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

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Aakash IIT-JEE

### -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 = 2z y 3 x× = z 2 xy 3 2. Answer (2)

Number of moles = molecularweight weight

= 256 2.56 Number of moles = 10–2 Number of molecules = 10–2 N 0

Q one molecule contain 16 lone pair electrons ∴ 10–2 N

0 molecule will contain = 10–2 N0 × 16 = 0.16 N0 3. Answer (2) Moles of CaO = 56 62 . 1 = moles of CaCl2 Mass of CaCl2 = 56 62 . 1 × 111 = 3.21 gm % = 10 21 . 3 × 100 = 32.1% 4. Answer (3) 3O2(g) 2O3 1000 – 3x 2x 1000 – 3x + 2x = 888

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Aakash IIT-JEE -x = 112 ml Volume of O3 at STP = 224 ml moles of O3 = 22400 224 = 0.01 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 = 2 1 1 y y x

x2 = final volume of solution Volume of H2O added = x2 – x1 = 1 2 1 1 x y y x − = ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −1 y y x 2 1 1 6. Answer (1) 2IClx⎯→ I2 + 2 x Cl2 moles of Cl2 = 5 10 3 22400 112 = × − moles of IClx = x 5 . 35 127 625 . 1 + ∴ moles of Cl2 = 3 10 5 x 5 . 35 127 625 . 1 2 x = × − ⎠ ⎞ ⎜ ⎝ ⎛ + 1.625x = 10 × 127 × 10–3 + 10 (35.5 10–3 x) x (1.625 – 0.355) = 1.27 x = 1 27 . 1 27 . 1 = 7. Answer (4) 2KClO3⎯→ KCl + KClO4 + O2

Molecular mass of KClO3 = (39 + 35.5 + 16 × 3) gm

= 122.5 moles of KClO3 = 0.1 5 . 122 25 . 12 =

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Aakash IIT-JEE

-moles of pure KClO3 = 0.1 ×

100 75 = 0.075 mass of residue = 9.1875 – 0.32 = 8.85 gm 8. Answer (2) 2 F 2 x Xe + ⎯⎯→ XeF x 1 mole 1 mole 131 gm (131 + 19x) gm ∴ 2 gm (131131+19x)2 gm of Xe Fx 158 . 3 131 ) x 19 131 ( 2 + = 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 equivalent Mass = 20 × 10–3 equivalent mass = 100020 50 10 20 1 3 = = × − 10. Answer (4) Z2O3 + 3H2⎯→ 2Z + 3H2O (2x + 48) gm 6 gm

Q (2x + 48) gm metal oxide requires = 6 gm H2 gas

∴ 0.1596 metal oxide requires = 2x+648 × 0.1596 gm H2 gas.

3 10 6 48 x 2 ) 1596 . 0 ( 6 = × − + 2x + 48 = 159.6 x = 55.8

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Aakash IIT-JEE -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 = 3 500 ml 12. Answer (1) K2Cr2O7 + 14 HCl ⎯→ 2KCl + 2 CrCl3 + 7H2O + 3Cl2

Q 14 mole HCl produces = 3 moles Cl2

∴ 1 mole HCl produces = 143 moles Cl2

MnO2 + 4 HCl ⎯→ MnCl2 + 2H2O + Cl2 1 mole 1 mole ∴ moles of MnO2 = 14 3 Mass of MnO2 = 14 3 × 87 gm = 18.642 gm 13. Answer (4) MnO2 + 4HCl ⎯→ MnCl2 + Cl2 + 2H2O equivalent mass of Cl2 = 2 71 = 35.5

6NaOH + Cl2⎯→ 5NaCl + NaClO3 + 3H2O

equivalent mas of Cl2 = 10 71 6× = 42.6 14. Answer (3) n factor of HCl = 7 3 146 = equivalent mass of HCl = 36.5 × 3 7 = 85.16 15. Answer (3) 2Mg + O2⎯→ 2MgO moles of O2 = 0.01 100 20 22400 1120 × = moles of Mg reacted = 0.02 mass of Mg = 0.02 × 24 = 0.48 gm

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Aakash IIT-JEE -initial moles of Mg = 0.1 24 4 · 2 =

moles of Mg reacted with nitrogen is 0.1 – 0.02

3 Mg + N2 ⎯⎯→ Mg3N2 3 mole 1 mole 0.08 3 08 . 0 ∴ mass of Mg3N2 = 3 2.6gm 8 100 3 08 . 0 × = = 16. Answer (1)

Let the equivalents of Na2CO3 is X

equivalents of NaHCO3 is Y Phenolphthalein indicator 2 X = 2.5 × 0.1 × 2 × 10–3 X = 1 × 10–3 (in 10 mL) ∴ In one litre = 1 × 10–1 mass of Na2CO3 = 5.3 gm

methyl orange indicator

Y 2

X + = 2.5 × 0.2 × 2 × 10–3

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)

Number of equivalent of KMnO4 =

1000 4 101 ×

= 4 × 10–4

Q 5 ml contains 4 × 10–4 equivalent of oxalate ion (equivalents of KMnO

4 = equivalents of oxalate ion) ∴ 200 ml contains = 5 10 4 200× × −4 = 16 × 10–3 weight of oxalate = 16 × 10–3 × 44 = 704 × 10–3 % of oxalate = 100 5 . 1 10 704 3 × × − = 47%

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Aakash IIT-JEE -18. Answer (2)

Molecular weight of lewsite

= 24 22 24 22 10 66 . 1 10 25 . 1 12 2 10 66 . 1 10 78 . 1 1 2 − − − × × + × + × × + × = 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 = 1 2 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 + 4H

2O

8moles of HNO3 exchange 6 moles of electrons

1 moles of HNO3 exchange

8 6 or 4 3 mole of electrons. n factor of HNO3 = 4 3

equivalent mass of HNO3=

4 / 3 63 = 3 63 4× = 84 gm. 21. Answer (3)

XZ and YZ planes are nodal planes.

22. Answer (3) ΔX = ΔP (ΔX)2 π 4 h π = Δ 4 h X

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Aakash IIT-JEE -ΔX · ΔV = 4πhm m 4 h V 4 h π = Δ π π = Δ h m 2 1 V 23. Answer (2) Angular momentum (mvr) = π 2 nh = π =1.π5h 2 h 3 24. Answer (3) hν = hν0 + eVstop ν = ν0 + ⎟⎠ ⎞ ⎜ ⎝ ⎛ h e Vstop θ Stopping potential ν ν0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = θ h e tan So the given graph will be a straight line with slope equal to

14 34 19 10 414 . 2 10 626 . 6 10 6 . 1 h e = × × × = − 25. Answer (1) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∞ − = λ 2 2 1 1 1 109678 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

Energy of single photon = 3.3 10 J

10 600 10 3 10 6 . 6 hc 19 9 8 34 − − × = × × × × = λ

No. of photon ejected per second = 3.3 10 19 6.06 1019

20 = × × − 4 4 4 19 10 10 10 10 06 . 6 × × = − = NAV 28. Answer (3) Angular momentum mvr = π 2 nh Angular momentum ∝ n r n∝ Angular momentum ∝ r

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Aakash IIT-JEE -29. Answer (1) 10 2 ) 1 n ( n − =

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

P h

Δx =

P h

From Heisenberg’s uncertainty principle

π ≥ Δ Δ 4 h P · x π ≥ Δ ⇒ × π ≥ Δ 4 1 P P h P 4 h P

Minimum percentage error in measuring velocity would be

8 ~ 96 . 7 4 100 100 P P V V 100 = π = × Δ = Δ × . 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 = 2 1 mv2 = 4.55 × 10–25 v2 = 31 25 10 1 . 9 10 55 . 4 2 − − × × × = 1 × 106 v = 103 m/s 3 31 34 10 10 1 . 9 10 626 . 6 mv × × × = λ = λ = 7.28 × 10–7 m 34. Answer (2)

Let the no. of photons required to be n

17 10 nhc = − λ 8 34 9 17 17 10 3 10 626 . 6 10 550 10 hc 10 n × × × × × = λ = − − = 27.6 = 28 photons

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Aakash IIT-JEE -⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − = λ 2 2 2 1 2 n 1 n 1 Rz 1 2 Z 1 ∝ 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 × 2 2 1 2 eV = 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, 3d6

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 × 4 1 = 3.4 eV 39. Answer (4) KE = hν – hν0 0 h h h 4 3 ν= ν ν ν = ν 4 1 0 16 10 2 . 3 4 1× × = = 8 × 1015 Hz 40. Answer (2) E hc = λ ⎟⎠ ⎞ ⎜ ⎝ ⎛ λ = ∴ p h ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = p E c

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Aakash IIT-JEE -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)

Shortest wavelength in Lyman series X

R1 = = λ

Longest wavelength in Balmer series for He+ =

2 RZ 5 36 R 4 5 36 × = (∴ Z = 2) R 5 9 = = X 5 9 44. Answer (4)

Kinetic energy in first excited state = 3.4eV

2 6 . 13 2 = + …(i) Difference in P.E. between n = 2 and n = 1 level

U2 – U1 = 20.4eV 2 6 . 13 2 1 6 . 13 2 2 2 − × = ×

Potential energy in the first excited level 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) ⎥⎦ ⎤ ⎢⎣ ⎡ × = λα 2 2 2 2 1 1 1 Z R 1 ⎥⎦ ⎤ ⎢⎣ ⎡ = λβ 2 2 2 3 1 1 1 RZ 1 8 9 4 3 × = λ λ α β 32 27 = 32 27 = λβ × 0.32 Å = 0.27Å

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Aakash IIT-JEE

-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 =

π + 2 h ) 1 (l l 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+ O 22–.

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

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Aakash IIT-JEE -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+2μ1μ1cos60º = 2 1 ) 36 . 0 ( 2 ) 36 . 0 ( ) 36 . 0 ( 2+ 2+ 2× = 0.36× 3 = 0.36 × 1.732 = 0.62 D 60. Answer (4) Cl Cl μ = μ12+μ12+2μ12cos60º = 2 1 ) 5 . 1 ( 2 ) 5 . 1 ( ) 5 . 1 ( 2+ 2+ 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)

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Aakash IIT-JEE -5 2 2 8 2 N = + = 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 P P P 60º 68. Answer (2) NH3 hybridisation is sp3, 4 2 3 5 2 N = + = sp3 PCl5 2 5 5 2 N= + = 5 → sp3d. BCl3 2 3 3 2 N= + = 3 → sp2 In [PtCl4]2– hybridisation is dsp2. 69. Answer (1)

Due to H-bonding H2O has higher boiling point than others.

70. Answer (4) , 5 2 4 6 2 N SF4 = + = lone pairs = 5 – 4 = 1 , 4 2 4 4 2 N CF4 = + = lone pairs = 4 – 4 = 0 , 6 2 4 8 2 N

XeF4 = + = lone pairs = 6 – 4 = 2

71. Answer (1)

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Aakash IIT-JEE -72. Answer (2)

Species Bond order

Cl – O– 1 O = Cl – O– 1.5 Cl O O– O 1.66 O = Cl – O– O O 1.75

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 = 8 18 10 275 . 1 10 03 . 1 − − × ×

Percentage ionic character = 10 8

18 10 10 8 . 4 275 . 1 100 10 03 . 1 − − − × × × × × 74. Answer (4)

sp3d and dsp3 have same geometry but d-orbitals that takes part in hybridisation are different.

75. Answer (2) 5 2 1 3 6 2 N= + + = , attached atoms = 3 ∴ T-shaped. 76. Answer (2) KO2⎯→ K+ + O 2 –

In O2 unpaired electrons = 1 → paramagnetic

Na2O2⎯→ 2Na+ + O

22–

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

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Aakash IIT-JEE -P1 = RT V n1 P2 = RT V n2 1 2 1 2 P P n n = ⇒ n2 = 100 01 . 0 120 12 P P n 1 2 1× = ×

Number of molecules left = n2 × N0 = 6 × 1018.

79. Answer (2)

At low pressure the volume is high

RT ) b V ( V a P 2⎟ − = ⎞ ⎜ ⎝ ⎛ + V−b~V RT ) V ( V a P 2⎟ = ⎞ ⎜ ⎝ ⎛ + RT V a PV+ = 80. Answer (2) RT ) b V ( V a P 2⎟ − = ⎞ ⎜ ⎝ ⎛ + Z < 1, V – b ~ V Vr = 2 V a 1 RT + 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= E 3 2 E = PV 2 3

For 1 mole gas PV = RT

E = RT

2 3

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Aakash IIT-JEE -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 = M dRT d = RT PM d ∝ P, d ∝ T 1 87. Answer (2) PV = nRT 2 × 3 = nAR × 273 nA = R 273 6 for vessel B 4 × 1 = nBR × 300 nB = R 300 4

After the connection

P × 7 = R 300 R 300 4 R 273 6 × ⎠ ⎞ ⎜ ⎝ ⎛ + P = 1.51 atm. 88. Answer (4) PV = nRT 10 × V = R 320 32 1 × V = R litre After leakage 320 R n R 8 5 10× = × × 48 1 300 8 5 10 n = × × = ∴ mass of gas = gm 3 2 48 32 =

Mass of gas leaked out 1 –

3 2

gas = gm 0.33gm.

3

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Aakash IIT-JEE

-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 2α

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) A B B A M M r r = rate of diffusion = t 50 time diffused volume = A B M M t 40 t 50 = 64 M 4 5 = B 64 M 16 25 B = MB = 100 93. Answer (3) moles X 2 SO H moles X H O(g) CO(g) HCOOH⎯⎯2⎯⎯4→ + COOH COOH y moles ) g ( O H ) g ( CO ) g ( CO 2 moles Y moles Y 2 SO H2⎯⎯4 + + ⎯ → ⎯

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Aakash IIT-JEE -6 1 Y 2 X Y = + 6Y = X + 2Y X = 4Y 1 : 4 YX = 94. Answer (2) CH4 + 2O2 → CO2 + 2H2O volume of C2H2 is X mL Pressure (63 – X)mm (63 – X)mm 2 ) mm ( X 2 2 2 ) mm ( X essure Pr 2 2 H CO 2 O 2 5 H C + ⎯⎯→ + total pressure of CO2 = (63 –X) + 2X = 63 + X 63 + X = 69, X = 6 mm fraction of methane = 0.9 63 57 63 X 63− = = 95. Answer (2) 6 . 2 n n n n P 4 2 2 2 CH He H H H = + + × ⇒ 1.6 atm 96. Answer (2) 1 : 2 2 2 4 4 1 2 1 m m n n r r 2 2 2 H He He H He H = = = 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

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Aakash IIT-JEE

-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 0 E 3 EO 4 2 O 3− < 2 O O 4 3 E 3 <

This in equality valid only

3 O 2 O E E > . 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 = R

2 7 , Cv = R 2 5 Only 7 5

= 0.71 of energy supplied increases the temperature of gas. The rest is used to do work against external pressure

0.71 × 60 = 42.6 kcal. 104. Answer (1) 1 2 2 1 1 1V T V T γ− = γ− ∴ 1 1 2 1 1 2 2 V V T T γ− = γ− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =

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

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Aakash IIT-JEE -106. Answer (2)

BaCl2·2H2O + aq → Ba2+(aq) + 2Cl(aq) + 2H

2O; 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,

2 1 H2C2O4 + NaOH → 2 1 Na2C2O4 + H2O; ΔH = –53.35 kJ 2 1 H2C2O4 + OH– 2 1 C2O42– + H 2O; ΔH = –53.35 kJ …(1) H+ + OH→ H 2O; ΔH = –57.3 kJ …(2)

Subtracting eq (1) from eq (2) we get

2 1 H2C2O4→ 2 1 C2O42– + H+; ΔH = 3.95 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 + 2 1 (O = O) → N=N– + = O(g) ΔHf = 2 498 (607 418) 1 946 ⎟− + ⎠ ⎞ ⎜ ⎝ ⎛ + × = 170 kJ mol–1

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 2 1 F2(g) → F(g); ΔH = 80 kJ/mol

Therefore heat of formation = Bond energy of reactants – Bond energy of products –1100 = (275 + 6 × 80) – 6 × (S – F)

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Aakash IIT-JEE

-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 = 0.095N 200 ) 1 18 ( + = . 112. Answer (3)

CH3COONa + HCl → CH3COOH + NaCl

initially moles 0.1 0.2 0 0

after reaction moles 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+ + Cl– 0.1 0.1 0.1 [H+] = 0.1 M pH = – log[H+] = – log 0.1 = 1 113. Answer (1)

For CaCO3 (s) CaO(s) + CO2(g)

Kp = 2

CO

P = 0.0095 atm

Since atmospheric pressure is 1 atm so percentage of CO2 in air = 100 0.95%

P P

total

CO2 × =

.

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

114. Answer (1) pH = + ⇒ acid salt a C C log pK pH = pK a + log acid salt C C ∴ Csalt = Cacid. pKa + pKb = 14 pKa = 14 – 4.7 = 9.3 a pK pH= pH=9.3.

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Aakash IIT-JEE -115. Answer (2) BCl B+ + Cl 0.2 0.2 0.2 BOH B+ + OH initially moles 0.1 0 0 after equilibrium 0.1 – x (x + 0.2) x Kb = 10 5 x 1 . 0 x ) 2 . 0 x ( = − − + ∴ x + 0.2 ~ 0.2 0.1 – x ~ 0.1 x = 10 5 5 10 6 2 1×= × − ∴ degree of dissociation = 6 5 10 5 1 . 0 10 5× − = ×. 116. Answer (2) NH4OH NH4+ + OH– C2(1 – α) C2α C2α C2 = [NH4OH] = 0.15 M NaOH –→ Na+ + OH C1 C1 Kb = = α +α α = α − + 1 2 2 1 2 4 4 C ) 1 ( C ) C C ( C ] OH NH [ ] OH ][ NH [ α = 4 1 b 1.8 10 C K = × 117. Answer (3) Qp = PCO2.PNH2 3 = (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 = x ) x a ( kf

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Aakash IIT-JEE

-PCl5(g) 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–] = 21 1.0 10–19 05 . 0 10 5× − = × 14 7 2 2 10 1 10 1 ] S H [ ] S ][ H [ + − × × × = [H+] = 19 21 10 1 1 . 0 10 1 − − × × × pH = 1.50. 121. Answer (3)

Its equilibrium constant Keq=

w b a K K K × = 14 10 10 10 24 . 3 − − × = 1.8 × 1.8 × 104 122. Answer (2) Solubility of PbSO4 = Ksp = 1.44×10–4=1.2×10–4M 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 =

36 . 36 1000 = 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 O H BaCO ) OH ( Ba CO 2 white 3 2 2+ → ↓+

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Aakash IIT-JEE -moles of BaCO3 = 1.2 197 4 . 236 = moles of CO2 at equilibrium ⇒ 1 + x = 1.2 x = 0.2 Kc = 2.25 8 . 0 2 . 1 x 1 x 1 2 2= ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + 124. Answer (2) Kp = PCO2 = 2.25 Number of moles of CO2 = 600 0821 . 0 1 25 . 2 × ×

Min. moles of CaCO3 required = 0.0457

Min. weight of CaCO3 required = 0.0457 × 100

= 4.57 gm. 125. Answer (2) NH3 + H2O NH4+ + OH– Kb = 5 3 3 3 3 4 1.8 10 ] NH [ 10 5 . 1 10 5 . 1 ] NH [ ] OH ][ NH [ + − = × − × × − = × − [NH3] = 0.125 M total [NH3] required = 0.125 + 1.5 × 10–3 = 0.1265 M 126. Answer (1) 50 . 0 X XCl Cl 2 = = Kp = (0.5 1) ) 1 5 . 0 ( P X ) P X ( P P 2 T Cl 2 T Cl Cl 2 Cl 2 2 × × = × × = = 0.5 M 127. Answer (2) H2(g) + S(s) H2S(g) At t = 0 0.2 1 0 At eq. 0.2 – x 1 – x x ) x 2 . 0 ( x 10 8 . 6 ] H [ ] S H [ K 2 2 2 c = = × − = x = 1.27 × 10–2 moles/litre atm 38 . 0 1 363 082 . 0 10 27 . 1 V nRT P 2 S H2 = × × × = = −

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Aakash IIT-JEE -pH = 11.27 – log[H+] = 11.27 [H+] = 5.37 × 10–12 [OH–] = 3 12 – 15 – 10 322 . 1 10 37 . 5 10 1 . 7 = × − × × Kb = –5 2 3 2 10 75 . 1 1 . 0 ) 10 322 . 1 ( C ) C ( × = × = α − 129. Answer (3) Since PCO < 2 CO P 2000 10 500 1 P P K 6 CO CO p 2 = × = = Since 2.303 RTlogKp = −ΔGop 2.303 × 8.314 × T log2000 = 20700 + 12T T = 404.3 K 130. Answer (2) K = [H BO[Complex][Glycerine] ] 0.9 3 3 = 5 . 1 40 60 ] BO H [ ] Complex [ 3 3 = = K = [Glycerine1.5 ] =0.9 [Glycerine] = 1.7M 9 . 0 5 . 1 = 131. Answer (1) % of [In–] = 100 91% 1 10 10 100 ] HIn [ ] In [ ] In [ × = + = × + − − 132. Answer (1) M(OH)2(s) M2+ + 2OH– x 2x pH = 10.6 pOH = 14 – 10.6 = 3.4 [OH–] = antilog(–3.4) = 3.98 × 10–4 M M 10 99 . 1 2 10 98 . 3 x 4 4 − − × = × = Ksp = [M2+][OH]2 = 4x3 = 4 × (1.99 × 10–4)3 = 3.15 × 10–11

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Aakash IIT-JEE -133. Answer (4)

Mg(OH)2(s) Mg2+ (aq) + 2OH (aq.)

[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⎯→Fe2++2e At cathode Pb2++2e⎯→Pb Net reaction Pb2++Fe⎯→Fe2++Pb º anode º cathode º cell E E E = − = –0.13 V – (–0.44) = + 0.31V

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 Na O — S — O Na+ — – + S –1 –1 –2 –2O +2 +1 +1 +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

4 4 . 22 56 . 0 n 52 6 . 2 × = × n = 2 i.e. Cr2+⎯⎯→Cr 137. Answer (2)

cell = E°cathode – E°anode

= E°H2O2(aq)|H2O(l)−E°Fe3+(aq)|Fe2+(aq)

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Aakash IIT-JEE

-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 2++ ⎯⎯→ 2+

2 e AmO

AmO

Second electrode AmO22++4H++2e⎯⎯→Am4++2H2O

Third electrode 4++ ⎯→ 2+

Am e

2 Am

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) Anode H2⎯⎯→2H++2e 10–6 M Cathode 2H++2e⎯⎯→H2 Ecell = 0.118 = cathode6 2 2 ) 10 ( ] H [ log 2 059 . 0 − + On solving we get [H+] cathode = 10– 4 M 142. Answer (2) 4 2 2 2 Zn 2e ZnMn O MnO 2 + ++ ⎯⎯→ 1 87 E 2 MnO = t = 51.35days 24 3600 87 10 2 96500 8 E i 96500 w 3× × × = × × = × × − 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.

S HO HO O O S HO HO O O S HO HO O O

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Aakash IIT-JEE -144. Answer (1) 2 H e 2 H 2 ++ ⎯⎯→ 2 H / H 2 H H / H [H ] 1 log 2 059 . 0 E ] H [ P log 2 059 . 0 E E 2 2 2 − + = ° − + ° = + + 2 H / H [H ] 100 log 2 059 . 0 E E 2 + − ° = ′ + ∴ [H ] 0.059V ] H [ 100 log 2 059 . 0 E E 2 2 H / H H / H 2 2 = × = ′ − + + + + 145. Answer (1)

For the following electrochemical cell

Pt | ) atm 1 ( Cl | ) C ( Cl || ) C ( Cl | atm) (1 Cl | Pt 212 2

The half cell reactions are

Anode − ⎯⎯→ Cl (g)+e− 2 1 ClA 2 Cathode Cl2(g)+e⎯⎯→Cl−c 2 1

The Ecell is given by

2 1 1 2 A C cell C C log 059 . 0 C C log 1 059 . 0 ] Cl [ ] Cl [ log 1 059 . 0 E =− =− = −

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

Hgnn+C+ne⎯⎯→nHg

Anode nHg⎯⎯→

Hgnn+ A+ne−

Net reaction

Hgnn+C⎯⎯→

### ( )A

Hgnn+

Thus the given cell is electrolyte concentration cell

A n n C n n cell ] Hg [ ] Hg [ log n 0591 . 0 E + + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 20 1 2 1 log n 0591 . 0 0295 . 0 n = 2

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Aakash IIT-JEE

-Above half cell is metal-metal insoluble salt-anion electrode when it acts as cathode half cell reaction will be

) aq ( Cl ) s ( Ag e ) s ( AgCl + −⎯⎯→ + −

i.e. AgCl is consumed. So during reaction quantity of AgCl decreases.

148. Answer (1) Q + 2H+ + 2e QH 2 2 cell cell ] H [ 1 log 2 0591 . 0 E E = ° − + pH 0591 . 0 E Ecell= °cell− intercept = Eocell=0.699V pH = 3.5 0591 . 0 492 . 0 699 . 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 149. Answer (1) H2O H+ + OH; w 1 RTlnK G =− Δ o H+ + e 2 1 ⎯→ ⎯ H2 ; o o 2 H / H 2 nFE G =− + Δ H2O + e– 2 1 ⎯→ ⎯ H2 + OH– ; Δ o =− o OH , H / O H 3 nFE 2 2 G SinceEH /H 0 So Go2 0 2 = Δ = + o Thus o 3 o 1 G G =Δ Δ –RTln Kw = oHO/H,OH 2 2 E F n − − ) 1 n ( ce sin K ln F RT EoHO/H,OH w 2 2 = = − 150. Answer (4) 8 4 2 o Mn / MnO Mn / MnO ] H ][ MnO [ ] Mn [ log 5 059 . 0 E E 2 4 2 4 − + + − = − + + −

Let the initial conc. of H+ be x. When it is reduced to x/2 the electrode potential is given by

+ − ′ 2 4/Mn MnO E = 8 4 8 2 o Mn / MnO x ] MnO [ ] 2 ][ Mn [ log 5 059 . 0 E 2 4 − + − + − = 8 8 4 2 o Mn / MnO 5 log2 059 . 0 x ] MnO [ ] Mn [ log 5 059 . 0 E 2 4 − − + + − = 0.02846 x ] MnO [ ] Mn [ log 5 059 . 0 E 8 4 2 o Mn / MnO4 2 − − − + + −

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Aakash IIT-JEE -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) Cu2+ + 2e → Cu; o 1 G Δ = –2F(0.337) Cu+ + e → Cu; o 2 G Δ = –1F(0.153) Cu2+ + e → Cu+; o 3 G Δ = –1FE° o 3 G Δ = 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) 242 125 239 128 3 3 4NO KOH KNO NH +Λ −Λ = + − = Λ = Λ∞ ∞ ∞ ∞ Degree of dissociation = 0.10 242 24 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Λ Λ ∞ . 155. Answer (2) 5 . 19 01 . 0 10 5 . 19 1000 N K 1000 = × × 5 = = Λ − Degree of dissociation = 0.05 390 5 . 19 = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Λ Λ ∞ 156. Answer (1) o o o anode RP cathode RP cell E E E = − = 0.13 – (–0.34) = 0.47 2Tl + Sn4+→ Sn2+ + 2Tl+ Ecell = 0.47 – ] Sn [ ] Sn [ ] Tl [ log 2 059 . 0 4 2 2 + + + = 0.47 – log[10] 0.47 0.059 0.411V 2 059 . 0 2 = − = 157. Answer (2) ) 44 . 0 ( 23 . 1 Ecello = − − = 1.23 + 0.44 = 1.67 ΔG° = – o cell nFE = –2 × 96500 × 1.67 × 10–3 kJ = –322 kJ 158. Answer (4)

Ni(s) + Cu2+(aq) → Cu(s) + Ni2+(aq)

V 59 . 0 ) 25 . 0 ( 34 . 0 E E

Ecello = RPo cathodeRPo anode= − − =

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Aakash IIT-JEE

-Since oxidation potential of Zn is high therefore Zn will be oxidised ∴ Zn(s)|Zn2+(aq)||H+(aq)|H

2(g), Pt

160. Answer (2)

Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)

E1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 1 01 . 0 log 2 059 . 0 Ecello

when the [Zn2+] = 1 M[Cu2+] = 0.01 M

E2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 01 . 0 1 log 2 059 . 0 Ecello

It is clear from both equation E1 > E2 161. Answer (4) Fe2+ + 2e → Fe; o 1 G Δ = –2F(–0.44) Fe3+ + 3e → Fe; o 2 G Δ = –3F(–0.036) Fe3+ + e → Fe2+; o 3 G Δ = –1FE o 3 G Δ = ΔGo2−ΔGo1 –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

⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = 22++ cell Cu Zn log 2 059 . 0 Eo 1.1 = ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + 2 2 Cu Zn log 2 059 . 0 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + 2 2 Cu Zn log = 37.3 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + 2 2 Cu Zn = 1037.3 163. Answer (2)

Since the cell reactions proceed in the standard condition and Eocell is negative therefore the electricity cannot be produced.

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Aakash IIT-JEE -164. Answer (2)

AgCl(s) Ag+(aq) + Cl(aq) ; ΔG

1° = –2.303 RT log Ksp

Ag+(aq) + e → Ag(s) ; ΔG

2° = −1FE°Ag+/Ag

AgCl(s) + e → Ag(s) + Cl–(aq) ; ΔG

3° = −1FE°Cl−/AgCl/Ag

From the given equations

ΔG3° = ΔG1° + ΔG2° Ag / AgCl / Cl E° − = E°Ag+/Ag + 2.303 RT log Ksp –0.15 = 0.80 + 2.303 RT log Ksp log Ksp = 16.101 059 . 0 95 . 0 = − ∴ 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 =

96500 7200 I

At. anode 2OH–

2 1

O2 + H2O + 2e At. cathode 2H+ + 2e → H

2

Moles of O2 released at anode =

4 1 96500

I 7200 ×

Moles of H2 released at cathode =

2 1 96500

7200 ×I

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

672 4 3 I 96500 7200 22400× × = I = 0.536 amp. 166. Answer (3) Copper is oxidised to Cu2+ 167. Answer (1) 2Cl–→ Cl 2 + 2e (anode) Cu2+ + 2e → Cu (cathode)

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Aakash IIT-JEE -3 3 8 23 3 AV Cm / gm 2 . 3 46 . 5 023 . 6 780 4 ) 10 46 . 5 ( 10 023 . 6 ) 78 ( 4 a N M Z d = × × = × × = × × = 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 = 3 r3 3 16 r 3 4 4× π = π

For fcc unit cell, a = 2 2r

Volume of cube = a3 =(2 2r)3 =16 2r3 Packing fraction = 2 3 r 2 16 r 3 16 cube of volume sphere of volume 3 3 π = π = 174. Answer (4) d = 3 23 8 3 3 AV Cm / g 75 . 411 023 . 6 620 4 ) 10 ( 10 023 . 6 62 4 a N M Z = × = × × × = × × −

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 = 2 a a = 2.3 × 2 = 4.6 Å d = 3 23 3 24 AV 6.023 10 (4.6) 10 78 4 a N M Z − × × × × = × × = 3 ) 6 . 4 ( 023 . 6 780 4 × × = 5.32 gm/Cm3. 177. Answer (4)

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Aakash IIT-JEE -178. Answer (1) In bcc, 3a=4r r = 4 297 3 4 a 3 × = In fcc, 2a=4r a = 2 2r 2 r 4 = = 297 4 3 2 2 × × = 297 2 3 × = 363.79 pm. 179. Answer (2) d = 3 AV a N M Z × × 10.5 = 6.023 1023 (4.09 10 8)3 108 Z − × × × × 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 =

2 5

Effective number of ions of A =

2 7 2 1

4− =

∴ New formula of compound A7/2B5/2. 181. Answer (2) 732 . 0 r r = − + r– = 136.61pm. 732 . 0 100 732 . 0 r+ = = 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)

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Aakash IIT-JEE

-In simple cubic unit-cell the packing fraction = 3 3 ) r 2 ( r 3 4 π = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π 6 . 186. Answer (3)

Number of alternate corners = 4 Number of alternate edges = 4 Number of alternate faces = 2 Hence, Number fo A atoms =

2 1 4 8 1 = × Number of B atoms = 2 1 2 1× = Number of C atoms = 4 1 2 4 1 = + ×

So, the formula of compound per unit cell is A1/2BC2, simplest formula of compound is AB2C4. 187. Answer (2) 4 4 2 3 2 10 3 10 2 2 3 dt ] H [ d dt ] NH [ d 2 1 dt ] H [ d 3 1 − − = × × × = − + = − So, 2 3 3 10 4 2 10 4 dt ] NH [ d dt ] H [ d × = ×× × − − = 6 × 10–8 188. 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 = 3 P Pt0

For first order reaction,

) x a ( a log t 303 . 2 k − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = 3 P P P P log t 303 . 2 k 0 t 0 0 ) P P 4 ( P 3 log t 303 . 2 k t 0 0 − =

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Aakash IIT-JEE

-190. Answer (3)

U

238

92 is disintegrated through (4n + 2) series.

191. Answer (1) Total time = 2 1 t n× 69.2 = n × 138.4 n = ½ N = 2 1 n 0 2 1 1 2 1 N ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ N = 1.414 0.7 1 2 1 = = Disintegrated amount = 1 – 0.7 = 0.3 g Po Po He 20682 210 84 4 2⎯ ⎯ → ⎯−

Volume of helium accumulated = 0.3g

210 22400 × = 32 ml. 192. Answer (2) 2 2 5 2B 2AB ½B A ⎯⎯→ + dt ] B [ d 2 dt ] AB [ d 2 1 dt ] B A [ d 2 5 =+ 2 =+ 2 − ] B A [ k 2 ] B A [ k 2 1 ] B A [ k1 2 5 = × 2 2 5 = × 3 2 5

So, the relation becomes

2k1 = k2 = 4k3 193. Answer (3) t 0 N N log t 303 . 2 = λ 303 . 2 t N N log t 0 = λ×

Comparing with y = mx + C then Slope = 303 . 2 λ + 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

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Aakash IIT-JEE -) g ( C 2 ) g ( B ) g ( A 2 + ⎯⎯→ 2 10 100 120 dt ] C [ d 2 1 == + min / mm 4 dt ] C [ d = + 197. Answer (1) 1 n ½ ) a ( 1 t ∝ (n = order of reaction) 198. Answer (2)

Overall order of reaction is 2.

199. Answer (2) 1 2 . 0 206 238 1 log t 303 . 2 k ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × = 1 . 0 t 303 . 2 10 5 . 4 693 . 0 9 = × × t = 1.5 × 109 years. 200. Answer (3) 2A + 3B ⎯⎯→4C + 5D dt ] C [ d 4 1 dt ] B [ d 3 1 =+ − dt ] C [ d dt ] B [ d 3 4 =+ − 201. Answer (4) For ‘B M = 100 1 1 K 1000 b × × × (M = molar mass of B) Kb = 10 M For ‘A’ M1 = 100 10 1 2 M 1000 × × × × (M1 = molar mass of A) M1 = 2 M

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Aakash IIT-JEE -202. Answer (2)

From Roult’s law, Ps = PAo·XA+PBo·XB

Ps = PAo·XA+PBo(1−XA) (Q XA + XB = 1) Ps = PAo·XA+PBo −PBo·XA

Ps = PBo−XA(PBo −PAo) …(i)

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

o B P = 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 = 55.5 18 1000 = xurea = 5 . 56 1 5 . 55 1 1 n n n water urea urea = + = + 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.

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Aakash IIT-JEE

-In case of Ca(NO3)2

20 M

and 0.05 M Na2SO4, effective molarity are same.

208. Answer (1) M = W T w K 1000 f × Δ × × 62 = W 3 . 9 50 86 . 1 1000 × × × W(unfreezed water) = 161.29 g Ice separated = 200 – 161.29 = 38.71 g 209. Answer (3) m = 0.5m 1000 200 95 5 . 9 = (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) HX H+ + X– 1 0 0 before dissociation (1 – 0.2) 0.2 0.2 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 ∝ power g coagulatin 1 .

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Aakash IIT-JEE -214. Answer (3)

As2S3 sol is negatively charged.

215. Answer (1)

The potential required to stop electro-osmosis is known as Dorn potential.

216. Answer (2)

K is highly reactive towards water.

217. Answer (1)

Gold number is equal to the number of milligram of substance required to prevent the coagulation of 10 ml gold sol before adding 1 ml of 10% NaCl solution.

218. Answer (2)

van der Waal’s adsorption occurs at low temperature and high pressure.

219. Answer (1)

Higher is the value of van der Waal constant ‘a’ more is the adsorption on charcoal.

220. Answer (3)

Fe(OH)3 is positively charged sol due to adsorption of Fe3+ ions.

### 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 = 100 52% 1 16 5 24 24 = × + + + %C in C6H12O6 = 100 40% 96 12 72 72 × = + + %C in CH3COOH = 100 40% 1 32 12 3 12 24 × = + + + + %C in C2H5NH2 = 100 53% 2 14 5 24 24 = × + + + 3. Answer (1, 4) O H NaF NaOH HF+ ⎯⎯→ + 2 moles of HF = 0.01 1000 100 1 . 0 × = moles of NaOH = 0.01 1000 100 1 . 0 × =

number of moles of NaF formed = 0.01

M 05 . 0 1000 200 01 . 0 ] NaF [ = =

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Aakash IIT-JEE -KCrO3Cl = 1 + x + 3 × –2 + (–1) = 0 x = +6 Cr O O O O O CrO5 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 × 2 1 + 30 × 3 1 = N4(1000) 4 N 100025 = N 40 1 N4 = Resultant [H+] = 40 1 = 0.025

Normality = mass/equivalentvol(litre) mass

40 x

401 × molecular mass of NaOH = 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 = 1 2 2 = Volume of SO2 at STP = 22400 × 10–3 = 22.4 ml. Concentration of SO2 in air is 22.4 pm

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Aakash IIT-JEE -9. Answer (1, 3, 4) O H 9 NO 2 ) NO ( Cu 3 HNO 8 Cu 30 + 3⎯→ +2 3 2+ + 2

In the above balance equation is it clear that only two moles of NO3 undergo change in oxidation state while

six moles remain in same oxidation state 2HNO3 + 6H+ + 6e → 2NO + 2H

2O

Total 8 moles of HNO3 exchange 6 mole of electrons

1 mole of HNO3 exchanges

8 6 or 4 3 mole of electrons ∴ n factors of HNO3 = 4 3 Cu is oxidised of Cu2+

equivalents mass of HNO3 = 84

4 / 363 = gm. 10. Answer (1, 3, 4) O H 2 Cl MnCl HCl 4 MnO 2 0 2 2 2 1 4 2+ → + + + − + n factor of HCl = 2 1 4 2 = n factor of MnO2 = 2

equivalent mass of MnO2 =

2 mass molecular . 11. Answer (1, 2) 2 2 7 ) 10 n ( 4 2 Mn ) MnO ( Ba + + + = ⎯⎯→ milliequivalents of Ba(MnO4)2 = 10 100 10 1 100× × = Fe2+→ Fe3+ + e n = 1 milliequivalents of FeSO4 = 100 × 1 = 100 4 2 3 3 ) 3 n ( 4 3 2 2 CO Fe O C Fe + + + = + + + → milliequivalents of FeC2O4 = 3 100 3 100× =

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 = 2 98 = 49.

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Aakash IIT-JEE

-Radial node for 1s

### ⇒

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 n 2 = (m + 1) ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + − = λ 2 2 2 m (m 1) 1 1 1 Rz 1 for nth line n 2 = (n +1) ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + − = λ 2 2 2 n (n 1) 1 1 1 Rz 1 ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ − + − + + + = λ λ 1 ) 1 m ( 1 ) 1 n ( ) 1 n ( ) 1 m ( 2 2 2 2 n m 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.

### ∫

ψ2dxdydz=1

∞ ∞ − 18. Answer (3, 4) n = 4, l = 0, 1, 2, 3 for l = 2, m = –2, –1.0 + 1, +2 s = 2 1 + (correct) n = 4, l = 0, 1, 2, 3 l = 2, m = –2, –10 + 1 + 2 s = 2 1 − (correct)

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Aakash IIT-JEE -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 = n2r 1 ) n ln( 4 n ln r r ln A A ln 4 2 1 2 n 1 n = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ π π = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ln ln(n) A A n 1 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) pyramidal trigonal ion Hybridisat ; 1 pair lone 4 2 3 5 2 N NH ion Hybridisat ; 1 pair lone 4 2 1 3 4 2 N CH ion Hybridisat ; 1 pair lone 4 2 1 3 6 2 N O H 3 3 3 3 3 3 ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎬ ⎫ = ∴ = = + = = ∴ = = + + = = ∴ = = − + = − + sp sp sp 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+ + Br

3–.

(4) Sodium sulphate is soluble in water but BaSO4 is sparingly soluble because hydration energy of BaSO4 is less than its lattice energy.

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Aakash IIT-JEE -Rb+ – ionic radius 1.48 Å O2– – ionic radius 1.40 Å and Li+ – ionic radius – 0.68 Å Mg2+ – ionic radius – 0.60 Å 28. Answer (2, 3)

Electron affinity of O(g) and S(g) are negative therefore involve emission of energy. 29. Answer (1, 2, 3)

Alkali metal have lowest I.E. energy because after releasing one electron these acquires noble gas configuration.

Electron affinity of nitrogen is less than oxygen because nitrogen has half filled p-orbital therefore it is more stable.

F– is weakest reducing agent among halide due to maximum stability due to highest hydration energy.

30. Answer (1, 2, 4)

Tl+ is more stable than Tl3+

Ga3+ is more stable than Ga+

Pb4+ is less stable than Pb2+

Bi3+ is more stable than Bi5+

These are due to inert pair effect. 31. Answer (2, 3, 4)

Electronegativity, ionisation energy and oxidizing power increases from iodine to fluorine. 32. Answer (1, 2, 3)

Low ionisation energy, high electron affinity and high lattice energy favours the ionic bond formation. 33. Answer (1, 3, 4)

Increasing metallic character increase the electron donating tendency of metal. 34. Answer (1, 4)

PCl5 in solid form exists as [PCl4+][PCl

6–] and PBr5 exists as [PBr4+][Br–] in solid state. 35. Answer (1, 4) 5 2 2 8 2 N

XeOF2 = + = attached atoms = 3; ∴ T-shaped

4 2 1 2 5 2 N NH2 = + + = −

attached atoms = 2; ∴ Angular 36. Answer (2, 3) B – C C – B≡ F F F F sp2 sp sp sp2 ∴ planar N – SiH3 SiH3 SiH3

Lone pair of nitrogen is involve in pπ-dπ bonding there fore delocalised therefore nitrogen is sp2 hybridsed and planar.

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Aakash IIT-JEE -37. Answer (2, 4) – 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 CH3 (μ ≠ 0) N N O O O O (μ 0)= Br Cl (μ 0)~ C C H CH3 H C H2 5 (μ ≠ 0) 41. Answer (2, 3, 4)

Statement 2 and 3 are facts

Unit of P = Unit of 2

2

V a n

∴ 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) RT PV Z= Average RT 2 3 KE= 44. Answer (2, 3, 4)

Above critical temperature gas cannot be liquified. 45. Answer (1, 3, 4) ) V ( ) V ( V V Z id m ideal molar =

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Aakash IIT-JEE

-m id

Vm = Vid Z = 1

Vm < Vid Z < 1

If force of attraction dominates then Z < 1

46. Answer (2, 3)

Mg2C3⎯⎯→ 2 Mg2+ + C 3

4–

2 Mg2+ + [2–C C C2–] Two sigma and two pi bonds

CaCN2 ⇒ Ca2+ + CN 22–

[–N C N] Two sigma and two pi bonds 47. Answer (1, 2, 3)

In solid state N2O5 exist as [NO2+][NO3−] therefore hybridisation of each nitrogen is sp and sp2 respectively.

In gaseous state N2O5 exists as N O N O

O

O

O

therefore hybridisation of each nitrogen is sp2.

N2O5 is called anhydride of nitric acid because in reaction with H2O, N2O5 forms nitric acid H2O + N2O5⎯⎯→ 2 HNO3 48. Answer (1, 2, 3, 4) KK (σ2s2) (σ*2s2) (π 2p x 2 = π 2p y 2)

Four electrons are present in 2π molecular orbitals that’s why double bond contains both π bonds. 49. Answer (1, 3, 4) (1) Cl P Cl Br Br Cl dipole moment 0μ ≠ (2) Cl P Br Br Br Cl dipole moment = 0μ (3) O O CH3 CH3

dipole moment is not zero due to following structure

(4) N

H

H O

H dipole moment is not zero

50. Answer (1, 3, 4)

Order of acidic strength H3PO2 > 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.

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Aakash IIT-JEE

-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 = RT 2 3

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 = 2 b 27 a Vc = 3b. 56. Answer (1, 3, 4) 224 . 1 : 128 . 1 : 1 M RT 3 : M RT 8 : M RT 2 u : u : ump average rms π 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

dQ dP

is intensive property.

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Aakash IIT-JEE

-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 → BCl + H2O x x 4 3 0 0 x 4 3 x− 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 4 x 3 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 4 x 3 pOH = x 4 4 x 3 log 5 ] BoH [ ] Salt [ log pKb × × + = + pH = 14 – 5 – log 3 = 8.523.

68.(a). Answer (3) (IIT-JEE 2008)

At equivalence point (acid)2 2 (base)1 1 V N V N = 2.5 × 5 2 = 15 2 × V V = 7.5 ml

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Aakash IIT-JEE -pH = 7 – 2 1 pkb – 2 1 log C = 7 – 6 – 2 1 log 10 1 = 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 K log n 0591 . 0 Ecell° = . 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– + H 2O HCN + OH – (basic) CH3COONa CH3COO– + Na+ CH3COO– + H 2O CH3COOH + OH– (basic) Na2CO3 2Na+ + CO 32– CO32– + 2H 2O 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)

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