Hexagonal a = b

≠c

α = β = 90° γ = 120°

51. Answer - A(p, q), B(p, q, r, s), C(q, r), D(q, s) (A)

Rock salt : Cl– constitutes fcc and Na+ is present in all oh voids

(B)

Zinc blends : S2– constitute ccp and Zn2+ is present in alternate td voids

(C)

Na2O : O2– constitute ccp and Na+ is present in all td voids

(D)

CsCl ⇒ Cl– constitutes primitive cube and Cs+ is present at body centre.

52. Answer - A(q), B(p), C(s), D(r) A

(B)

A4 B4 + 4 = AB2

(C)

A

(D)

A

1 8

B

= AB 3

(A)

12 ×

1 4

B 1 1 8× +1 2× 8 2 4×

1 8

B

1 4

C

⇒ A 2B

1 2× + 1 2

⇒ A 1 B C2 2

53. Answer - A(r), B(p), C(q), D(s) For Rock salt distance between nearest ions = For Fluorite distance between nearest ions = For CsCl distance between nearest ions =

a 2

3 a 4

3 a 2

54. Answer - A(q, r, s), B(p), C(q, r, s), D(r) (A) Rate = K[H2O2] first order reaction (B)

Rate =

K1[NH3 ] 1 + K 2 [NH3 ]

High concentration of NH3, 1 can be neglected w.r.t. [NH3] K 1[NH3 ] K1 = =K K 2 [NH3 ] K 2

Zero order

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(C) For very low concentration of NH3, K2[NH3] can be neglected w.r.t. 1. Thus K1[NH3] first order. (D) Rate = K[CH3CHO]2 Second order 55. Answer - A(q), B(p), C(s), D(q, r) (A)

CCl

4 2N2O 5 ⎯⎯⎯ → 4NO 2 + O 2

K=

2.303 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. K=

r − r∞ 2.303 log 0 t rt − r∞

r0, rt and r∞ are polarimeter readings after time 0, t and ∞ respectively. H+

⎯→ CH COOH + C H OH (C) CH3COOC2H5 + H2O ⎯⎯ 3 2 5 K=

V − V0 2.303 log ∞ t V∞ − Vt

Where V0, Vt and V∞ are volume of NaOH used at time 0, t and ∞ respectively. (D) 2H2O2 → 2H2O + O2 K=

2.303 V∞ log t V∞ − Vt

Vt, V∞ are volume of O2 obtained at time t and ∞ respectively K=

V 2.303 log 0 t Vt

V0, Vt are volume of KMnO4 used at time 0 and t respectively 56. Answer - A(p), B(r), C(q), D(s) x t

(A)

For zero order reaction K =

(B)

2O3

(C)

Hydrolysis of ester in basic medium is 2nd order reaction

3O2 is first order reaction

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(D)

Physical Chemistry

1 x(2a − x ) for third order reaction 2 t a 2 (a − x ) 2

K=

57. Answer - A(p), B(q, r), C(p, s), D(p, s) α-emission increase

n ratio p

β-emission decrease

n ratio p

1 0n

→ 1p1 +

0 –1 e

(β-particle)

n ratio p

Positron emission increases 1 1p

→ 0n1 +

0 1e (positron )

Electron capture increases 1 1p

+

–1e

0

n ratio p

→ 0n 1

58. Answer - A(p, r), B(p, r), C(r), D(q) (A)

Hydrolysis of ester in acidic medium is pseudounimolecular reaction

(B)

Inversion of cane sugar is pseudounimolecular reaction

(C)

Decomposition of H2O2 is first order reaction

(D)

Hydrolysis of ester in alkaline medium is 2nd order reaction

59. Answer - A(q), B(r), C(s), D(p) (A)

82Pb

207

207 = 4n + 3 – actinium series 4

(B)

208 82Pb

208 = 4n – thorium series 4

(C)

83Bi

(D)

82Pb

209

206

→ →

209 = 4n + 1 – neptunium series 4

206 = 4n + 2 – uranium series 4

60. Answer - A(q), B(s), C(p), D(q) 4K+ + [Fe(CN)6]4–

(A) K4[Fe(CN)6] 1

0

0 before dissociation

(1 – α)

α after dissociation

i=

1 + 4α (Q α = 0.4) 1

1 + 4 × 0.4 = 2.6 1 (B) NaCl Na+ + Cl– i=

1

0

0

before dissociation

(1 – α)

α

α

after dissociation

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i=

1+ α (Q α = 0.9) 1

i=

1 + 0.9 = 1.9 1

(C) AlCl3

Al3+ + 3Cl–

1

0

0

(1 – α)

α

3α after dissociation

i=

1 + 3α (Q α = 0.6) 1

i=

1 + 3 × 0.6 = 2.8 1

(D) CaF2

before dissociation

Ca2+ + 2F–

1

0

0

(1 – α)

α

2α after dissociation

i=

1 + 2α (Q α = 0.8) 1

i=

1 + 2 × 0.8 = 2.6 1

before dissociation

61. Answer - A(r), B(q), C(r), D(p) The ratio of effective molarity is equal to the ratio of osmotic pressure. 62. Answer - A(q, s), B(q, r), C(q), D(p, q) (A) Elevation in boiling point is a colligative property and the elevation constant is also known as ebullioscopic constant. (B) Osmotic pressure is a colligative property and it is measured by Berkley Hartley method. (C) Relative lowering in vapoure pressure is a colligative property. (D) Depression in freezing point is a colligative property and the depression constant is also known as cryoscopic constant. 63. Answer - A(p, q), B(p, s), C(r, s), D(r, q) Solid sol : When solid is dispersed in solid named solid sol. Sol : When solid is dispersed in liquid. Emulsion : When liquid is dispersed in liquid Gel : When liquid is dispersed in solid 64. Answer - A(p, r, s), B(p, r, s), C(p, s), D(q) a and b : Brownian movement and Tyndall effect shown by colloidal solution, suspension and emulsion, not by true solutions. Emulsion are colloidal solutions in which dispersed phase as well as dispersion medium are liquids. For colloidal system particle size is 10–9 – 10–8 m. True solutions are homogenous. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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65. Answer - A(q), B(p), C(s), D(r) (A) Argyrol is the colloidal solution of silver. (B) Aquadag is the colloidal solution of graphite in water. (C) Purple of cassius is the colloidal solution of gold. (D) Colloidion is the colloidal solution of cellulose nitrate in ethanol. 66. Answer - A(p), B(p, s), C(q, r), D(q, r) (A) Physical adsorption is exothermic. (B) Chemisorption is exothermic and specific in nature. (C) Desorption is endothermic and removal of adsorbed material. (D) Activation of adsorbent is endothermic and removal of adsorbed material. 67. Answer - A(r), B(s), C(p), D(q) (A) Tyndall effect is due to Scattering of light by colloidal particles. (B) Brownian movement is the Zig-zag motion of colloidal particles. (C) Ultra filteration is the purification of colloids. (D) Electrophoresis is the movement of colloidal particles towards oppositely charged electrode.

Section - F : Subjective Type 1.

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. 2 x 3x Total moles of H2 and O2 = x + = 2 2 H2 +

Molar contraction after spark = 22400 × x=

3x 2

3x = 1500 2

10 224

Equivalent of Zn = equivalent of KMnO4 + equivalent of H2SO4

Amount of Zn =

=

20 + equivalent of H 2 1000

=

20 2 × 10 1 20 + = + 1000 224 50 224

65 ⎡ 1 20 ⎤ = 3.55 gm + 2 ⎢⎣ 50 224 ⎥⎦

So mass percentage of Zn in the sample =

3.55 × 100 = 35.5% . 10

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2.

Success Magnet (Solutions)

Let the normality of FeC2O4 and FeSO4 solutions be x and y respectively milliequivalent of FeC2O4 = 75x milliequivalent of FeSO4 = 75y m.eq. of FeC2O4 + m.eq. of FeSO4 = m.eq. of K2Cr2O7 75x + 75y = 20 × 0.04 × 6 x + y = 0.064

…(i)

Zn and dil HCl will reduce all the

Fe3+

ions to

millimoles of Fe3+ ions from FeC2O4 =

Fe2+

ions.

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

millimoles of Fe3+ ions from FeSO4 =

75 x + 75 y = 36 × 0.02 × 5 3

x + y = 0.048 3

…(ii)

On solving, x = 0.024 and y = 0.04 Normality of FeC2O4 = 0.024 N Normality of FeSO4 = 0.040 N. 3.

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 Equivalents of KI in 20 ml = 0.012 moles of KI in 20 ml =

0.012 [Q I– → I+] 2

moles of KI in 50 ml =

0.012 50 × = 0.015 . 2 20

equivalents of KIO3 reacting with excess of KI = 50 × 10–3 ×

1 × 4 = 0.02 10

equivalents of KI in excess = 0.02 moles of KI in excess =

0.02 = 0.01 2

moles of KI consumed by AgNO3 = 0.015 – 0.01 = 0.005 moles of AgNO3 = 0.005 mass of AgNO3 = 0.005 × 170 = 0.85 % AgNO3 = 0.85 × 100 = 85%. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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4.

Physical Chemistry

Let the mass of H3PO4 and HIO3 in the sample are x and y gm. At the methyl orange end point. Equivalents of NaOH = Equivalents of H3PO4 (n = 1) + Equivalents of HIO3 ⎞ ⎛ x ⎞ ⎛ y 66 × 10 −3 × 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 × 0.15 = ⎜ × 1⎟ × 2⎟ + ⎜ ⎝ 98 ⎠ ⎝ 176 ⎠

…(2)

Substracting equation (i) from (ii) ⎛ x ⎞ (85 × 10 −3 × 0.15 − 66 × 10 −3 × 0.15) = ⎜ ⎟ ⎝ 98 ⎠ 19 × 10 − 3 × 0.15 =

x 98

x = 0.2793 gm ∴

y = 47 × 10 − 3 × 0.15 176

y = 1.2408 g 0.2793 × 100 = 2.793 % 10

% of H3PO4 in the sample =

% of HIO3 in the sample = +2 −1

5.

+4

1.2408 × 100 = 12.408 % 10 2+

+

Mn H 2 O2 Fe S 2 ⎯⎯→ Fe 3+ + SO2 ⎯⎯ ⎯⎯→ Fe 2+ + Mn7 + ⎯⎯H⎯ ⎯⎯ ⎯→ Fe 3+ (n =1)

Normality of H2O2 solution =

" 40 mL of 10 vol"

10 = 1.785 5.6

Equivalents of H2O2 solution = 1.785 × 40 × 10–3 = 0.0714 = Equivalents of Fe2+ reacted Equivalents of Fe2+ produced = 0.0714 = Equivalents of Fe3+ reacted = moles of Fe3+ produced (n factor = 1) Moles of FeS2 = 0.0714 Mass of sulphur in FeS2 = 0.0714 × 2 × 32 = 4.57 Percentage of Sulphur in the sample =

4.57 × 100 = 45.7% 10

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According to Bohr's calculation, third ionisation energy of lithium (IP3) =

2.18 × 10 −18 × (3)2 12

J atom −1 = 1.962 × 10 −17

= 1.962 × 0–17 × 6.023 × 1023 J mol–1 = 11.82 × 106 J/mole Since the ratio of first, second and third ionisation potential is 1 : 4 : 9. Hence first ionisation potential (IP1) =

and second ionisation potential (IP2) =

11.82 × 10 6 J / mole 9 4 × 11.82 × 10 6 J / mole 9

Therefore the energy of the reaction Li(g) → Li3+(g) + 3e– = IP1 + IP2 + IP3 =

11.82 × 10 6 4 × 11.82 × 10 6 + + 11.82 × 10 6 9 9

= 1.84 × 107 J/mole.

7.

⎡ 1 1 1⎤ = R Hz 2 ⎢ 2 − 2 ⎥ λ ⎣⎢ n1 n 2 ⎥⎦ for n1 = 1, n2 = ?

1 30.4 × 10

−9

⎡1 1⎤ = 1.097 × 10 7 × 4 ⎢ 2 − 2 ⎥ n2 ⎥⎦ ⎢⎣1

n2 = 2 ∴ for n1 = 2 (first excited state) n2 = ?

1 108.5 × 10 −9

⎡ 1 1⎤ = 1.097 × 10 7 × 4 ⎢ 2 − 2 ⎥ n 2 ⎥⎦ ⎢⎣ 2

∴ n2 = 5. 8.

According to de Broglie matter wavelength λ =

h h = mv p

We have, p2 h2 1 2 = mv Kinetic energy E = = 2m λ2 (2 m) 2

λ=

h 2 mE

The rate of change of de Broglie wavelength λ with the kinetic energy E Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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dλ = dE

Physical Chemistry 1

d(E) 2 · dE 2m h

1 h λ3m =− 2 = −2× h 2 mE × E (1× 10 −10 )3 × 9.1× 10 −31 ⎛ dλ ⎞ =− = −2.07 × 10 6 mJ −1 . ⎜ ⎟ −34 2 dE (6.626 × 10 ) ⎝ ⎠at λ =1 Å

9.

1 8 = RH λ 9

λ = 105.14 nm Ethreshold =

Eincident =

K.E. =

hc 6.626 × 10 −34 × 3 × 10 8 = λ 230 × 10 −9 6.626 × 10 −34 × 3 × 10 8 105.14 × 10 −9

6.626 × 10 −34 × 3 × 10 8 ⎡ 1 1 ⎤ –18 ⎢105.14 − 230 ⎥ = 1.026 × 10 J. −9 10 ⎣ ⎦

10. Energy of photon =

hc 6.626 × 10 −34 × 3 × 10 8 = = 3.313 × 10–19 J λ 600 × 10 −9

E (required for melting) =

Number of photons =

500 × 6 = 166 .67 kJ 18

166.67 × 10 3 3.313 × 10 −19

Average energy required/molecule =

= 5.03 × 1023

6000 6000 = = 9.96 × 10 −21 J 23 NAV 6.023 × 10

11. Let XF and XH are the electronegativity of F and H atm then

[

]

1

XH ~ XF = 0.208 EH − F − E H − H × EF − F ] 2 1

1⎤2 ⎡ XH ~ XF = 0.208 ⎢134 .6 − (104 .2 × 36.6) 2 ⎥ ⎢⎣ ⎥⎦

XH ~ XF = 1.78 and XH < XF XF = 1.78 + XH = 1.78 + 2.1 = 3.88. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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12. 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 = 0.02845 % of Mg+ =

0.02845 × 100 = 68.28 1 24

% of Mg2+ = 31.72. 13. ΔH/molecule of Li+ and Cl– = IP1 + EA = 5.41 – 3.61 = 1.80 eV ∴ ΔH of reaction per mole = 1.80 × 6.023 × 1023 eV = 1.80 × 6.023 × 1023 × 1.6 × 10–19 × 10–3 kJ = 173.7 kJ. 14. pπ – pπ back bonding in BF3 gives some double bond character which is absent in BF4–.

F 15. (i)

O

O

O (ii)

Cl

F F Trigonal bipyramidal 3 with sp d hybridisation of Cl

(iii)

⎡Cl ⎢Cl ⎣

O

⎤ ⎥ Cl ⎦

F

(ii)

F

Square pyramidal 3 2 with sp d hybridisation of Xe

⎡ ⎢ ⎣Cl

(iv)

I

⎤ ⎥ Cl⎦

+

V-shaped with sp hybridisation of I

Square pyramidal 3 2 with sp d hybridisation of I 16. (i)

Xe

Cl I

F

F

3

NO2 is paramagnetic. NO2+ is linear with bond angle 180° NO2– is bent with bond angle slightly less than 120°.

(iii) The NO2+ ion has the shortest and strongest bond. The NO2– ion has the longest and the weakest bonds of the three. 17. The molecule BrF5 has bromine atom in centre surrounded by six electron pair, five of which are used to form bonds to fluorine atoms. Figure shows the structure of BrF5 molecule. 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. 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 short F

F long

85° 90°

Br F

F

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Physical Chemistry

18. IV = II < III < I. 19. Effective molar mass of the gas mixture =

dRT 0.6 × 0.082 × 273 = = 13.43 g / mol p 1

Let the mole fraction of methane in the gaseous mixture be x x(16) + (1 – x)4 = 13.43 x = 0.78 Therefor, partial pressure of methane in the mixture = 0.78 × 1 = 0.78 atm Partial pressure of helium in the mixture (1 – 0.78) × 1 = 0.22 atm. 20. Let the formula of nitrogen hydride be NxHy and the initial volume of nitrogen hydride by aml

initial

NxHy ⇒

x y N2 (g) + H2 (g) 2 2

a

0

0

0

ax 2

ay 2

ax ay + = 2a 2 2

Thus

x+y=4

…(i)

H2 (g) +

1 O 2 (g) → H2O(l) 2

Since the gaseous mixture was needed to pass through alkaline pyrogallol solution. The oxygen was left in excess and hydrogen would have been consumed completely. 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. 21. Let the volume of each vessel be V L moles of Cl2 =

V ×1 300 R

moles of Cl2O = Cl2O → Cl2 +

V × 1 .5 300 R

1 O 2 2

After reaction total moles of Cl2 = moles of O2 =

V 1.5 V 2.5 V 5V + = = 300 R 300 R 300 R 600 R

1.5 V 1.5 V = 2 × 300 R 600 R

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5V 1.5 V 6.5 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. Let the moles of gases left in one of vessel maintained at 27°C ⎛ 6 .5 V ⎞ − x ⎟⎟ . be x. Thus moles of gases in another vessel maintained at 52°C would be ⎜⎜ ⎝ 600 R ⎠ ⎞ R × 325 x × R × 300 ⎛ 6.5 V = ⎜⎜ − x ⎟⎟ V ⎝ 600 R ⎠ V −3 x = 5.65 × 10

−3 P = 5.65 × 10

V R

V R × 300 × = 1.695 ~ 1.7 atm . R V

22. 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 = 22.4 L ∴ Pr < Pi (= 1 atm) If pressure correction is ignored P(V – b) = RT at STP Pr = Pi = 1 atm ∴ Vr > Vi (22.4 L) At low pressure volume is so high that b >

a V2

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Physical Chemistry

P(V – b) = RT PV – Pb = RT PV = RT + Pb PV Pb = 1+ RT RT

∴ Z > 1. 23. (i)

rv = rO 2

32 = 1.33 M

32

M=

(1.33)2

= 18.09 g / mol

(ii) Molar volume V = 18 .09 = 50 .25 × 10 −3 m 3 0.36 (iii) Compression factor (Z) =

PV 50.25 × 10 −3 × 1.013 × 10 5 = = 1.224. RT 8.314 × 500

(iv) Repulsive forces dominate since the actual density is less than the density if it were ideal. 24. Work done for irreversible process

⎛ 1 ⎞ − 0 ⎟ × 2 × 373 = –41.44 cal/gm W = –P(V2 – V1) = –ΔnRT = − ⎜ ⎠ ⎝ 18 Now q = 540 cal/gm

ΔE = Δq + W = 540 – 41.44 = 498.56 cal/gm For isothermal process ΔG = 2.303 nRT log

1 V1 = 2.303 × 4 × 8.314 × 400 log = –30635.4 J. 10 V2

25. Since T is constant ΔU = 0 W = 2.303nRT log

P2 ⎛5⎞ = 2.303 × 3 × 8.314 × 400 log⎜ ⎟ = 1.61 × 104 J P1 ⎝ 1⎠

also q = –W = –1.61 × 104 J of course work done on the gas W is positive for compression. 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. 26. The vapour pressure of mercury (in atm) is equal to Kp for the reaction Hg(l)

Hg(g); Kp = PHg

Note that Hg(l) is omitted from equilibrium constant expression because it is pure liquid. Because the standard state for elemental mercury is the pure liquid ΔGf° = 0 for Hg(l) and ΔG° for the vapourisation reaction is simply equal to ΔGf° for 1 mol of Hg(g)

− ΔG° − 31.85 × 10 3 = −5.58 = log Kp = 2.303 RT 2.303 × 8.314 × 298 Kp = antilog(–5.58) = 2.63 × 10–6 Since Kp is defined in units of atmosphere therefore the vapour pressure of mercury at 25°C is 2.63 × 10–6 atm. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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27. The enthalpy change for the given reaction is calculated as ClO2(g) +

1 1 O(3(g) → Cl2O7(g) ; ΔH1 = –121.85 kJ 2 2

1 1 Cl2O7(g) → ClO3 + O 2 2

; ΔH2 = 143.5 kJ

1 1 1 O (g) + O(g) → O3(g) 2 2 2 2

; ΔH3 = –53.35 kJ

1 O (g) 2 2

; ΔH4 = –249.17 kJ

ClO2(g) + O(g) → ClO3(g)

; ΔH = –280.87 kJ

O(g) →

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

ΔH = 102.5 = 170.5 + 498.34 – 2 ECl = 0 ∴ ECl = 0 = 283.17 kJ 2ClO2(g) + O3(g) → Cl2O7(g) ΔH = –243.7 = 605.04 – 566.34 – 2 ECl – 0 ECl – O = 141.2 kJ. 28. The given buffer being an acidic buffer its pH is given as (pH)1 = pKa + log

[salt ]1 [acid]1

[salt ]1 0.5 4 = pKa + log[salt]1 + log2

…(i)

pH2 = 6 = pKa + log[salt]2 + log2

…(ii)

4 = pKa + log

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 =

0.5 + 0.5 = 0.5 M [acid]1 2

[Salt ] f =

[salt]1 + [salt] 2 2

[Salt ] f =

101 [salt ]1 2

(pH)f = pK a + log

101 [salt ]1 2 [acid]1

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(pH)f = pK a + log

Physical Chemistry

[ salt]1 101 101 + log = (pH)1 + log [acid]1 2 2

(pH)f = 5.703. 29. V.D. =

dRT 4.3 × 0.0821× 400 = = 70.6 2p 2×1

(CF3COOH)2(g)

2CF3COOH(g)

1

2

1–X

2X

initial vapour density equilibrium moles = equilibrium vapour density initial moles 114 1 + x = 70.6 1

X = 0.61

Kp =

PCF3COOH P( CF3COOH)2

⎛ 2 × 0.61 ⎞ × 1⎟ ⎜ 1.61 ⎝ ⎠ = 2.4 atm = 0.39 ×1 1.61

We know, Kp = Kc(RT)Δn, Here Δn = 1 Kc =

Kp RT

=

2.4 = 7.3 × 10 −2 mol / litre . 0.0821× 400

30. 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 = 0.362 atm 760

Total pressure in atm =

PNH3 = PHI =

0.362 = 0.181 atm 2

(Kp)1 = pNH3 × PHI = (0.181)2 ~ 0.033 atm 2 But when HI starts dissociating the pressure will increase further when both the equilibria are established simultaneously NH4I(s)

NH3(g) + HI(g) x

x–y

2HI(g)

H2(g) + I2(g)

x–y

y 2

y 2

(Kp)1 = x(x – y) = 0.033

…(i)

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⎛ y ⎞⎛ y ⎞ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ = 0.015 (Kp)2 = ( x − y )2

y = 0.015 2 ( x − y)

…(ii)

On solving equation (i) and (ii) we get x = 0.202 and y = 0.04 atm Ptotal = PNH3 + PHI + PH2 + PI2 = 0.202 + (0.202 − 0.04 ) +

0.04 0.04 + 2 2

Ptotal = 0.404 atm = 307.04 mm of Hg. 31. Solubility of CH3COOAg = 8.35 g/litre = 0.05 mol/L Hence Ksp(CH3COOAg) = 0.05 × 0.05 = 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 CH3COO– + Ag+

CH3COOAg

(x – y) CH3COO– + H+

x

CH3COOH

x

10–3

0

x–y

10–3

y

(buffer) Solubility of CH3COOAg in buffer x =

61 .8 = 0.37 M 167

Ksp = (x – y)x (0.37 – y)0.37 = 25 × 10–4 ∴ y = 0.3632 and

1 y = K a ( x − y )10 −3

∴ Ka = 1.87 × 10–5 M. 32. Ag2S

2Ag+ + S2– ; Ksp = 1.6 × 10–49

H2S

H+ + HS– ; K1 =

HS–

H+ + S2– ; K2 =

[H+ ] [HS − ] = 1× 10 −7 [H2 S] [H + ] [S 2− ] [HS − ]

= 1× 10 −14

Neglecting second ionisation let x is the amount of HS– produced then x2 = 1× 10 −7 0.1 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

x = 1 × 10–4

K2 =

[H+ ] [S 2− ]

= 1× 10 −14

[HS ]

[S2–] = 1 × 10–14

[Ag2+]2 =

[Ag+] =

K sp

=

[S 2 − ]

Q ([H+) ~ [HS–]) 1.6 × 10 −49 1× 10 −14

1.6 × 10 −49 1× 10 −14

= 4 × 10–18 M.

33. Mole fraction of oxygen in air = 0.21 So partial pressure of oxygen in air

PO2 = 0.21× 720 mm of Hg = 1.99 × 10–1 atm

Kp =

=

(PO 2 )3 (PO 3 )2

= 1.3 × 10 57

(1.99 × 10 −1 )3 (PO3 )2

= 1.3 × 10 57

PO 3 = 2.46 × 10 −30 atm

1 m3 = 1000 litre Volume = 1010 litre, T = 298 K, P = 2.46 × 10–30 atm. Number of molecules of ozone present in 10 million cubic metre of air at 25°C =

2.46 × 10 −30 × 1010 = 1× 10 −21 mole 0.082 × 298

Number of ozone molecules = 1 × 10–21 × 6.023 × 1023 = 6.02 × 102. 34. (i)

The given cell representation can be reduced to

⎛ K sp ⎞ + ⎟ H (10 −12 M) H2 (1 atm) Pt Cd | Cd2 + ⎜⎜ − 2⎟ ⎝ (OH ) ⎠ The cell reaction is Cd + 2H+ → Cd2+ + H2 Ecell = E °cell −

[Cd2 + ] PH2 0.0591 log 2 [H+ ] 2

0 = 0 − ( −0.4) −

⎡ K sp × PH2 ⎤ 0.0591 log ⎢ + 2 − 2⎥ 2 ⎣⎢ [H ] [OH ] ⎦⎥

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0 = 0.4 −

Success Magnet (Solutions)

0.0591 ⎛⎜ K sp × 1 ⎞⎟ log ⎜ K2 ⎟ 2 w ⎠ ⎝

⎛ K sp ⎞ 0.4 × 2 = 13.536 log ⎜ 2 ⎟ = ⎜ K ⎟ 0.0591 ⎝ w ⎠ Ksp = 3.44 × 1013 × 10–28 = 3.44 × 10–15 M3. (ii) ΔG = –nFE = 0 ⎛ ∂E ⎞ ⎟ = 2 × 96500 × 0.002 = 386 JK −1 ΔS = nF ⎜ ⎝ ∂T ⎠ P

ΔH = ΔG + TΔS = 0 + 298 × 386 = 115028 J = 115.028 kJ. 35. Total amount of charge passed =

5 × 2 × 60 × 60 = 0.373 F 96500

Sodium nitrite will be formed as 2H+ + NO3– + 2e → NO2– + H2O 0.386 g of NaNO2 =

0.386 0.386 mole NaNO2 = mole NO2– 69 69

0.386 × 2 [n-factor of NO2– = 2] 69 0.386 ×2F Amount of charge used to obtain 0.386 g of NaNO2 = 69

∴ Equivalent of NO2– reduced from NO3– =

∴ Current efficiency for NaNO2 =

0.011 × 100 = 2.95 % 0.373

Ammonia will be formed by the reduction of nitrate following the equation 9H+ + NO3– + 8e– → NH3 + 3H2O 0.095 g of ammonia =

0.095 mole NH3 17

Equivalent of NH3 obtained =

0.095 × 8 [n-factor of NH3 is 8] 17

Charge used for this purpose =

0.095 × 8 F = 0.045 F 17

0.045 × 100 = 12.06% 0.373 Hydrogen gas will evolve through reaction

Current efficiency for NH3 =

2H+ + 2e → H2 3.55 litre of hydrogen =

3.55 22.4

Equivalent of H2 evolved =

3.55 × 2 [n-factor for H2 = 2] 22.4

Charge used to liberate hydrogen = 0.3169 F Current efficiency for H2 =

0.3169 × 100 = 84 .96% . 0.373

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Physical Chemistry

IIT-JEE 2008

W it = E F

W 10 × 10 −3 × t = 0.01× 2 = E 96500

t = 19.3 × 104 s 36. The degree of dissociation of acetic acid is given by α=

λc 5 .2 = = 0.013 λ ∞ 390.7

CH3COO– + H+

CH3COOH initially At equilibrium

C

0

0

C(1 – α)

Dissociation constant for acetic acid Ka = Cα2 = 0.1 × (0.013)2 = 1.69 × 10–5 M. 37. 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. The reaction occuring in the cell is Zn(s) + 2H+ → Zn2+ + H2(g) ∴ Ecell = E° +

H / H2

− E°

Zn 2 +/ Zn

0.701 = − ( −0.76) −

PH2 [ Zn 2 + ] 0.059 log – 2 [H+ ] 2

0.059 1× 0.1 log + 2 2 [H ]

[H+] = 0.0316 M 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. Let the weight of NaOH added be x gm x × 1 = 0.0316 × 1 40

x = 1.264 g. After the addition of sufficient NaOH the solution becomes neutral and we have [H+] = 10–7 M ∴ Ecell = 0.76 −

38. H2(g) +

0.059 0.1 log = 0.3765 volt. 2 (10 −7 )2

1 O (g) → H2O(l); ΔG1 = –237.23 kJ 2 2

H2O(l) → H+(aq) + OH–(aq); ΔG = 79.71 kJ Cell reaction H2O(l) +

1 O (g) + H2(g) → 2H+ (aq) + 2OH–(aq) 2 2

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The given cell reaction can be obtained from equation (i) are (ii) as 1 O (g) → H2O(l); ΔG1 = –237.23 kJ 2 2

H2(g) +

2[H2O(l) → H+(aq) + OH+(aq)]; ΔG2 = 2 × 79.71 kJ

1 O (g) + H2(g) → 2H+(aq) + 2OH–(aq); ΔG3 = –237.23 + (2 × 79.71) = –77.81 kJ 2 2

H2O(l) +

ΔG3 = –nFEcell = –2 ×96500 Ecell ∴ Ecell = 0.403 V H2(g) +

1 O (g) → H2O(l); ΔH1 = –285.85 kJ 2 2

H2O(l) → H+(aq) + OH–(aq); ΔH2 = 56.9 kJ ΔH3 of cell reaction = –285.58 + 2(56.9) = –172.05 kJ. ΔH3 − ΔG 3 − 172.05 + 77.81 ⎛ ∂E ⎞ = = −0.00164 VK −1 ⎜ ⎟ = ∂ T nFT 2 × 96500 × 298 ⎝ ⎠p

39. Cell reaction is Zn + Cu2+ → Zn2+ + Cu Ecell = E°cell –

2.303 × RT [ Zn2+ ] log 2F [Cu2+ ]

1 = 0.34 − ( −0.763) − [ Zn2+ ] [Cu2+ ]

0.059 [ Zn2 + ] log 2 [Cu2+ ]

= 3101

Let x be the amount of Cu2+ that is converted into Cu when the cell potential drops to 1.0 V. ∴

[ Zn2+ ] 2+

[Cu ]

Hence

=

1+ x 1− x

1+ x = 3101 , x = 0.9993 mole 1− x

Each half cell reaction requires 2 electrons. Therefore quantity of electricity delivered = 2 × 0.9993 × 96500 = 1.92 × 105 coulomb. 40. Hg2Br2(s)

Hg22+(aq) + 2Br–(aq); Ksp = [Hg22+][Br –]2

[Hg22+]anode =

K sp − 2

[Br ]

=

K sp (0.1) 2

reduction reaction of mercury half cell is Hg22+(aq) + 2e → 2Hg(l) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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E

Hg22 + / Hg

= 0.80 −

Physical Chemistry

(0.1)2 0.059 log K sp 2

reduction reaction of manganese half cell MnO4– + 8H+ + 5e → Mn2+ + 4H2O E

MnO4 – , H + / Mn2 +

E cell = EMnO

= 1.51 −

− + 2+ 4 , H / Mn

0.059 (0.1) log = 1.42 V 5 (0.1)(0.1)8

− EHg2+ / Hg 2

⎡ 0.059 (0.1)2 ⎤ 1.21 = 1.42 − ⎢0.80 − log ⎥ 2 K sp ⎥⎦ ⎢⎣

log

(0.1)2 = 20 K sp

K sp =

41. (i)

(0.1) 2 10 20

A=8×

B = 12 ×

= 1× 10 −22 M3 .

1 1 +6× =4 8 2

1 +4=3+4 =7 4

Formula = A4B7 (ii) oh voids occupied

=

3 (by edge centre)

Fraction occupied

=

3 = 0.75 4

=

4

=

4 = 0.5 8

(iii) td void occupied Fraction occupied

(iv) Total fraction occupied = 42. Edge length of unit cell

7 = 0.58 12

= 290 × 2 = 580 pm

V of unit cell

= 195.112 × 10–24 cm³

Density

=

Z ×M a3 × NA

=

4 × 239 195.112 × 10 −24 × 6.023 × 10 23

= 8.135 g/cm3 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

43.

rsimple :

rbcc

Success Magnet (Solutions)

:

rfcc

cubic

a 2

:

2 :

3 a a : 4 2 2

3 :

2

44. Either oh or td voids are occupied. Maximum radius of atom (rA) is possible in case of oh & Minimum radius of atom (rA) is possible in case of td ∴

rA = 0.414 (for maximum) r rA = 0.225 (for minimum) r

(r is radius of molecules constituting fcc) In fcc 4r =

2a

r =

2 a = 4

2 × 432 = 152.7 pm 4

∴ Max. radius possible = 0.414 × 152.7 ≈ 63.`21 pm Min. radius possible = 0.225 × 152.7

≈ 34.35 pm

45. 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. of cationic vacancies

= 6.02 × 10–5 × 1023 = 6.02 × 1018

Ea 46. log K = log A – 2.303 RT

log K = log(1.26 × 1013) –

58.5 × 10 3 2.303 × 1.987 × 800

log K = 13.1003 – 15.9799 log K = – 2.8796 K = 1.3194 × 10–3 sec–1 t1 / 2 =

0.693 0.693 = = 525 seconds. K 1.3194 × 10 − 3

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Physical Chemistry

47. For the (i) equation K eq =

[NOBr2 ] [NO][Br2 ]

or [NOBr2] = Keq[NO][Br2] Since, N0, 3r2 is reaction intermediate rate = K·[NOBr2][NO] rate = K·Keq[NO][Br2][NO] rate = K′[NO]2[Br2]. 48. Total time = n × t1/2 69.2 = n × 138.4 n=

1 2 n

1

⎛ 1⎞ ⎛ 1⎞2 N = N0 ⎜ ⎟ = 1⎜ ⎟ ⎝2⎠ ⎝2⎠ 1 = 0.7 gm N= 2 Disintegrated amount = 1 – 0.7 = 0.3 gm −α 210 ⎯→ 206 84 Po ⎯⎯ 82 Pb

So, volume of helium = 49.

22400 × 0.3 = 32 cm2 at STP. 210

2N2O5(g)

⎯⎯→

4NO2(g) + O2(g)

At t = 0

a

0

0

At t = 30 min

(a – x)

2x

x 2

At t = ∞

0

2a

a 2

Number of moles ∝ pressure At t = ∞

At t = 30 minutes

K=

2.303 a log t (a − x )

K=

2.303 233.8 log 30 200

a = 584 .5 2 584.5 × 2 = 233.8 a= 5 2a +

a+

3x = 284.5 2

x=

284 .5 − 233.8 × 2 = 33.8 3

K = 5.206 × 10–3 minute–1. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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50. For AB2,

mAB2 =

1000 × k f × w ΔT × W

mAB2 =

1000 × 5.1× 1 2.3 × 20

mAB2 = 110.87 For AB4, m AB 4 =

1000 × 5.1× 1 1.3 × 20

mAB4 = 196.15 Suppose, atomic weight of A is x and atomic weight of B is y So,

x + 2y = 110.87

..........(1)

x + 4y = 196.15

..........(2)

By solving (1) & (2) Molecular mass of A, x = 25.59 Molecular mass of B, y = 42.64

51.

2 × 10 −3 × 10 3 2 1000 60 m= = × 800 60 800 1000 m = 0 .0417

CH3COO − + H 0 0 α α

CH3COOH 1 (1− α )

Given,

α = 0.2

i=

1 + α 1 + 0.2 = = 1.2 1 1

ΔTf = i × kf × m ΔTf = 1.2 × 1.86 × 0.0417 ΔTf = 0.093 Freezing point of solution = 0 – 0.093 = – 0.093°C 52. For KCN, ΔTf = i × kf × m 0.704 = 1 × 1.86 × 0.1892 i=2 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

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Physical Chemistry

This show that KCN is completely dissociated.

+

Hg(CN)2

nCN−

Hg(CN)nn−+ 2

At t = 0

0.095

0.1892

0

At t = t

(0.095 – x)

(0.1892 – nx)

x

x may be taken as approx. 0.095 due to the less amount of Hg(CN)2 & K+ is also present due to complete dissociation. So, total molarity = K + + nCN− + Hg(CN)nn −+2 = 0.1892 + (0.1892 – 0.095 n) + 0.095 ΔTf m= k f

0.1892 + (0.1892 – 0.095 n) + 0.095 =

0.53 1.86

0.4734 – 0.095 n = 0.2849 n≈2 So, the complex becomes Hg(CN)24− 53. There is formation of an electrical double layer of opposite charges on the surface of colloidal particles. [AgI] I–

K+

[AgI] Ag+

NO3–

[As2S3] S2–

2H+

+ + solid + +

– – – –

Fixed layer

Diffused layer

Potential difference across this electrical double layer is called zeta potential or electrokinetic potential, Z and is given by

Z=

4 πημ D

Where η is called coefficient of viscosity, D is dielectric constant of the medium and μ is the velocity of the colloidal particles when an electric field is applied. 54. Whenever a mixture of gases is allowed to come in contact with a particular adsorbent under the same conditions, the more strongly adsorbable adsorbate is adsorbed to a greater extent irrespective of its amount present. e.g. air besides moisture contains a number of gases such as N2, O2 etc., yet moisture is adsorbed more strongly on silica than the other gases of the air. 55. Gold number is the weight of protective colloid in milligrams which prevents the coagulation of 10 cm3 of gold sol. Gold number of starch = 0.025 × 1000 = 25 (mg)

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