unit-i - physical chemistry - solution(final)

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Section A : Straight Objective Type1. 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 = z2

y3x × =

z2xy3

2. Answer (2)

Number of moles = weightmolecularweight

= 2562.56

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 = 5662.1 = moles of CaCl2

Mass of CaCl2 = 5662.1 × 111 = 3.21 gm

% = 1021.3 × 100 = 32.1%

4. Answer (3)

3O2(g) 2O3

1000 – 3x 2x

1000 – 3x + 2x = 888

Physical Chemistry UNIT 1


Success Magnet (Solutions) Physical Chemistry

x = 112 ml

Volume of O3 at STP = 224 ml

moles of O3 = 22400224 = 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

11yyx

x2 = final volume of solution

Volume of H2O added = x2 – x1

= 12

11 xyyx

− = ⎟⎟⎠

⎞⎜⎜⎝

⎛−1

yy

x2

11

6. Answer (1)

2IClx ⎯→ I2 + 2x

Cl2

moles of Cl2 = 310522400112 −×=

moles of IClx = x5.35127625.1

+

∴ moles of Cl2 = 3105x5.35127

625.12x −×=⎟

⎠⎞⎜

⎝⎛

+

1.625x = 10 × 127 × 10–3 + 10 (35.5 10–3 x)

x (1.625 – 0.355) = 1.27

x = 127.127.1 =

7. Answer (4)

2KClO3 ⎯→ KCl + KClO4 + O2

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

= 122.5

moles of KClO3 = 1.05.122

25.12 =


Physical Chemistry Success Magnet (Solutions)

moles of pure KClO3 = 0.1 × 10075

= 0.075

mass of residue = 9.1875 – 0.32

= 8.85 gm

8. Answer (2)

2F2xXe + ⎯⎯→ XeFx

1 mole 1 mole

131 gm (131 + 19x) gm

∴ 2 gm 1312)x19131( +

gm of Xe Fx

158.3131

)x19131(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

massequivalentMass

= 20 × 10–3

equivalent mass = 5020

100010201

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 = 48x26+ × 0.1596 gm H2 gas.

310648x2

)1596.0(6 −×=+

2x + 48 = 159.6

x = 55.8



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 = 3500 ml

12. Answer (1)

K2Cr2O7 + 14 HCl ⎯→ 2KCl + 2 CrCl3 + 7H2O + 3Cl2Q 14 mole HCl produces = 3 moles Cl2

∴ 1 mole HCl produces = 143 moles Cl2

MnO2 + 4 HCl ⎯→ MnCl2 + 2H2O + Cl21 mole 1 mole

∴ moles of MnO2 = 143

Mass of MnO2 = 143 × 87 gm

= 18.642 gm

13. Answer (4)

MnO2 + 4HCl ⎯→ MnCl2 + Cl2 + 2H2O

equivalent mass of Cl2 = 271

= 35.5

6NaOH + Cl2 ⎯→ 5NaCl + NaClO3 + 3H2O

equivalent mas of Cl2 = 10716× = 42.6

14. Answer (3)

n factor of HCl = 73

146 =

equivalent mass of HCl = 36.5 × 37

= 85.16

15. Answer (3)

2Mg + O2 ⎯→ 2MgO

moles of O2 = 01.010020

224001120 =×

moles of Mg reacted = 0.02

mass of Mg = 0.02 × 24

= 0.48 gm



initial moles of Mg = 1.024

4·2 =

moles of Mg reacted with nitrogen is 0.1 – 0.02

3 Mg + N2 ⎯⎯→ Mg3N2

3 mole 1 mole

0.08 308.0

∴ mass of Mg3N2 = gm6.238100

308.0

==×

16. Answer (1)

Let the equivalents of Na2CO3 is X

equivalents of NaHCO3 is Y

Phenolphthalein indicator

2X = 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

Y2X + = 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 = 10004

101

×

= 4 × 10–4

Q 5 ml contains 4 × 10–4 equivalent of oxalate ion (equivalents of KMnO4 = equivalents of oxalate ion)

∴ 200 ml contains =5

104200 4−××

= 16 × 10–3

weight of oxalate = 16 × 10–3 × 44

= 704 × 10–3

% of oxalate = 1005.110704 3

×× −

= 47%



18. Answer (2)

Molecular weight of lewsite

= 24

22

24

22

1066.11025.1122

1066.11078.112 −

−

−

−

××+×+

××+×

= 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 = 122 =

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 86 or 4

3 mole of electrons.

n factor of HNO3 = 43

equivalent mass of HNO3 = 4/363

= 3634×

= 84 gm.

21. Answer (3)

XZ and YZ planes are nodal planes.

22. Answer (3)

ΔX = ΔP

(ΔX)2 ≥ π4h

π=Δ

4hX



ΔX · ΔV = m4hπ

m4hV

4h

π=Δ

π

π=Δ h

m21V

23. Answer (2)

Angular momentum (mvr) = π2nh

= π=

πh5.1

2h3

24. Answer (3)hν = hν0 + eVstop

ν = ν0 + ⎟⎠⎞⎜

⎝⎛

he Vstop

θ

Stopping potential

ν ν0

⎟⎠⎞⎜

⎝⎛=θ

hetan

So the given graph will be a straight line with slope equal to

1434

1910414.2

10626.6106.1

he

×=×

×=

−

−

25. Answer (1)

⎟⎠⎞

⎜⎝⎛

∞−=

λ 221

111096781

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 intransition 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 = J103.310600

103106.6hc 199

834 −−

−×=

××××=

λ

No. of photon ejected per second = 19

19 1006.6103.3

20 ×=× −

44

41910

10101006.6 −=××= NAV

28. Answer (3)Angular momentum mvr = π2

nh

Angular momentum ∝ n

rn ∝

Angular momentum ∝ r



29. Answer (1)

102

)1n(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 Ph

Δx = Ph

From Heisenberg’s uncertainty principle

π≥ΔΔ

4hP·x

π≥

Δ⇒×

π≥Δ

41

PP

hP

4hP

Minimum percentage error in measuring velocity would be

8~96.74

100100PP

VV100 =

π=×

Δ=

Δ× .

31. Answer (4)

It has highest number of orbitals among all mentioned ones hence maximum orientation is possible forf-orbitals.

32. Answer (3)

Cl(17) – 1s2, 2s2, 2p6, 3s2, 3p5

n = 3, l = 1, m = 1

33. Answer (2)

KE = 21 mv2 = 4.55 × 10–25

v2 = 31

25

101.91055.42−

−

××× = 1 × 106

v = 103 m/s

331

34

10101.910626.6

mv ××

×=

λ=λ

−

−

= 7.28 × 10–7 m

34. Answer (2)

Let the no. of photons required to be n

1710nhc −=λ

834

91717

10310626.61055010

hc10n

×××××=λ= −

−−−

= 27.6 = 28 photons



35. Answer (1)

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

λ 22

21

2

n1

n1Rz1

2Z1∝

λSince 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

12 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 × 41 = 3.4 eV

39. Answer (4)

KE = hν – hν0

0hhh43

ν−ν=ν

ν=ν41

0

16102.341 ××=

= 8 × 1015 Hz

40. Answer (2)

Ehc=

λ⎟⎠⎞

⎜⎝⎛

λ=∴

hp

⎟⎟⎠

⎞⎜⎜⎝

⎛=

pEc



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 XR1 ==λ

Longest wavelength in Balmer series for He+ = 2RZ536

R4536×

= (∴ Z = 2)

R59=

= X59

44. Answer (4)

Kinetic energy in first excited state = eV4.32

6.132 =

+…(i)

Difference in P.E. between n = 2 and n = 1 level

U2 – U1 = eV4.202

6.1321

6.132 22 =×−×

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)

⎥⎦⎤

⎢⎣⎡ −×=

λα22

2

21

11ZR1

⎥⎦

⎤⎢⎣

⎡ −=λβ

222

31

11RZ1

89

43

×=λ

λ

α

β

3227=

3227

=λβ × 0.32 Å = 0.27Å



46. Answer (4)

The lines in the Balmer series are emitted when the electrons jumps from n = 3, 4, 5...... orbits to the secondallowed 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

+2h)1(ll

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 feasibleto 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 extrascreening 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



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 = º60cos2 1121

21 μμ+μ+μ

= 21)36.0(2)36.0()36.0( 222 ×++

= 336.0 ×

= 0.36 × 1.732

= 0.62 D

60. Answer (4)

ClCl

μ = º60cos2 21

21

21 μ+μ+μ

= 21)5.1(2)5.1()5.1( 222 ++

= 35.1

= 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 BeCl2120º 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.



64. Answer (3)

52

282N =+=

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

PP60º

68. Answer (2)

NH3 hybridisation is sp3, 42

352N =+= sp3

PCl5 255

2N += = 5 → sp3d.

BCl3 233

2N += = 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)

,52

462NSF4 =+= lone pairs = 5 – 4 = 1

,42

442NCF4 =+= lone pairs = 4 – 4 = 0

,62

482NXeF4 =+= lone pairs = 6 – 4 = 2

71. Answer (1)

Greater electronegativity when bonding through axial position.



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

10275.11003.1

−

−

××

Percentage ionic character = 810

18

10108.4275.11001003.1

−−

−

×××××

74. Answer (4)

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

75. Answer (2)

52

1362N =++= , attached atoms = 3

∴ 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.



78. Answer (2)

P1 = RTVn1

P2 = RTVn2

1

2

1

2

PP

nn

=

⇒ n2 = 100

01.012012

PPn

1

21 ×=×

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

79. Answer (2)

At low pressure the volume is high

RT)bV(VaP 2 =−⎟

⎠⎞

⎜⎝⎛ + V~bV −

RT)V(VaP 2 =⎟

⎠⎞

⎜⎝⎛ +

RTVaPV =+

80. Answer (2)

RT)bV(VaP 2 =−⎟

⎠⎞

⎜⎝⎛ +

Z < 1, V – b ~ V

Vr = 2V

a1

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= E32

E = PV23

For 1 mole gas PV = RT

E = RT23

therefore E represent here translational kinetic energy.



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 expansiontemperature 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 = RTPM

d ∝ P, d ∝ T1

87. Answer (2)

PV = nRT

2 × 3 = nAR × 273

nA = R2736

for vessel B

4 × 1 = nBR × 300

nB = R3004

After the connection

P × 7 = 300RR300

4R273

6×⎟

⎠⎞

⎜⎝⎛ +

P = 1.51 atm.

88. Answer (4)

PV = nRT

10 × V = 320R321

×

V = R litre

After leakage

320RnR8510 ××=×

481

3008510n =

××

= ∴ mass of gas = gm32

4832

=

Mass of gas leaked out 1 – 32

gas = .gm33.0gm31 =



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 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

MM

rr

=

rate of diffusion = t50

timediffusedvolume

=

A

B

MM

t40t

50

=

64M

45 B=

64M

1625 B=

MB = 100

93. Answer (3)

molesX2

SOH

molesX)g(CO)g(OHHCOOH 42 +⎯⎯⎯ →⎯

COOH

COOHy moles

)g(OH)g(CO)g(CO 2molesYmolesY

2SOH 42 ++⎯⎯⎯ →⎯



61

Y2XY

=+

6Y = X + 2Y

X = 4Y

1:4YX

=

94. Answer (2)

CH4 + 2O2 → CO2 + 2H2O volume of C2H2 is X mL

Pressure (63 – X)mm (63 – X)mm

2)mm(X2

22)mm(XessurePr

22 HCO2O25HC +⎯→⎯+

total pressure of CO2 = (63 –X) + 2X = 63 + X

63 + X = 69, X = 6 mm

fraction of methane = 9.06357

63X63

==−

95. Answer (2)

6.2nnn

nP

42

22

CHHeH

HH ×

++=

⇒ 1.6 atm

96. Answer (2)

1:2224

4121

mm

nn

rr

2

22

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 gaseouscarbon 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.



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 beenformed from its constituent elements.

101. Answer (1)

3O2(g) → 2O3(g) is endothermic

0E3EO42O3 <−

2O O43E

3<

This in equality valid only 3O2O EE > .

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 decompositionΔH > 0 further more entropy increases ΔS > 0.

103. Answer (3)

For a diatomic gas Cp = R27

, Cv = R25

Only 75

= 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)

122

111 VTVT −γ−γ =

∴ 11

2

1

1

2 2VV

TT −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. Thuscurve C should respond to helium.



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,

21

H2C2O4 + NaOH → 21

Na2C2O4 + H2O; ΔH = –53.35 kJ

21

H2C2O4 + OH– → 21

C2O42– + H2O; ΔH = –53.35 kJ …(1)

H+ + OH– → H2O; ΔH = –57.3 kJ …(2)

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

21

H2C2O4 → 21

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 + 21

(O = O) → –NN =

+ = O(g)

ΔHf = )418607(49821946 +−⎟

⎠⎞

⎜⎝⎛ ×+ = 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

21

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)

Thus bond energy of 6 × (S – F) = 309 kJ/mol.



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 phenolphthaleinend point, the NaOH will use 18 m. mol of acid and Na2CO3 will use 1 m.mol of acid, Hence

Normality = N 095.0200

)118(=

+.

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 beneglected

∴ 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 = 2COP = 0.0095 atm

Since atmospheric pressure is 1 atm so percentage of CO2 in air = %95.0100PP

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

salta C

ClogpK pH = pKa + log acid

saltCC

∴ Csalt = Cacid.

pKa + pKb = 14

pKa = 14 – 4.7 = 9.3

apKpH = .3.9pH =



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 = 510x1.0

x)2.0x( −=−

+

∴ x + 0.2 ~ 0.2

0.1 – x ~ 0.1

x = 65 1051021 −− ×=×

∴ degree of dissociation = 56

1051.0

105 −−

×=× .

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=

−+

12

212

4

4 C)1(C

)CC(C]OHNH[

]OH][NH[

α = 4

1

b 108.1CK −×=

117. Answer (3)

Qp = 2NHCO 32

P.P = (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 pressurein the chamber would be equal to 30 atm.

118. Answer (2)

CKf

KbD

conc. at t = 0 a 0

conc at equi. a – x x

when equilibrium is achieved

kf (a – x) = kb.x

kb = x)xa(kf −



119. Answer (3)

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 movefor backward direction. As a result the equilibrium concentration of Cl2 will be less. So the equilibrium constantof 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 presentat new equilibrium will be less

120. Answer (2)

[S2–] = 19–21

100.105.0

105 ×=× −

147

2

2101101

]SH[]S][H[ −−

−+

×××=

[H+] = 19

21

1011.0101

−

−

×××

pH = 1.50.

121. Answer (3)

Its equilibrium constant Keq =w

ba

KKK ×

= 14

10

101024.3

−

−×

= 1.8 × 1.8 × 104

122. Answer (2)

Solubility of PbSO4 = M102.11044.1K 4–4–sp ×=×=

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 = 36.361000

= 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

OHBaCO)OH(BaCO 2white

322 +↓→+



moles of BaCO3 = 2.11974.236 =

moles of CO2 at equilibrium ⇒ 1 + x = 1.2

x = 0.2

Kc = 25.28.02.1

x1x1 22

=⎟⎠⎞⎜

⎝⎛=⎟

⎠⎞⎜

⎝⎛

−+

124. Answer (2)

Kp = 2COP = 2.25

Number of moles of CO2 = 6000821.0125.2

××

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

33

3

4 108.1]NH[

105.1105.1]NH[

]OH][NH[ −−−−+

×=×××

=

[NH3] = 0.125 M

total [NH3] required = 0.125 + 1.5 × 10–3

= 0.1265 M

126. Answer (1)

50.0XX ClCl2 ==

Kp = )15.0()15.0(

PX)PX(

PP 2

TCl

2TCl

Cl

2Cl

22×

×=

××

=

= 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

)x2.0(x108.6

]H[]SH[K 2

2

2c −

=×== −

x = 1.27 × 10–2 moles/litre

atm38.01

363082.01027.1V

nRTP2

SH2=

×××==

−



128. Answer (3)

pH = 11.27– log[H+] = 11.27[H+] = 5.37 × 10–12

[OH–] = 312–

15–10322.1

1037.5101.7 −×=

××

Kb = 5–232

1075.11.0

)10322.1(C

)C(×=

×=

α −

129. Answer (3)

Since PCO < 2COP

200010500

1PP

K 6CO

COp

2 =×

==−

Since 2.303 RTlogKp = opGΔ−

2.303 × 8.314 × T log2000 = 20700 + 12TT = 404.3 K

130. Answer (2)

K = 9.0]Glycerine][BOH[

]Complex[

33=

5.14060

]BOH[]Complex[

33==

K = 9.0]Glycerine[

5.1=

[Glycerine] = M7.19.05.1

=

131. Answer (1)

% of [In–] = %91100110

10100]HIn[]In[

]In[=×

+=×

+−

−

132. Answer (1)

M(OH)2(s) M2+ + 2OH–

x 2xpH = 10.6pOH = 14 – 10.6 = 3.4

[OH–] = antilog(–3.4) = 3.98 × 10–4 M

M1099.12

1098.3x 44

−−

×=×

=

Ksp = [M2+][OH–]2 = 4x3 = 4 × (1.99 × 10–4)3 = 3.15 × 10–11



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 e2FeFe 2 +⎯→⎯ +

At cathode Pbe2Pb2 ⎯→⎯++

Net reaction PbFeFePb 22 +⎯→⎯+ ++

ºanode

ºcathode

ºcell EEE −= = –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+ areformed.

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

44.22

56.0n52

6.2×=×

n = 2 i.e. CrCr 2 ⎯→⎯+

137. Answer (2)

E°cell = E°cathode – E°anode

= )aq(Fe|)aq(Fe)l(OH|)aq(OH 23222EE ++°−°

= 1.763 – 0.769 = 0.994 V



138. Answer (3)

As the Cu2+ ions lost from the solution are compensated by copper anode therefore concentration of the solutionremain same

At anode e2Cu)s(Cu 2 +⎯→⎯ +

At cathode )s(Cue2Cu2 ⎯→⎯++

139. Answer (2)

As the surface area of contact of an electrode with electrolyte increases. Conductance of electrolyte increasethereby time rate of electrolysis increases.

140. Answer (2)

Half cell reactions of the given electrodes are

First electrode ++ ⎯→⎯+ 222 AmOeAmO

Second electrode OH2Ame2H4AmO 242

2 +⎯→⎯++ +++

Third electrode ++ ⎯→⎯+ 24 Ame2Am

Thus it is evident that half cell reaction of only second electrode involves H+ ions so its reduction potentialwill change with varying pH value.

141. Answer (1)

Anode e2H2H2 +⎯→⎯ +

10–6 M

Cathode 2He2H2 ⎯→⎯++

Ecell = 0.118 = 26cathode

2

)10(]H[log

2059.0

−

+

On solving we get

[H+]cathode = 10–4 M

142. Answer (2)

422

2 OZnMne2ZnMnO2 ⎯→⎯++ +

187E

2MnO =

t = days35.5124360087102

965008Ei

96500w3 =

×××××

=×

×−

143. Answer (4)

In pure state sulphuric acid makes cyclic ring type of structure in the absence of water so it cannot give offH2 gas to react with metals.

SHO

HO

O

OS

HO

HO

O

OS

HO

HO

O

O



144. Answer (1)

2He2H2 ⎯→⎯++

2H/H2H

H/H ]H[1log

2059.0E

]H[

Plog

2059.0EE

2

2

2 ++−°=−°= ++

2H/H ]H[100log

2059.0EE

2 +−°=′ +

∴ V059.0]H[]H[

100log2059.0EE 2

2H/HH/H 22=×=′− +

+++

145. Answer (1)

For the following electrochemical cell

Pt|)atm1(Cl|)C(Cl||)C(Cl|atm) (1Cl|Pt 2212−−

The half cell reactions are

Anode −− +⎯→⎯ e)g(Cl21Cl 2A

Cathode −⎯→⎯+ c2 Cle)g(Cl21

The Ecell is given by

2

1

1

2

A

Ccell C

Clog059.0CClog

1059.0

]Cl[]Cl[log

1059.0E =−=−=

−

−

For Ecell to be positive C1 > C2

146. Answer (2)

Let the formula of mercury ion is +nnHg then the formula of mercury nitrate would be Hgn(NO3)n. The reactions

occuring at two electrodes are

Cathode ( ) nHgneHg Cnn ⎯→⎯++

Anode ( ) −+ +⎯→⎯ neHgnHg Ann

Net reaction ( ) ( )AnnC

nn HgHg ++ ⎯→⎯

Thus the given cell is electrolyte concentration cell

Ann

Cnn

cell ]Hg[]Hg[log

n0591.0E +

+

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

20121

logn

0591.00295.0

n = 2

It means mercury ion exists as +22Hg



147. Answer (1)

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

)aq(Cl)s(Age)s(AgCl −− +⎯→⎯+

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

148. Answer (1)

Q + 2H+ + 2e QH2

2cellcell ]H[1log

20591.0EE

+−°=

pH0591.0EE cellcell −°=

intercept = V699.0Eocell =

pH = 5.30591.0

492.0699.0=⎟

⎠⎞

⎜⎝⎛ −

149. Answer (1)

H2O H+ + OH– ; w1 KlnRTG −=Δ o

H+ + e 21

⎯→⎯ H2 ;oo

2H/H2 EFnG +−=Δ

H2O + e– 21

⎯→⎯ H2 + OH– ; oo−−=Δ OH,H/OH3 22

EFnG

Since 0GSo0E o2H/H 2

=Δ=+o

Thus o3

o1 GG Δ=Δ

–RTln Kw = oOH,H/OH 22

EFn −−

)1n(cesinKlnF

RTE wo

OH,H/OH 22==−

150. Answer (4)

84

2o

Mn/MnOMn/MnO ]H][MnO[]Mn[log

5059.0EE 2

42

4 +−

+

−= +−+−

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

+−′ 24 Mn/MnO

E = 84

82o

Mn/MnO x]MnO[]2][Mn[log

5059.0E 2

4 −

+

−+−

= 8

84

2o

Mn/MnO 2log5059.0

x]MnO[]Mn[log

5059.0E 2

4−−

−

+

+−

= 02846.0x]MnO[]Mn[log

5059.0E 8

4

2o

Mn/MnO 24

−−−

+

+−

Electrode potential decreases by 28.46 mV



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; o1GΔ = –2F(0.337)

Cu+ + e → Cu; o2GΔ = –1F(0.153)

Cu2+ + e → Cu+; o3GΔ = –1FE°

o3GΔ = oo

21 GG Δ−Δ

–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)

242125239128334 KNOKOHNONH =−+=Λ−Λ+Λ=Λ ∞∞∞∞

Degree of dissociation = 10.024224

=⎟⎠⎞

⎜⎝⎛=

ΛΛ

∞ .

155. Answer (2)

5.1901.0

105.191000N

K1000 5=

××==Λ

−

Degree of dissociation = 05.0390

5.19==⎟

⎠⎞

⎜⎝⎛

ΛΛ

∞

156. Answer (1)

oooanodeRPcathodeRPcell EEE −= = 0.13 – (–0.34) = 0.47

2Tl + Sn4+ → Sn2+ + 2Tl+

Ecell = 0.47 – ]Sn[

]Sn[]Tl[log2059.0

4

22

+

++

= 0.47 – V411.0059.047.0]10log[2059.0 2 =−=

157. Answer (2)

)44.0(23.1Ecell −−=o = 1.23 + 0.44 = 1.67

ΔG° = – ocellnFE = –2 × 96500 × 1.67 × 10–3 kJ = –322 kJ

158. Answer (4)Ni(s) + Cu2+(aq) → Cu(s) + Ni2+(aq)

V59.0)25.0(34.0EEE anodeRPcathodeRPcell =−−=−= ooo

Q ocellE is positive therefore reaction will be spontaneous.



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 = ⎟⎠⎞

⎜⎝⎛−

101.0log

2059.0Ecell

o

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

E2 = ⎟⎠⎞

⎜⎝⎛−

01.01log

2059.0Ecell

o

It is clear from both equation

E1 > E2

161. Answer (4)

Fe2+ + 2e → Fe; o1GΔ = –2F(–0.44)

Fe3+ + 3e → Fe; o2GΔ = –3F(–0.036)

Fe3+ + e → Fe2+; o3GΔ = –1FE

o3GΔ = oo

12 GG Δ−Δ

–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

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

+

+

2

2

cell CuZnlog

2059.0Eo

1.1 = ⎥⎥⎦

⎤

⎢⎢⎣

⎡+

+

2

2

CuZnlog

2059.0

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

+

2

2

CuZnlog = 37.3

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

+

2

2

CuZn

= 1037.3

163. Answer (2)

Since the cell reactions proceed in the standard condition and ocellE is negative therefore the electricity cannot be

produced.



164. Answer (2)

AgCl(s) Ag+(aq) + Cl–(aq) ; ΔG1° = –2.303 RT log Ksp

Ag+(aq) + e → Ag(s) ; ΔG2° = Ag/AgEF1 +°−

AgCl(s) + e → Ag(s) + Cl–(aq) ; ΔG3° = Ag/AgCl/ClEF1 −°−

From the given equations

ΔG3° = ΔG1° + ΔG2°

Ag/AgCl/ClE −° = Ag/AgE +° + 2.303 RT log Ksp

–0.15 = 0.80 + 2.303 RT log Ksp

log Ksp = 101.16059.0

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 = 965007200 I

At. anode 2OH– → 21

O2 + H2O + 2e

At. cathode 2H+ + 2e → H2

Moles of O2 released at anode = 41

96500I7200

×

Moles of H2 released at cathode = 21

965007200

×I

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

67243I

96500720022400 =××

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



168. Answer (3)

338233

AVCm/gm2.3

46.5023.67804

)1046.5(10023.6)78(4

aNMZd =

××

=××

=×

×=

−

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 = 33 r3

16r344 π=π×

For fcc unit cell, a = r22

Volume of cube = 333 r216)r22(a ==

Packing fraction = 23r216

r3

16

cubeofvolumesphereofvolume

3

3π

=π

=

174. Answer (4)

d = 3

38233AV

Cm/g75.411023.66204

)10(10023.6624

aNMZ

=×

=××

×=

××

−

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. effectiveone 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– = 2a

a = 2.3 × 2 = 4.6 Å

d = 243233AV 10)6.4(10023.6

784aN

MZ−×××

×=

××

= 3)6.4(023.67804×

× = 5.32 gm/Cm3.

177. Answer (4)

Total number of effective atoms in unit cell = 4 + 4 = 8.



178. Answer (1)

In bcc, r4a3 =

r = 42973

4a3 ×

=

In fcc, r4a2 =

a = r222r4

=

= 2974322 ××

= 29723

×

= 363.79 pm.

179. Answer (2)

d = 3AV aN

MZ×

×

10.5 = 3823 )1009.4(10023.6108Z

−×××

×

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 edgecentre ions of A will be removed

therfore effective number of ions of B = 4 – 1.5 = 25

Effective number of ions of A = 27

214 =−

∴ New formula of compound A7/2B5/2.

181. Answer (2)

732.0rr

=−

+

r– = .pm61.136732.0

100732.0r ==

+

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°.



185. Answer (1)

In simple cubic unit-cell the packing fraction = 3

3

)r2(

r34

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 = 214

81

=×

Number of B atoms = 1221

=×

Number of C atoms = 21441

=+×

So, the formula of compound per unit cell is A1/2BC2, simplest formula of compound is AB2C4.

187. Answer (2)

442

32

10310223

dt]H[d

dt]NH[d

21

dt]H[d

31

−− ×=××=−

+=−

So, 4432 102103dt

]NH[ddt

]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

PP 0t −

For first order reaction,

)xa(alog

t303.2k

−=

⎟⎠⎞

⎜⎝⎛ −−

=

3PPP

Plogt303.2k

0t0

0

)PP4(P3log

t303.2k

t0

0−

=



190. Answer (3)

U23892 is disintegrated through (4n + 2) series.

191. Answer (1)

Total time = 21tn×

69.2 = n × 138.4

n = ½

N = 21n

0 211

21N ⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

N = 7.0414.11

21

==

Disintegrated amount = 1 – 0.7 = 0.3 g

PoPo 20682

He21084

42 ⎯⎯ →⎯−

Volume of helium accumulated = g3.0210

22400×

= 32 ml.

192. Answer (2)

2252 B½AB2BA +⎯→⎯

dt]B[d2

dt]AB[d

21

dt]BA[d 2252 +=+=−

]BA[k2]BA[k21]BA[k 523522521 ×=×=

So, the relation becomes

2k1 = k2 = 4k3

193. Answer (3)

t

0NNlog

t303.2

=λ

303.2t

NNlog

t

0 ×λ=

Comparing with y = mx + C then

Slope = 303.2λ

+

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



196. Answer (2)

)g(C2)g(B)g(A2 ⎯→⎯+

210

100120dt

]C[d21

=−

=+

min/mm4dt

]C[d=+

197. Answer (1)

1n½ )a(1t −∝ (n = order of reaction)

198. Answer (2)

Overall order of reaction is 2.

199. Answer (2)

1

2.02062381

logt303.2k

⎟⎠⎞

⎜⎝⎛ ×+

=

1.0t303.2

105.4693.0

9 ×=×

t = 1.5 × 109 years.

200. Answer (3)

2A + 3B ⎯→⎯ 4C + 5D

dt]C[d

41

dt]B[d

31

+=−

dt]C[d

dt]B[d

34

+=−

201. Answer (4)

For ‘B

M = 1001

1K1000 b

×××

(M = molar mass of B)

Kb = 10M

For ‘A’

M1 = 1001012M1000

××××

(M1 = molar mass of A)

M1 = 2 M



202. Answer (2)

From Roult’s law,

Ps = BBAA X·PX·P oo +

Ps = )X1(PX·P ABAA −+ oo (Q XA + XB = 1)

Ps = ABBAA X·PPX·P ooo −+

Ps = )PP(XP ABABooo −− …(i)

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

oBP = 210 and o

AP = 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 = 5.5518

1000=

xurea = 5.561

5.5511

nnn

waterurea

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 alsoincreases 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.



207. Answer (2)

In case of 23 )NO(Ca20M

and 0.05 M Na2SO4, effective molarity are same.

208. Answer (1)

M = WT

wK1000 f

×Δ××

62 = W3.95086.11000

×××

W(unfreezed water) = 161.29 g

Ice separated = 200 – 161.29 = 38.71 g

209. Answer (3)

m = m5.0

100020095

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 = 2.112.1

=

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 andhigher is the colligative properties.

212. Answer (1)

Emulsifying agent stabilised the emulsion of oil in water by forming an interfacial film between suspended particlesand the medium.

213. Answer (2)

A negatively charged sol is formed due to adsorption of I–.

Coagulation value ∝ powergcoagulatin

1 .



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 solbefore 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 Questions1. Answer (2, 4)

The substances having same composition of atoms and similar crystal structure are isomorphous.

2. Answer (2, 3)

%C in C2H5OH = %52100116524

24=×

+++

%C in C6H12O6 = %40100961272

72=×

++

%C in CH3COOH = %4010013212312

24=×

++++

%C in C2H5NH2 = %53100214524

24=×

+++

3. Answer (1, 4)

OHNaFNaOHHF 2+⎯→⎯+

moles of HF = 01.010001001.0 =×

moles of NaOH = 01.010001001.0 =×

number of moles of NaF formed = 0.01

M 05.0

1000200

01.0]NaF[ ==



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

x = +6

CrO

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 highestoxidation state.For Cl, range of oxidation number is from –1 to 7.In HCl, Cl is present in lowest oxidation stateIn HClO4, Cl is present in highest oxidation state.

6. Answer (1, 2, 3)MnO2 + (NH4)2SO4 → MnSO4 + (NH4)2S2O8

n = 2 n = 1equivalents of MnO2 = equivalents of (NH4)2SO4

1 × 2 = 1xx = 2

7. Answer (1, 2, 3)Resultant normality of solutionN1V1 + N2V2 + N3V3 = N4V4

5 × 1 + 20 × 21 + 30 × 3

1 = N4(1000)

4N1000

25 =

N401N4 =

Resultant [H+] = 401 = 0.025

Normality = )litre(volmassequivalent/mass

40x

401 × molecular mass of NaOH = 40

x = 1 gm8. 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 = 122 =

Volume of SO2 at STP = 22400 × 10–3 = 22.4 ml.Concentration of SO2 in air is 22.4 pm



9. Answer (1, 3, 4)

OH9NO2)NO(Cu3HNO8Cu3 2232

30

++⎯→++

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 86 or 4

3 mole of electrons

∴ n factors of HNO3 = 43

Cu is oxidised of Cu2+

equivalents mass of HNO3 = 844/3

63 = gm.

10. Answer (1, 3, 4)

OH2ClMnClHCl4MnO 2

0

2

2

2

14

2 ++→++−+

n factor of HCl = 21

42

=

n factor of MnO2 = 2

equivalent mass of MnO2 = 2

massmolecular.

11. Answer (1, 2)

22

7

)10n(24 Mn)MnO(Ba

++

+

=⎯→⎯

milliequivalents of Ba(MnO4)2 = 10010101100 =××

Fe2+ → Fe3+ + e

n = 1

milliequivalents of FeSO4 = 100 × 1 = 100

4

2

33

)3n(4

3

2

2COFeOCFe++

+

=

+++→

milliequivalents of FeC2O4 = 10033

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.



13. Answer (1, 3, 4)

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 n2 = (m + 1)

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−=

λ 222

m )1m(1

11Rz1

for nth line n2 = (n +1)

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−=

λ 222

n )1n(1

11Rz1

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

−+

+

+=

λλ

1)1m(1)1n(

)1n()1m(

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 energyof 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. 1dxdydz2 =ψ∫∞

∞−18. Answer (3, 4)

n = 4, l = 0, 1, 2, 3

for l = 2, m = –2, –1.0 + 1, +2

s = 21

+ (correct)

n = 4, l = 0, 1, 2, 3

l = 2, m = –2, –10 + 1 + 2

s = 21

− (correct)



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

)nln(4nlnrrln

AAln 4

21

2n

1

n ==⎟⎟⎠

⎞⎜⎜⎝

⎛

π

π=⎟⎟

⎠

⎞⎜⎜⎝

⎛

ln

ln(n)

AA

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 statetherefore (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)

pyramidaltrigonal

ionHybridisat;1pairlone42

352NNH


1342NCH


1362NOH

33

33

33

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=∴==+

=

=∴==++

=

=∴==−+

=

−

+

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+ + Br3–.

(4) Sodium sulphate is soluble in water but BaSO4 is sparingly soluble because hydration energy of BaSO4is less than its lattice energy.



27. Answer (1, 2)

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 gasconfiguration.

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

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+][PCl6

–] and PBr5 exists as [PBr4+][Br–] in solid state.

35. Answer (1, 4)

52

282NXeOF2 =

+= attached atoms = 3; ∴ T-shaped

42

1252NNH2 =

++=− 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 andplanar.



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)

OCH3

OCH3

(μ ≠ 0)

N

N

OO

OO(μ 0)=

Br

Cl

( μ 0)~–

C

C

HCH3

H C H2 5

(μ ≠ 0)

41. Answer (2, 3, 4)

Statement 2 and 3 are facts

Unit of P = Unit of 2

2

Van

∴ 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)

RTPVZ =

Average RT23KE =

44. Answer (2, 3, 4)

Above critical temperature gas cannot be liquified.

45. Answer (1, 3, 4)

)V()V(

VVZ

id

m

ideal

molar=



When Vm > Vid Z > 1

Vm = Vid Z = 1

Vm < Vid Z < 1

If force of attraction dominates then Z < 1

46. Answer (2, 3)

Mg2C3 ⎯⎯→ 2 Mg2+ + C34–

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

CaCN2 ⇒ Ca2+ + CN22–

[–N C N–] Two sigma and two pi bonds

47. Answer (1, 2, 3)

In solid state N2O5 exist as ][NO][NO 32−+

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 2px2 = π 2py

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

BrBr

Cl

dipole moment 0μ ≠

(2)

Cl

P

Br

BrBr

Cl

dipole moment = 0μ

(3)O O

CH3 CH3

dipole moment is not zerodue to following structure

(4) NH

HO

Hdipole 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. InH3PO3, 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 zerotemperature the kinetic energy of gas molecules will be zero.



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 = RT23

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 thereforekinetic energy of gas molecule remain same.

55. Answer (1, 2)

Pc = 2b27a

Vc = 3b.

56. Answer (1, 3, 4)

224.1:128.1:1

MRT3:

MRT8:

MRT2

u:u:u rmsaveragemp

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 dQdP

is intensive

property.



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 arestate function.

66. Answer (2, 3)

During the streching of rubber band the long flexible macromolecules get uncoiled the uncoiled arrangementhas 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 x43 0 0

x43x − 0 ⎟

⎠⎞

⎜⎝⎛

4x3

⎟⎠⎞

⎜⎝⎛

4x3

pOH = x44x3log5

]BoH[]Salt[logpKb ×

×+=+

pH = 14 – 5 – log 3 = 8.523.

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

At equivalence point

(acid)22

(base)11 VNVN =

2.5 × 52

= 152

× V

V = 7.5 ml

∴ Milli equivalents of salt = 1



pH = 7 – 21

pkb – 21

log C

= 7 – 6 – 21

log 101

= 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 °cellE = – 2.303 RT log K

At 25°C

Klog n

0591.0Ecell =° .

70. Answer (1, 2, 3, 4)

Pressure favours the forward reaction. The temperature at which atmospheric pressure is equal to vapourpressure 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 favoursthe 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 CH3COOH + OH– (basic)

Na2CO3 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 forwardreaction 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.



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)

Hln H+ + In–

Ka = ]HIn[

]In][H[ −+

[H+] = ]base[]acid[Ka

for 75% red [H+] = 55 103257510 −− ×=⎥⎦

⎤⎢⎣⎡

pH = 4.52

for 75% blue [H+] = 55 1031

752510 −− ×=⎟

⎠⎞

⎜⎝⎛ = 5.47

80. Answer (2, 3)

On increasing the temperature ionic product of water increases so pH and pOH decreases but water willremain neutral.

81. Answer (1, 2, 4)

⎥⎦

⎤⎢⎣

⎡−

Δ=

211

2T1

T1

R303.2H

KKlog

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 notΔH = 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



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., 3t1 and

2t1 because initial molar ratio is 1 : 3.

⎟⎠⎞

⎜⎝⎛=

−−

=3

142)x39(

28)x3()H(W)N(W

2

2 remain same at 3tt 1= as well as

2tt 1= .

83. Answer (1, 2, 4)

Upto phenolphthalein NaOH is fully neutralised and Na2CO3 will be converted to NaHCO3. In next step NaHCO3coming from Na2CO3 neutralised by HCl using methyl orange indicator. So y ml should be less than x mlthat 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)

Pb(s) (Pb2+)A(aq)

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

for this cell E°cell = 0

Ecell = C

2A

2

]Pb[]Pb[log

2059.0

+

+

−

Ksp = [Pb2+][SO42–] = s2

s = )PbSO(K 4sp

Ksp = [Pb2+][I–]2 = 4s3

s = 31

2sp )PbI(4

K⎥⎦

⎤⎢⎣

⎡

Ecell = 21

4sp

31

2sp

]PbSO(K[

4)PbI(K

log2059.0

⎥⎦

⎤⎢⎣

⎡

86. Answer (2, 3)

Ecell = E°cell – ]Cu[]Cd[log

2059.0

2

2

+

+

on increasing the concentration of [Cu2+] and decreasing the concentration of [Cd2+] Ecell will be more positive.



87. Answer (1, 2, 3)

H — O — S — O — O — H–1 +2 +1 –1

O–2

O–2

+1 +2 total charge at s = +6

Cr

O–2

O

O

O

O

–1–1

–1–1

+1 +1

+1 +1+2CrO5 total charge on Cr = +6.

88. Answer (1, 2)

Molar conductance of an electrolyte depends upon its degree of dissociation with increase in dilution the molarconductance increases due to increase in dissociation specific conductance decreases upon dilution becausenumber of current carrying ions per unit volume of solution decreases.

89. Answer (1, 2)

1

2n

5.2

6422n

0

21n

3

2

22 aIN2OSNaIOSNa2−

=

+

==

++⎯→⎯+

90. Answer (1, 3)

Ecell = 23

25

)10()10(log

2059.00

−

−

− = positive

0E2H/H

=° +

Cr ⎯→⎯ Cr3+(aq) + 3e

Cu2+(aq) + 2e ⎯→⎯ Cu

2Cr + 3Cu2+ ⎯→⎯ 2Cr3+ + 3Cu

Ecell = E°cell – 32

23

)Cu()Cr(log

6059.0

+

+

= E°cell – 3

2

)2.0()1.0(

6059.0

= E°cell + 2log36059.0

91. Answer (2, 3)

The value of the constant A for a given solvent and temperature depends on the type of electrolyte i.e., chargeon cation and anion produced on the dissociation of the electrolyte in the solution.



92. Answer (1, 2, 3)

]Alog[nF

RT303.2EE nA/AA/A nn

+−°= ++ from this equation it is clear that

+nA/AE decreases with increasing [An+]

]A[1log

nFRT303.2EE

nA/AA/A nn +−°= ++

93. Answer (1, 2, 3)

If a given metal ion has negative reduction potential H+ will be reduced by metal. Similarly if reduction potentialis positive metal will be reduced by H2. If metal ion with negative potential is coupled H-electrode the hydrogenhalf cell should function as cathode thus metal electrode will be negative half cell (anode). In aqueous solutioncontaining 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 thestate of equilibrium.

95. Answer (1, 2, 3)

Mole of Fe3+ = 0.1 × 1 = 0.1

Mole of electron = mole149.096500

43600=

×

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 = Femole025.0049.021

=×

96. Answer (1, 2)

H – O – S – O – O – H

O

O(peroxy linkage)

H – O – S – O – O – S – OH

O

O(peroxy linkage)

O

O

97. Answer (2, 3)

240

4n

22SOCuSCu

+

=

−++⎯→⎯ equivalent mass =

4M

2

24n2

0OHO−

=⎯→⎯ equivalent mass =

4M



98. Answer (1, 2, 4)

Factual type

99. Answer (3, 4)

Original Formula

Number of A atoms = 214

81

=×

Number of B atoms = 1221

=×


+× = 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.


812–

21

=×

Number of B atoms = 1 – 1 = 0

Number of C atoms = ⎟⎠⎞

⎜⎝⎛ +× 12

41–4 =

212

Hence, formula of compound is 25

41 CA or AC10

Possibility II : Suppose 2 face atoms are not removed.


81–

21

=×

Number of B atoms = 1 – 0 = 1


41–4 =⎟

⎠⎞

⎜⎝⎛ +×

Hence, formula of compound is 212

41 BCA 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 verylarge.



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 = 812–4 × = 3.75

Hence, formula per unit cell is Na4Cl3.75.

IInd Combination : Two face Cl– ions are removed


Number of Cl– ions = 221–4 × = 3

Hence, formula per unit cell is Na4Cl3IIIrd Combination : One corner and one face Cl– ions are removed.


Number of Cl– ions = 1211

81–4 ×+× = 3.375

Hence, formula per unit cell is Na4Cl3.375.

103. Answer (1, 3)

86.095.169.1

rr

==−

+ i.e. coordinate no. = 8

∴ bcc structure

∴ a3)rr(2 =+ −+

∴732.1

)95.169.1(2a +=

104. Answer (1, 4)

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 whereasfor 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 packingcoordination number = 4

Hexagonal close packingcoordination number = 6



109. Answer (1, 3)

Zn2+ is present in alternate tetrahedral voids therefore its C.N is 4. In rock salt structure there is 4Na+ and4Cl– 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 14elements 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)

303.2t·k

]R[]R[log 0 =

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)

1n½ )a(1t −∝ (n = order of reaction) and the unit of frequency factor ‘A’ is equal to the unit of ‘k’’

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. SPMallow the movement of solvent only not the solute. So, no ppt. of BaSO4 is formed in right side.



125. Answer (1, 2)

When CuSO4 is dissolved in NH4OH then association takes place instead of dissociation.

CuSO4 + 4NH4OH → [Cu(NH3)4]SO4

Hence, freezing point of solution is raised and boiling point of solution is lowered.

126. Answer (1, 3)

Acetone and chloroform shows negative deviation from Raoult’s law while ethanol and water shows positivedeviation from Raoult’s law.

127. Answer (3, 4)

Statement 1 and 2 are incorrect because sol particles neither move toward anode nor cathode.

128. Answer (2, 3)

An anion caused the precipitation of a positively charged sol and vice versa. The higher the valency of theeffective ion, greater is the penetrating power.

129. Answer (2, 3)

Size of suspension particles are > 10–5 cm in diameter.

Size of colloidal particles are 10–7 –10–5 cm in diameter.

Size of true particle < 10–7 cm in diameter.

130. Answer (1, 2)

When dispersion medium is gas, the colloidal system is called aerosol, smoke, dust are example of aerosolsof solids whereas fog, clouds are example of aerosol of liquids.

131. Answer (1, 3, 4)

Starch, gum and protein in water are examples of lyophilic sols. Sulphur in water is an example of lyophobicsol.

Sulphur in water is an example of ryophobic sol.

132. Answer (2, 4)

Chemisorption is specific in nature and it is shown by the gases which can react with adsorbent. Chemisorptionis unimolecular not multimolecular and favourable at high temperature not at low temperature.

133. Answer (2, 4)

Peptization is the preparation method of colloids, electrophoresis is the property of colloids. While ultrafilterationand electrodialysis are the purification method of colloids.

134. Answer (2, 3)

In homogeneous catalysis, the physical state of reactants and catalyst are same

N2(g) + 3H2(g) Fe(s)

2NH3 (Haber’s process)

2SO2(g) + O2(g) Pt(s)

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 isnegatively charged sol.



Section - C : Linked ComprehensionC1. 1. Answer (2)

Let the % of B-10 = Xthen % of B-11 = (100 – X)

Average atomic mass 2.10100

11)X100(X10=

−+=

10X + 1100 – 11X = 1020X = 80

% of B–10 = 802. Answer (3)

Average atomic mass = 5.354

371353=

×+×

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

Z < Y < XC2. 1. Answer (2)

Let the moles of water = 1 moleMoles of urea will also be 1.

Mass percentage of water 1006018

18×

+=

781800

=

= 23.072. Answer (1)

1 kg water has 11.11 moles of solute

mole fraction of solute 167.055.5511.11

11.11

18100011.11

11.11=

+=

+=

3. Answer (3)106 gm water contains = 300 gm CaCO3

∴ Molarity L/mol1000

100/300=

M003.01000

3==

C3. 1. Answer (1)

π=Δ⋅Δ

4hpx (Δx = Δp)

π=Δ

4hx



π=Δ⋅Δ

4hvmx

xm4hv

Δ⋅π=Δ

h4

m4h π

×π

=

m1

4h

×π

=

1.910

14.3410626.6 3134

××

×=

−

= 8 × 1012 m/s2. Answer (2)

Δx = vm4h

Δπ

23

34

1021014.3410626.6

−−

−

××××

×=

= 2.64 × 10–30 m3. Answer (1)

λ = v (given)

λ = mVh

v = mvh

v2 = mh

v = mh

C4. 1. Answer (2)

34

23

EEEE

−−

= ⎥⎦⎤

⎢⎣⎡ −

⎥⎦⎤

⎢⎣⎡ −

91

161Z6.13

Z41

916.13

2

2

= 720

7144

365

=×

2. Answer (2)

Zn529.0r

2

n ×=

1n529.0

292.16 2

×=

n = 4Maximum no. of electrons = 2n2

= 2 × 42 = 32



3. Answer (4)For the visible region transition belongs to Balmar series. For the Balmar series the electron should jumpsfrom higher level to 2nd energy level.

C5. 1. Answer (3)Mn2+ – 1s2, 2s2, 2p6, 3s2, 3p6, 3d5

3d5

Total spin = 25

215 =×

2. Answer (4)Spin quantum no. is not derived from Schrondiger wave equation.

3. Answer (3)Maximum number of electrons in any subshell having same value of spin quantum number is (2l + 1)

C6. 1. Answer (2)For the maximum wavelength energy should be minimum. In the option (2) there is no shell change.Energy emission is minimum in this transition. So the wavelength will be maximum.

2. Answer (2)Energy electron for H-atom depends only on the value of n not the value of l. Therefore

4s > 3d = 3p = 3s3. Answer (1)

Radial nodes = (n – l – 1)For 3s ; n = 3, l = 0Radial nodes = 3 – 0 – 1 = 2For 2p ; n = 2, l = 1Radial nodes = 2 – 1 – 1 = 0

C7. 1. Answer (1)For the first excited state n = 2

Energy = 2

2

nZ6.13 ×−

eV4

6.13216.13 2

−=×−=

= –3.4 eV2. Answer (4)

1n529.0r

2

H =

= 0.529 × 1 Å= 0.529 Å

42529.0r

2

Be3 ×=+

= 0.529 Å3. Answer (4)

1s electronic level allow the H-atom to absorb a photon but not to emit a photon. If it emit a photon itwill drop into nucleus, that is not possible.



C8. 1. Answer (1)N(7) – 1s2, 2s2, 2p3 – Half filled (more stable)

O(8) – 1s2, 2s2, 2p4 – Less stable∴ Electron Affinity of Nitrogen is less than Oxygen.

2. Answer (4)

Electron Affinity of Br is less than Chlorine.3. Answer (1)

Na(g) ⎯→ Na+ (g) + e ΔH = ionisation energy …(1)Na+(g) + e ⎯→ Na ΔH = Electron affinity …(2)Since (1) and (2) process are opposite therefore ionisation energy of Na is equal to electron affinity of Na+.

C9. 1. Answer (4)In He+ ions the electrons are tightly held up by the nucleus therefore its ionisation energy is more thanHe.

2. Answer (1)

Be (4) ⎯→ 1s2, 2s2 – full filled (more stable)B (5) ⎯→ 1s2, 2s2, 2p1 – Less stable

∴ Be has more first I.E. than I.E. of B.3. Answer (1)

Due to half filled P-orbital nitrogen electronic configuration is more stable. Therefore its ionisation energyis more.

C10. 1. Answer (2)Since IE of element B is less therefore it is most reactive metal amongst given elements.

2. Answer (4)

Element D has high IE but less than IE of A. Therefore (D) is non metal.3. Answer (1)

The first ionisation potential is highest for element A therefore A is noble gas.

C11. 1. Answer (2)

B — F

F

N

H H H

..FN

F FF

..

μ = 0 Addition Subtraction

∴ BF3 < NF3 < NH3.2. Answer (3)

Cl

Cl

Cl μ = 0

3. Answer (2)% ionic character = 16 (ΔEN) + 3.5 (ΔEN)2

= 16(0.2) + 3.5(0.2)2 = 3.34



C12. 1. Answer (2)

N = O.. ... ..

N..

O O.... ....

2. Answer (2)

–O — C

O

O–Bond order = 33.1

311 =+

3. Answer (3)CO molecule

Total no. of electrons = 14.

Arrangement – σ1s2 σ*1s2 σ2s2 σ* 2s2 ⎥⎥⎦

⎤

⎢⎢⎣

⎡

π

π2z

2y

p2p2

σ2px2.

B.O. = 21 (10 – 4) = 3.

In CO+ ion, one electron is released from antibonding molecular orbital.

∴ B.O. = 21 [10 – 3] = 3.5.

C13. 1. Answer (2)

º5.104cos2 21

21

21R μ+μ+μ=μ

2. Answer (3)μ = q × d1.03 × 3.33 × 10–30 = charge × 1.27 × 10–10

Charge = 2.7 × 10–20

Percentage ionic character = %87.16100106.1107.2

19

20=×

××

−

−

~ 17%

3. Answer (1)

O = C = O+ μ = 0

C14. 1. Answer (4)

−22O →

⎥⎥⎦

⎤

⎢⎢⎣

⎡

π

π

⎥⎥⎦

⎤

⎢⎢⎣

⎡

π

πσσσσσ

2

2

2

222222

2*

2*

2

222*21*1

z

y

z

yx

p

p

p

ppssss

There is no unpaired electrons∴ paramagnetism is not shown.

2. Answer (2)Bond order of N2 is 3.Bond order of O2 is 2.Bond order of F2 and Cl2 are 1.

∴ Bond length ∝ orderBond1



3. Answer (1)In NO molecule there are total no. of 15 electrons.

Arrangement - σ1s2 σ* 1s2 σ2s2 σ*2s2 σ2px2

⎥⎥⎦

⎤

⎢⎢⎣

⎡

π

π2z

2y

p2p2

π* 2py1.

∴ Unpaired electrons = 1.C15. 1. Answer (3)

2

2

H

mix

mix

H

MM

rr

=

2M4 mix=

2M16 mix=

Mmix = 32Let the molar ratio of C2H4 and CO2 is a : b.

32ba

b44a28Mmix =++

=

32a + 32b = 28a + 44b4a = 12b

1:313

ba

==

2. Answer (2)

252.0284

32

rr

He

N2 ==

Let the moles of He is coming out to be xMoles of N2 coming out is 0.252x

2.0x252.1x252.0

nn

total

N2 ==

8.0nn

total

He =

Mmix = 0.2 × 28 + 0.8 × 4 = 8.8

52.032

8.8rr

mix

O2 ==

3. Answer (1)

Cl2(g) 2Cl(g)

Initial moles 1 0At equilibrium 1 – x 2xTotal no. of moles at equilibrium = 1 – x + 2x = 1 + x = 1 + 0.14 = 1.14



The composition of gas mixture is coming out would be

17.271

5.3528.086.0

rr

Cl

Cl2 ==

Let the moles of Cl atom coming out be x, then moles of Cl2.Coming out would be 2.17 x

68.0x17.3x17.2X

2Cl == XCl = 0.32

Mmix. = (0.68 × 71) + (0.32 × 35.5) = 59.64

18.164.598.83

MM

rr

mix

Kr

kr

mix ===

C16. 1. Answer (2)

Partial pressure of hydrogen totalP

16w

2w

2w

×+

=

totalP

16ww8

2w

×+

=

totalPw9w8

×=

totalP98

×⎟⎠⎞

⎜⎝⎛=

2. Answer (3)Dalton’s law of partial pressure is not valid for reacting gases

fumeswhite43 )g(ClNH)g(HCl)g(NH ↑⎯→⎯+

3. Answer (4)

Volume percent of H2 100mixture of pressure TotalH of Pressure 2 ×=

100100200300150

150×

+++=

%20100750150

=×=

C17. 1. Answer (1)

3O2(g) 2O3(g)

Initial volume 1 litre 0After reaction 1 – 3x 2xTotal volume of gaseous mixture = 1 – 3x + 2x = 0.8

x = 0.2

Mole fraction of O3 = 5.08.04.0

volume TotalO of volume 3 ==



2. Answer (2)

mL402

oilturpentine

mL10023 O)OO( ⎯⎯⎯⎯⎯ →⎯+

Therefore volume of O3 is 60 mL

2O3 3O2

2 mL 3 mL60 mL 90 mL

Total volume now = 40 + 90= 130 mL

Change (increase) in volume = 130 – 100= 30 mL

3. Answer (1)Let the volume of NH3 be x mLVolume of H2 = (50 – x) mL

223 H23N

21NH +⎯→⎯

Since 40 mL of O2 is added and sparked it must have reacted with H2 to form liquid water. Moreover since6 mL contraction occurs with alkaline pyrogallol, 34 mL is the volume of O2 is used up.Total volume of H2 is 68 (Q 2H2 + O2 → 2H2O)

( ) 68x23x50 =+−

x = 36% NH3 = 72

C18. 1. Answer (2)PV = K(constant)

2. Answer (4)PV = nRTn is not directly proportional to T∴ Its graph will not be straight line.

3. Answer (2)

M(average) = 20yx

y28x16=

++

16x + 28y = 20x + 20y 4x = 8y

Maverage = yxx28y16

++

24372

y3)y2(28y16

==+

=

C19. 1. Answer (4)At point A near low pressure region volume is very high thus

(V – b) ~ V

RTVVap 2 =⎟

⎠⎞

⎜⎝⎛ +



RTVaPV =+

RTV

a1ZRTPV

−==

2. Answer (3)

At point B at high pressure volume is low

⎟⎠⎞

⎜⎝⎛ + 2V

aP ~ p

P(V – b) = RT

RTPb1

RTPVZ +==

3. Answer (1)

RT)bV(VaP 2 =−⎟

⎠⎞

⎜⎝⎛ +

RTPbVab

VaPV 2 =−−+

PVVaPbRT −+=

Va

VRTbRT −+=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+=

RTab

V11RT

Comparing to equation of state

⎟⎠⎞

⎜⎝⎛ +++= ......

VB1RPV

⎟⎠⎞

⎜⎝⎛ −=

RTabB

19(a). Answer (1, 3, 4) (IIT-JEE 2008)

Because V is very large, so in Van der Waal’s equation ⎟⎠⎞

⎜⎝⎛ + 2V

aP (V – b) = RT, 2Va

and b are

neglected and equation becomes PV = RT. Coefficients depends on the identity of the gas but areindependent of the temperature. Real gas exert lower pressure than the same gas behaving ideallydue to intermolecular force of attraction.

C20. 1. Answer (3)

ΔH = ΔU + Δ(PV)

= ΔU + V(P2 – P1), therefore volume is constant

–56 = ΔU + 1(40 – 70) × 0.1 1L atm = 0.1 kJ

ΔU = –56 + 3

= – 53 kJ/mol



2. Answer (2)Amount of heat required to lower the temperature of 500 g of water from 20º to 0ºC

= 4.2 × 500 × 20 = 42000 J

No. of moles of ice needed to melt to absorb this heat = 7106

420003 =

×Since each ice cube contains one mole water so at least 7 ice cube are required.

3. Answer (3)Enthalpy change ΔH = nCpΔT, since process is isothermal ΔT = 0

= 0C21. 1. Answer (4)

For spontaneous process ΔG should be negative.∴ ΔH < 0 and ΔS > 0

2. Answer (2)Entropy change in isothermal process.

ΔS1

2

VVlognR303.2=

110log314.82303.2 ××=

= 38.29 JK–1 mol–1

3. Answer (2)

ΔS = bTHΔ

K40075100030Tb =

×=

C21(a). Answer (4) (IIT-JEE 2008)ΔG = ΔG° + RT ln Qat equilibriumΔG = 0Q = Keq

∴ ΔG° = –RT ln Keq

C22. 1. Answer (4)Work done in AB process = –PΔV

W1 = 3 (30 – 10) = –60 atm×litreWork done in B → C process = 0Work done in C → A process

W2 = –2.303 nRT log 1

2

VV

= 3010log)301(303.2 ×−

= +2.303 × 30 × 0.5= 34.54 atm × litre

Net work done in the process= –60 + 34.54 = –25.45 atm × litre



2. Answer (1)

At point A → PV = nRT

K375108.0

103T =×

×=

Temperature at point C = K37508.01

30nRpVT =

×==

3. Answer (2)

Since process is cyclic therefore ΔH and ΔU will be zero because in cyclic process state functions willbe 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)

∴ )COOHCH(pK)HCN(pK 3aa >

2. Answer (3)

Na+(aq) + OH– (aq) → NaOH (aq)

Heat of ionisation of strong base = 0

0)OH(H)Na(H)NaOH(HH fff =Δ−Δ−Δ=Δ −+

⇒ 0)8.228()Na(H7.470 f =−−Δ−− +

kJ9.241)Na(Hf −=Δ +

3. Answer (2)

ΔQ = nSΔT

Molar heat capacity = Tn

QΔ

Δ

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

∴ Molar heat capacity = ∞

C24. 1. Answer (1)

For acidic buffer

pH = pKa + log ]acid[]salt[

= 4.74 + log 1.0500

1000592

⎥⎦⎤

⎢⎣⎡ ×

= 4.74 + log 5940

= 4.74 – 0.1739

= 4.57



2. Answer (3)


[salt] = [acid]

pH = pKa

= 4.74.

3. Answer (3)

pH = 4.74 + log ]acid[]salt[

6.74 = 4.74 + log ]acid[]salt[

]acid[]salt[

= 1100

% of acid in the mixture = 1011 × 100

= 1 %

∴ % dissociation of acid = 99%

C25. 1. Answer (3)

Keq = r

fKK

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)

Since reaction is endothermic on increasing the temperature equilibrium constant increases ⎟⎠

⎞⎜⎝

⎛ =r

fc K

KK

but Kf increases more than Kr.

C26. 1. Answer (1)

CH3COOH CH3COO– + H+

1 – α α α + 0.01

Ka = 5–108.11

)01.0( ×=α−

+αα

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.



3. Answer (3)

Kw = [H+] [H–] = 10–14

Ka = 55.5510

]OH[]OH][H[ 14

2

−−+=

= 0.018 × 10–14

= 18 × 10–17.

C27. 1. Answer (4)

CH3COONa → CH3COO– + Na+

CH3COO– + H2O CH3COOH + OH– (basic).

2. Answer (3)

H2S 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)

CH COOH + NaOH3 CH COONa + H O3 2

10 4 0 0

6 0 4 4

m.moles (initial)

After reactionbuffer solution


pH – pKa = log 64

= log ⎟⎠⎞⎜

⎝⎛

32

.

C28. 1. Answer (3)

There is no effect at equilibrium at constant volume.

2. Answer (4)

NaNO3(s) NaNO2(s) + 21 O2(g)

∴ since no. of gaseous moles decrese in backward direction thererfore on increasing the presure reversereaction 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 molesdecreases in forward direction therefore on increasing the pressure forward reaction favours.



C29. 1. Answer (1)

H2(g) → 2H+ + 2e (anode)

2H+ + 2e → H2 (cathode)

Ecell = AH

CH

)P()P(

log2059.0–

2

2

= ⎟⎠⎞⎜

⎝⎛

21log

2059.0–

= 2log2059.0+

= mV90.83010.02059.0 =×

2. Answer (1)

H2 → (2H+)A + 2e

(2H+)C + 2e → H2

Ecell = [ ][ ]2

C

2A

H

Hlog2

059.00+

+

−

= – 0.059 log ⎟⎠

⎞⎜⎝

⎛

2

1CC

for non spontaneous process C1 > C2.

3. Answer (1)

Zn → Zn2+ + 2e

2H+ + 2e → H2

Ecell = [ ][ ]2

2

cellHZnlog

2059.0E

+

+° −

0.28 = – 0.78 – ( )22a

1

CK

Clog2059.0

– 36CKClogCK

Clog059.012.2

2a

1

2a

1 −=⎟⎠⎞

⎜⎝⎛⇒⎟

⎠⎞

⎜⎝⎛= .

36

2a

1 10CKC −=

36

2

1a 10C

CK ×=



C30. 1. Answer (2)

Specific conductivity is directly proportional to the concentration.

2. Answer (3)

K = ⎟⎠⎞⎜

⎝⎛× aR

1 l

1.2 = ⎟⎠⎞⎜

⎝⎛× a55

1 l

1m66552.1a

−=×=⎟⎠⎞

⎜⎝⎛ l

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 decreases8 times. Specific conductance is directly proportional to concentration. Hence, it decreases 8 times.

C31. 1. Answer (3)

Λ = 001028.01095.41000

NK1000 5–××=×

= 48.15

degree of dissociation = ⎟⎠⎞⎜

⎝⎛=⎟

⎠⎞

⎜⎝⎛

ΛΛ

∞ 5.39015.48

= 0.123.

2. Answer (3)

∞Λm = 119 + 160 = 279

degree of dissociation α = ∞ΛΛ

m

0.1 = 279Λ

Λ = 27.9.

3. Answer (1)

Λm = MK1000 ×

K = 10001.09.27 ×

= 2.79 × 10–3.



C32. 1. Answer (4)

Q Anion is bigger Q it will constitute lattice

294.021.265.0

rr

==−+

(i.e. range of coordination number = 4)

2. Answer (1)

414.0rr

=−+

(for maximum packing efficiency)

∴ pm23.193414.080r ==−

3. Answer (2)

oh voids in fcc is situated at all edge centres and body centre and td voids are present at each bodydiagonal.

Nearest oh and td void pair is oh void at body centre and td void at edge centre.

∴ a43distance =

ohvoid

td void

bodydiagonal

√3a

td void

√34 a

C33. 1. Answer (2)

4214

4112

212

818

CBA+××+××

⇒ AB4C6

2. Answer (1)

Tetrad axis will remove 2 atoms at face centre and 1 of body centre (which is not present in thiscompound) 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 = 58.0127

=



C34. 1. Answer (3)

Density of CsCl

Z = 1 (1 formula unit is present)

For bcc structure

pm3

)rr(2a

)rr(2a3

−+

−+

+=

+=

Volume = 3303

3cm10

)3()]rr(2[ −−+ ×

+

∴ 30A

3

3CsCl

10N)]rr(2[)3(M

−−+ ××+

×=ρ

2. Answer (1)

48.52

)8.2()2(

4 3

3bcc

fcc =×=ρρ

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

C35. 1. Answer (3)

body diagonal plane of fcc.

2. Answer (1)

Rectangular plane (fcc)

3. Answer (3)

body diagonal plane (Bcc)



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 = %2510012

)td(1)oh(2=×

+

C37. 1. Answer (1)

On increasing temperature, K increases but Ea and A remains same.

2. Answer (3)

⎥⎦

⎤⎢⎣

⎡⋅−

−=21

21a

1

2

TTTT

R303.2E

KKlog

⎥⎦⎤

⎢⎣⎡

×−

×−=

800700800700

314.8303.2E4log a

By solving,

Ea = 64 kJ

3. Answer (4)

According to Arrhenious equation,

RT303.2EAlogKlog a−=

When RT303.2Ea becomes zero then K = A.

C38. 1. Answer (3)

On β-emission, pn

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

ratio increases.

2. Answer (4)

stable

136

e

unstable

137 CN

01⎯⎯ →⎯ +



3. Answer (1)

Tritium (1T3) has highest p

n ratio i.e. 2.

38(a). Answer (2) (IIT-JEE 2008)

Z (A

t no.

)

no. of neutrons45o

Elements with higher atomic number are more stable if they have slight excess of neutron as thisincrease the attractive force and also reduces repulsion between protons.

C39. 1. Answer (3)

Rate ∝ [A]x

2 ∝ (4)x

(4)1/2 ∝ (4)x

x = 21

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)

)xa(alog

t303.2K

−=



20100log

t303.2

3.69693.0

=

t = 161 minutes

3. Answer (1)

60100log

10303.2K1 = …(1)

10100log

t303.2K2 = …(2)

K1 = K2

10100log

t303.2

60100log

10303.2

=

t = 45 minutes

C41. 1. Answer (3)

xbenzene = 51

= 0.2. xtoluene = 54

= 0.8

According to Roult’s law

PT = toluenetoluenebenzenebenzene xPxP ×+× oo

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 = 78x

, ntoluene = 92x

Total moles = 9278x170

92x

78x

×=+

xbenzene = 17092

9278x170

78x

=

×

xtoluene = 17078

170921 =−

PT = toluenetoluenebenzenebenzene xPxP ×+× oo

PT = 17078600

17092700 ×+×

PT = 378.82 + 275.29

PT = 654.11 mm



3. Answer (2)

Mole fraction of benzene in vapour phase = 225.0620140

PP

T

benzene ==

C42. 1. Answer (2)

m = 1.0500

1000603

1000500603

=×=

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 – α) 3α α after dissociation

131i α+

= (Q α = 0.6)

8.21

6.031i =×+

=

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, theratio of osmotic pressure is also 2 : 2 : 1.

3. Answer (2)

2CH3COOH (CH3COOH)21 0 before association

(1 – α)2α

after association

Here, α = degree of dimerisation

8.01

5.01i =α−

=



4.05.02.0

==α

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 isequal to gold number of lyophilic colloid.

3. Answer (3)

As2S3 is negatively charged sol.

C45. 1. Answer (3)

According to Freundlich adsorption isotherm,

Plogn1Klog

mxlog +=

Comparing the above equation with y = mx + c then we get, intercept = log K and slope = n1

.

2. Answer (2)

According to Langmuir adsorption isotherm,

aP)1(

ab

xm

+=

Comparing the above equation with y = mx + c then we get, slope = a1

, intercept = ab

.

3. Answer (1)

According to Langmuir adsorption isotherm,

bP1aP

mx

+=

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

aPmx

= .

Section - D : Assertion - Reason Type1. 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.



4. Answer (4)

In KClO3 the Cl is reduced and oxygen atom is oxidised both are different atom therefore it is not exampleof disproportionation reaction.

5. Answer (3)

Oxidation number of nitrogen in HNO2 is +3 therefore N can gain higher oxidation state +5 as well as loweroxidation state –3.

6. Answer (2)

Oxidation number of sulphur in H2SO4 is +6. It is in highest oxidation state therefore can acts as a oxidisingagent.

7. Answer (2)

1642

0

2n2322 NaI2OSNaIOSNa −

=+⎯→⎯+

8. Answer (2)

232

3n42

5n4 COFeMnOFeCKMnO ++⎯→⎯+ ++

==

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, 3d6, 4s2

Fe2+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d6 – unpaired electrons = 4

Fe3+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d5 – (Half filled d-orbital more stable) unpaired electrons = 5

11. Answer (2)

Cu(30) → 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1

Cu+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d10 – 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.



14. Answer (4)

Electrons are filled in the orbital from lower energy level to higher energy level 3d orbital has more energythan 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 itsvelocity will be more than proton.

It is clear from this expression mVh

=λ

17. Answer (4)

de Broglie wavelength mVh

=λ . It is applicable to moving particle.

18. Answer (2)

Zn(30) → 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2

Zn2+ → 1s2, 2s2, 2p6, 3s2, 3p6, 3d10

Since Zn2+ has no unpaired electrons therefore it is diamagnetic.

19. Answer (4)y

x

dx2 – y2

In 22 yxd − 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 2

2

nZ6.13−=

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 π

+=2h)1(ll

For s-subshell l = 0



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 fluorineis 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.

30. Answer (3)

P :

P:

: P

P

:

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

31. Answer (4)

NF F

F

:

Subtraction

NH H

H

:

Addition

therefore dipole moment of NH3 is more than NF3

32. Answer (2)

I :

:

:

⎡⎢⎣

⎤⎥⎦

I

I

Linear

Hybridisation of central atom is sp3d



33. Answer (3)

N

O–::: O: :

:N

O–:

::O:

:

:

32

152N

=+

= attached atoms = 2

∴ shape is angular

34. Answer (3)

Bond order of O2

O2(16) — ⎥⎥⎦

⎤

⎢⎢⎣

⎡

π

π

⎥⎥⎦

⎤

⎢⎢⎣

⎡

π

πσσσσσ

1

1

2

222222

2*

2*

2

222*21*1

z

y

z

y

p

p

p

ppxssss

B.O. = 22

610=

−Number of unpaired electrons = 2

35. Answer (3)

222 s1*s1)4(He σσ−

B.O. 02

22=

−=

)3(He2+ bond order =

21

212

=−

Since B.O. of He2 is zero therefore it does not exist.

36. Answer (2)

B.O. of −2N = 2.5 = B.O. of +

2N

But B.O. of −22N is 2. Hence it is least stable.

N2(14) – 2x2

z

2y2222 p2

p2

p2s2*s2s1*s1 σ

⎥⎥⎦

⎤

⎢⎢⎣

⎡

π

πσσσσ

37. Answer (2)

Due to non available vancant d-orbital NCl5 does not exist down the group atomic size increases thereforeN 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.



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 configurationtherefore 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 gasmolecules 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 attractnor 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. 23

RT)

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. Dueto 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 directlyproportional to the value of ‘a’

∴ Critical temperature of SO2 will be more than H2.

51. Answer (3)

⎟⎠

⎞⎜⎝

⎛2V

a represents the inter molecular force of attraction.

It can not be neglected because in real gas the molecules are closer to each other.

52. Answer (4)

nRT)nbV(V

anp 2

2=−⎟

⎟⎠

⎞⎜⎜⎝

⎛+

Unit of V = unit of nb

Unit of b = litre mol–1

‘a’ represents inter molecular force of attraction



53. Answer (4)

2

2

1

1

VV

TV

= ⇒ 400V

300V 2=

V34V2 =

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) arezero 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 veryhigh 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 vapouris condensed.

59. Answer (4)

Pressure and temperature are intensive properties whereas volume is extensive property. Extensive propertydepends 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 reactionwill be non spontaneous at low temperature. ΔH will be more than TΔS therefore ΔG will be negative andreaction 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 concentrationof CO2 equilibrium shifts in forward direction.



64. Answer (4)

CH3COOH CH3COO– + H+ on addition of CH3COONa the concentration of CH3COO– increases andequilibrium 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 = 2

pKw

66. Answer (1)

CuO + H2 Cu + H2O at 75ºC the water in form of liquid ∴ ΔH = –1

Kp = Kc(RT)–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)

43POH −+ + 42POHH 1Ka

−42POH −+ + 2

4HPOH Ka2

Ka1 > Ka2 ∴ H3PO4 is stronger acid then −42POH

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 ofreactants 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

V)x1(x

Vx1Vx

Vx

K2

c −=

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=



73. Answer (1)

CH COONa + HCl3 CH COOH + NaCl3

2M 1M 0 01M 1M 1M0

Acidic buffer solution

74. Answer (1)

NH4OH −+ + OHNH4 on addition of NH4Cl the concentration of +4NH increases and equilibrium will be

shifted in backward direction therefore [OH–] decreases as a result pOH increases and pH will decreases.

75. Answer (1)

Blood is a basic buffer solution of )HCOCOH( 332−+ .

76. Answer (2)

+4NH

H — N — H

H

H

+form of charge on nitrogen is +1.

77. Answer (3)

CrO

O

O

O–1

O

–1

–1 –1–2

+1

+1 +1

+2

+1 oxidation number of chromium is +6.

78. Answer (1)

23

3n42 COFeOFeC +⎯→⎯ +

=

232

)5n(4

)3n(42 COFeMnMnOOFeC ++⎯→⎯+ ++

=

−

=

Equivalents of FeC2O4 = equivalent −4MnO

1 × 3 = 0.6 × 5

79. Answer (2)

OH2S3SH2SO 222 +⎯→⎯+

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 reactionshould 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.



82. Answer (2)

At in finite dilution, molar conductivity of any electrolyte (strong or weak) is equal to the sum of molarconductivities 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)

For concentration cell 0Eºcell =

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

1ºcell C

Clog2

0591.0E

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)

732.0to414.0rr

=−

+ (for coordination no. = 6)

When 414.0rr

=−

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

but when 732.0rr

=−

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

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



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 = a22

Distance between 2 nearest spheres in bcc = a23

96. Answer (2)

Ferromagnetic substances are those paramagnetic substance which persists their magnetic moment (i.e. spinalignment) even in absence of magnetic field. They turn to paramagnetic substance when heated above curiestemperature.

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 Pb20682 .

105. Answer (1)

For second order, t1/2 is inversely proportional to concentration.

106. Answer (2)

The rate law is

reaction of Ratedt

]C[ddt

]B[ddt

]A[d21

=+=−=−



107. Answer (2)

During β-decay, neutron is converted into proton and the atomic number increases by one.

108. Answer (3)

Alkaline hydrolysis of ester is known as saponification and it is second order reaction.

109. Answer (3)

Rate constant of any order reaction is directly proportional to temperature.

110. Answer (3)

The effective molarity of 1 M CuSO4, 0.5 M AlCl3 and 2 M urea solution are same and elevation in B.P. is acolligative property. Boiling point is not a colligative property.

111. Answer (1)

In 1 M aqueous solution, solvent is less than 1000 g while in 1 m aqueous solution, solvent is 1000 g.

Hence, 1 M is more concentrated than 1m.

112. Answer (4)

When mercuric iodide is added in KI solution then association takes place and freezing point is raised.

HgI2 + 2KI → K2HgI4 potassium mercuric iodide

In aqueous solution, HgI2 dissociates as

HgI2 → Hg2+ + 2I–

113. Answer (3)

Molality and mole fraction is independent of temperature and molality is the number of gm moles of solutedissolved in 1 kg of solvent.

114. Answer (1)

The effective molarity of KCl is twice the sugar because KCl gives two ions. So, the vapour pressure of KClis less than sugar.

115. Answer (3)

In presence of more volatile solute, vapour pressure of solution is developed by solute and solvent both.

116. Answer (4)

In benzene, acetic acid dimerises. So, the van’t Hoff factor is less than 1.

117. Answer (4)

Micelles is formed at above CMC and at above Kraft temperature.

118. Answer (1)

The extent of adsorption of CO2 is much more higher than H2 because critical temperature and van der Waal’sforce of attraction of CO2 is much higher than H2.

119. Answer (4)

Solution of starch in water is a lyophilic sol it is reversible sol.

120. Answer (2)

For coagulation, higher is the charge on oppositely charged ions, greater will its coagulating power.



Section - E : Matrix-Match Type1. Answer - A(p, q, r), B(q, s), C(p, q), D(q)

(A) 1.7 gm NH3

moles = 1.017

7.1=

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 = 1.032

2.3=

molecules = 0.1 N0

atoms = 0.2 N0

volume = 2.24 L

(C) 2.6 gm C2H2

moles = 1.026

6.2=

molecules = 0.1 N0

atoms = 0.4 N0

volume = 2.24 L

(D) 6.4 gm SO2

moles = 1.064

4.6=

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 = 1240

608=

×

⇒ 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



∴ 8 gm oxygen reacts with = 5.146.35

4.648=

×

⇒ Equivalent weight of metal = 14.5Equivalent 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 = 5.1471

295.35=

×

⇒ Equivalent weight of metal = 14.5

(D) Equivalent weight of Mg = 122

24=

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

)dependent eTemperatur( (litre) Volume

solute of moles of .NoMolarity =

)tindependen eTemperatur( (kg) solvent of Mass

solute of moles of .NoMolality =

dependent) re(Temperatu (litre) Volume

solute of mass formula gram of .NoFormality =

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 possiblehence acts as a oxidising agent

HNO2 ⇒ Oxidation No. of Nitrogen = +3

∴ acts as a reducing as well as oxidising agent

+5

+3

–3

Oxidation

ReductionN2O – Oxidation No. of Nitrogen = +1

N3H is called hydrazoic acid

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

(A) 243

34

3

2

2COFeOCFe

+++

+++⎯→⎯

n factor of FeC2O4 = 1 + 1 × 2 = 3

(B) 242

221

2 SOCuSCu++

+−+

+⎯→⎯

n factor of Cu2S = 1 × 2 + 6 = 8

(C) 243

32

12SOFeSFe++

+−+

+⎯→⎯

n factor of FeS2 = 1 + 5 × 2 = 11

(D)22

233

53NOFe)NO(Fe

+++

+++⎯→⎯

n factor of Fe(NO3)3 = 1 + 3 × 3 = 10



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– )nicisoelectro(⎪⎭

⎪⎬

⎫

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

(B) 92U238 no. of neutrons = 238 – 92 = 146

(n – p) = 146 – 92 = 54

90U234 no. of neutrons = 234 – 90 = 144

(n – p) = 144 – 90 = 54

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

(C) 7N15 6N

14 8O16 — all have 8 no. of neutrons therefore isotones

(D) 7N14 6N

14 — since mass no. is same therefore isobars

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

No. of spectrum = 2

)1nn)(nn( 1212 +−−

(A) No. of spectrum = 62)125)(25( =+−− belongs to visible region

(B) No. of spectrum = 62)136)(36( =+−− belongs to infrared region

(C) No. of spectrum = 32)124)(24( =+−− belongs to visible region

(D) No. of spectrum = 102)148)(48( =+−− belongs to infrared region

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 ∴ Diamagnetic

Cr3+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d3

No. of unpaired electrons = 3 ∴ Paramagnetic

Co2+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d7


Cu2+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d 9


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

(A) No. of spectrum = 2

)1nn)(nn( 1212 +−−

= region Infrared 62

)136)(36(=

+−−



(B) No. of spectrum = region Infrared –102)137)(37( =+−−

(C) No. of spectrum = region – Visible62)125)(25( =+−−

(D) No. of spectrum = region – Visible102)126)(26( =+−−

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 = π

+2h)1l(l

for p-orbital l = 1

∴ Orbital angular momentum = π2h2

(B) Angular momentum π

=2nhmvr where n is principal quantum number

(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) Fr is radioactive element

(D) As is metalloid.



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) Metallic radius is bigger than covalent radius

(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 Subtractive, μR = 0

C = CH

CH3

F

H Additive, μR ≠ 0

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

B FF

F Substractive, μR = 0.

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)

O2(16) — ⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

π

π

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

π

πσσσσσ

1z

1y

2z

2

2y

22x

2222

p2*

p2*

p

pps2*s2s1*s1

Bond order of O2 = 22

6102

NN ab =−

=−

2 unpaired electrons ∴ Paramagnetic

Bond order of −2O = 5.1

2710

=−

1 unpaired electrons ∴ Paramagnetic

Bond order of −22O = 1

2810

=−

zero unpaired electron ∴ Diamagnetic

Bond order of +2O = 5.2

2510

=−

one unpaired electron ∴ Paramagnetic



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

(A) +2NO 2

215

2N

=−

= ∴ hybridisation sp

(B) −3NO 3

215

2N

=+

= ∴ hybridisation sp2

(C) +4NH 4

2145

2N

=−+

= ∴ hybridisation sp3

(D) NH3 42

352N

=+

= ∴ 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 62

662N

=+

= sp3d2

(B) XeF4 62

482N

=+

= sp3d2 two lone pairs on central atoms

(C) BrF5 62

572N

=+

= sp3d2 one lone pair on central atom

(D) ClF3 52

372N

=+

= 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

2 Na+ HO B BO

O

B O

OBOH

OH

OH

O · 8H2O

∴ Therefore hybridisation of boron is sp3 and sp2.

(B) NH4NO3 can be written as +4NH + −

3NO

∴ hybridisation of nitrogen in +4NH is sp3 and hybridisation of nitrogen in −

3NO is sp2

Both ionic and covalent bond are exist.



(C) 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+ + −4MnO

Hybridisation of Mn in −4MnO 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 sp3

(B) H3 BO3 + H2O ⎯⎯→ [B(OH)4]– + H+

sp2 sp3

(C) Cl F3 + Sb F5 ⎯⎯→ [Cl F2]+ [SbF6]

–

sp3d sp3

(D) P Cl5 + 4H2O ⎯⎯→ H3PO4 + 5HCl

sp3d sp3

24. 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

RbaTB =

RbaTi

2=

RbaTC 27

8=

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

RTab −

as b depends on molecular size and a depends on polarizability of molecule hence second virial coefficientdepends on these two factors.

(C) a depends on polarizability of molecule.

(D) b depends on molecular size.



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

Rate of diffusion = timediffused gas of Volume

Partial pressure = (mole fraction) × total pressure

Force is rate change of momentum t

mumvtP −

=Δ

Kinetic energy of ideal gas depends only on temperature i.e. ⎟⎠⎞

⎜⎝⎛ KT

23

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

(A) factorility CompressibRTPVZ ==

for ideal gas Z = 1

(B) When the pressure is low volume is high

(V – b) –~ V

RT)V(VaP 2 =⎟

⎠⎞

⎜⎝⎛ +

RTVaPV =+

VaRTPV −=

⎟⎠⎞

⎜⎝⎛ −==

RTVa1

RTPVZ

(C) At high pressure ⎟⎠⎞

⎜⎝⎛ + 2V

aP –~ P

RTPb1RT

PVPbRTPVRT)bV(P

+=

+==−

(D) Critical temperature Rb27a8TC =

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

Vrms = MRT3

VAV = MRT8π

VMP = MRT2

PV = 2mnc31



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 thevander Waal’s equation

(B) Pressure is not too low but volume is high

∴ (V – b) –~ V

VaRTPV

RTVaPV

RTVVaP 2

−=

=+

=⎟⎠

⎞⎜⎝

⎛ +

(C) Force of attraction is negligible 2va

–~ 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 ∝ Mass Molecular

1

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 CO2exists as a liquid

(B) Critical temperature Rb2789TC =

(C)C

CCRT

VP83 =

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 isused 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.



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

Ag2CrO4 2Ag+ + −24CrO

2s mol/L s mol/L

Ksp = ]CrO[]Ag[ 24

2 −+ = (2s)2 · s

4s3 = Ksp

s = 31

sp

4K

⎟⎟⎠

⎞⎜⎜⎝

⎛

AgCNS(s) Ag+ (aq) + CNS– (aq) s mol/L s mol/L

Ksp = [Ag+][CNS–] = (s)(s)

= (s)2

s = spK

Ca3(PO4)(s) 3Ca2+ (aq) + −34PO2 (aq)

3s mol/L 2s mol/L

Ksp = [Ca2+]3 [ −34PO ]2

= (3s)3 (2s)2

= 108 s5

s = 51

sp

108K

⎟⎟⎠

⎞⎜⎜⎝

⎛

Hg2Cl2 −2

2Hg + 2Cl–

2s mol/L 2s mol/L

Ksp = [ −22Hg ] [Cl–]2

= (s) (2s)2

= 4 s3

s = 31

sp

4K

⎟⎟⎠

⎞⎜⎜⎝

⎛

33(a). Answer (4) (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



(M3X) 27S4 = 2.7 × 10–15

S = 1 × 10–4

∴ Order is

MX > M3X > MX2

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

(A) Variation of vapour pressure with temperature

RTHv

AePΔ

−=

⎥⎦

⎤⎢⎣

⎡−

Δ=

21

v

1

2

T1

T1

RH

PPln

(B) Kirchhoff’s equation

pCT

)G(Δ=

∂Δ∂

(C) Gibb’s Helmholtz equation

⎥⎦⎤

⎢⎣⎡

∂Δ∂

+Δ=ΔT

)G(THG

(D) Vant Hoff isochore

2

0p

RTH

dTKlnd Δ

=

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 T

QS vRe=Δ

= T

VVlnnRT

1

2

= 1

2lnVVnR



(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) NO (g) + O3 (g) NO2(g) + O2(g) ΔH = –200 kJ

No. of moles in both sides is equal therefore no effect of pressure on decreasing the temperatureequilibrium shift in forward direction.

(B) 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2O(g) ΔH = –905.6 kJ

Δn = 1

So the reaction forwarded in forward direction by decreasing temperature, decreasing pressure andaddition of inert gas at constant pressure

(C) N2O4 (g) 2NO2(g) ΔH = 57.2 kJ

Δn = 1

Reaction is endothermic therefore by increasing temperature, decreasing the pressure and addition of inertgas at constant pressure equilibrium shift in forward direction

(D) N2 (g) + O2 (g) 2NO(g) ΔH = +180.5 kJ

Reaction is endothermic reaction shift in forward direction by increasing temperature. No effect of pressurebecause 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)

2211 CkCk]H[ +=+

(B) Mixture of strong acid and weak acid

2Ck4CC

]H[ 1a222 ++

=+

(C) Equivalent mixture of strong acid + weak base

b

w

kCk]H[ ⋅

=+

(D) Equivalent mixture of strong base + weak acid

Ckk]H[ aw ⋅

=+

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 = ]acid[]Salt[logpka +

= 3logp7525logpk ka

a −=+



(B) CH3COOH + NaOH CH3COONa + H2OInitial m. mol 100 50 0 0After reaction 50 0 50 50

pH = ]acid[]Salt[logpka +

= 5050logpka +

= pka

(C) CH3COOH + NaOH CH3COONa + H2O



pH = 3logpk2575logpk aa +=+

(D) CH3COOH + NaOH CH3COONa + H2O



pH of salt of weak acid with strong base is

pH = Clog21pk

21pk

21

aw ++

Conc. of salt = M21

200100

=

∴ pH = 2log21pk

21pk

21

aw −+

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)

Δn = 0

Kp = Kc (RT)0

= Kc



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 atconstant pressure favours the forward direction because Δng > 0.

(B) On addition of amount of reactant favours the forward direction and low temperature favours the forwarddirection because reaction is exothermic.

Since (Δng = –2) therefore high pressure favours forward direction and addition of inert gas at constantpressure favours backward reaction.

(C) Addition of amount of reactants favours forward direction there is no effect of pressure and addition of inertgases of constant pressure because Δng = 0.

(D) Addition of reactant favours forward direction and Δng = 1 therefore addition of inert gas at constantpressure 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 (y + x)

Since K1 >>> K2 ∴ x >>> y

[OH–] = M101.010C·K 471

−− =×=

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)

(A)Cr

O

O

O

O

–1 –1

–1 –1

O

+1+1

+1+1+2

–2

+1

Oxidation no. of Cr = +6 Peroxy linkage

(B) K2Cr2O7 ⇒ 2(+1) + 2x + 7 (–2) = 0

x = + 6

n factor of K2Cr2O7 = 6

Equivalent mass = 496

2946M

==

Used in chromyl chloride test



(C) 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

–2S

Reduction (Oxidising agent)

(B) K2Cr2O7 → 2(1) + 2 x + 7 (–2) = 0

x = +6

+6

0

Reduction (Oxidising agent)

(C) H2SO5O == S – O – O – H

O

O – H(two oxygen in form of peroxide)

Oxidation no. of sulphur = +6

(D) CrO5 CrO

O

O

O

–1 –1

–1 –1

O

+1+1

+1+1+2

–2

Oxidation no. of chromium = +6

Four oxygen in form of peroxide


(A) Conductance = siemenR1

cetanresis1

=⎟⎠⎞

⎜⎝⎛=

(B) Resistivity = ⎟⎠⎞

⎜⎝⎛

laR = ohm × m

(C) Conductivity = ⎟⎠⎞

⎜⎝⎛

al

R1

= siemen m–1

(D) Cell constant ⎟⎠

⎞⎜⎝

⎛al

= m–1



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

(A) Conductance = ⎟⎠⎞

⎜⎝⎛R1

= siemen

V = iR

R = ⎟⎠⎞

⎜⎝⎛

iV

(B) Resistivity = ⎟⎠⎞

⎜⎝⎛

laR = ohm × m

(C) Cell constant ⎟⎠⎞

⎜⎝⎛

al

= m–1

(D) Resistance = ampvolt

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

(A) H2O + KCl ⎯⎯⎯⎯⎯ →⎯ iselectrolys anode

2cathode

2 Cl)g(H ↑+↑

(K+ + OH– solution)

(B) AgNO3 + H2O ⎯⎯⎯⎯⎯ →⎯ iselectrolys anode

2cathode

O)s(Ag +

(H+ + NO3– solution)

(C) CuSO4 + H2O ⎯⎯⎯⎯⎯ →⎯ iselectrolys

anode2

cathodeO)s(Cu +

(H+ + SO4– solution)

(D) H2SO4 ⎯⎯⎯⎯⎯ →⎯ iselectrolys anode

2cathode

2 OH ↑+↑

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.



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)

Rhombohedral a = b = c α = β = γ ≠ 90°

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.


(A) 34112

818

ABBA =××

(B) A4 B4 + 4 = AB2

(C) BABA 22121

818

⇒×+×

(D) 2211

212

414

814

CBACBA ⇒+×××

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

For Rock salt distance between nearest ions = 2a

For Fluorite distance between nearest ions = a43

For CsCl distance between nearest ions = a23

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

(A) Rate = K[H2O2] first order reaction

(B)][NHK1

][NHKRate

32

31

+=

High concentration of NH3, 1 can be neglected w.r.t. [NH3]

KKK

][NHK][NHK

2

1

32

31 ==

⇒ Zero order



(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) 22CCl

52 ONO4ON2 4 +⎯⎯ →⎯

tVVVlog

t303.2K

−=

∞

∞

Vt = volume of O2 after time t

V∞ = volume of O2 after ∞ time

(B)fructosel

6126ecosglud6126

H2

sucrosed112212 OHGCOHCOHOHC +⎯⎯ →⎯+

+

−

After the reaction is complete the equimolar mixture of glucose and fructose obtained is leavorotatory.

∞

∞

−−

=rrrr

logt303.2K

t

0

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

(C) CH3COOC2H5 + H2O ⎯⎯ →⎯+H

CH3COOH + C2H5OH

t

0

VVVV

logt303.2K

−−

=∞

∞

Where V0, Vt and V∞ are volume of NaOH used at time 0, t and ∞ respectively.

(D) 2H2O2 → 2H2O + O2

tVVVlog

t303.2K

−=

∞

∞

Vt, V∞ are volume of O2

obtained at time t and ∞ respectively

t

0

VV

logt

303.2K =

V0, Vt are volume of KMnO4 used at time

0 and t respectively

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

(A) For zero order reaction K = tx

(B) 2O3 3O2 is first order reaction

(C) Hydrolysis of ester in basic medium is 2nd order reaction



(D) K = 22 )xa(a)xa2(x

t21

−

− for third order reaction

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

α-emission increase pn

ratio

β-emission decrease pn

ratio

0n1 → 1p

1 + –1e0 (β-particle)

Positron emission increases pn

ratio

1p1 → 0n1 +

)positron(

01e

Electron capture increases pn

ratio

1p1 + –1e

0 → 0n1

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) 82Pb207 → 4

207 = 4n + 3 – actinium series

(B) 82Pb208 → 4

208 = 4n – thorium series

(C) 83Bi209 → 4

209 = 4n + 1 – neptunium series

(D) 82Pb206 → 4

206 = 4n + 2 – uranium series

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

(A) K4[Fe(CN)6] 4K+ + [Fe(CN)6]4–


(1 – α) 4α α after dissociation

i = 141 α+

(Q α = 0.4)

i = 6.21

4.041=

×+

(B) NaCl Na+ + Cl–


(1 – α) α α after dissociation



11i α+

= (Q α = 0.9)

9.11

9.01i =+

=

(C) AlCl3 Al3+ + 3Cl–


(1 – α) α 3α after dissociation

131i α+

= (Q α = 0.6)

8.21

6.031i =×+

=

(D) CaF2 Ca2+ + 2F–


(1 – α) α 2α after dissociation

121i α+

= (Q α = 0.8)

6.21

8.021i =×+

=

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 ebullioscopicconstant.

(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 ascryoscopic 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 bytrue 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.




(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 Type1. The reaction are

6Zn + 2KMnO4 + 9H2SO4 → 6ZnSO4 + K2SO4 + 2MnSO4 + H2 + 8H2O

OHO21H 222 →+

Let x mole of H2 is evolved and 2x

mole of O2 is required.

Total moles of H2 and O2 = 2x3

2xx =+

Molar contraction after spark = 2x3

15002x322400 =×

22410x =

Equivalent of Zn = equivalent of KMnO4 + equivalent of H2SO4

= 2Hofequivalent1000

20+

= 22420

501

224102

100020

+=×

+

Amount of Zn = gm55.322420

501

265

=⎥⎦⎤

⎢⎣⎡ +

So mass percentage of Zn in the sample = %5.351001055.3

=× .



2. 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 Fe2+ ions.

millimoles of Fe3+ ions from FeC2O4 = 3x75

(n factor of FeC2O4 = 3)

millimoles of Fe3+ ions from FeSO4 = 1y75

(n factor of FeSO4 = 1)

final millimoles of Fe2+ = total millimoles of Fe3+ ions = y753

x75+

milliequivalent of Fe2+ ions = milliequivalent of KMnO4

⇒ 502.036y753

x75××=+

⇒ 048.0y3x

=+ …(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

= 310304101 −××× [Q I5+ → I+]

= 0.012

Equivalents of KI in 20 ml = 0.012

moles of KI in 20 ml = 2012.0

[Q I– → I+]

moles of KI in 50 ml = 015.02050

2012.0

=× .

equivalents of KIO3 reacting with excess of KI = 50 × 10–3 × 101

× 4 = 0.02

equivalents of KI in excess = 0.02

moles of KI in excess = 01.0202.0

=

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%.



4. 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

⎟⎠⎞

⎜⎝⎛ ×+⎟

⎠⎞

⎜⎝⎛ ×=×× − 1

176y1

98x15.01066 3

…(1)

At the phenolphthalein end point

Equivalents of NaOH = Equivalents of H3PO4 (n = 2) + equivalents of HIO3

⎟⎠⎞

⎜⎝⎛ ×+⎟

⎠⎞

⎜⎝⎛ ×=×× − 1

176y2

98x15.01085 3

…(2)

Substracting equation (i) from (ii)

⎟⎠

⎞⎜⎝

⎛=××−×× −−

98x)15.0106615.01085( 33

98x15.01019 3 =×× −

x = 0.2793 gm

∴ 15.01047176

y 3 ××= −

y = 1.2408 g

% of H3PO4 in the sample = %793.2100102793.0

=×

% of HIO3 in the sample = %408.12100102408.1

=×

5. ++++

=

+−+

⎯⎯⎯⎯⎯ →⎯+⎯⎯⎯ →⎯+⎯→⎯++ 3

vol" 10 of mL 40"

OH72HMn2

4

)1n(

31

2

2FeMnFeSOFeSFe 22

2

Normality of H2O2 solution 785.16.5

10==

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 = %7.451001057.4

=×



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

= 1712

21810962.1atomJ

1)3(1018.2 −−

−

×=××

= 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) = mole/J9

1082.11 6×

and second ionisation potential (IP2) = mole/J9

1082.114 6××

Therefore the energy of the reaction Li(g) → Li3+(g) + 3e–

= IP1 + IP2 + IP3

= 666

1082.119

1082.1149

1082.11×+

××+

×

= 1.84 × 107 J/mole.

7.⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

λ 22

21

2H n

1n1zR1

for n1 = 1, n2 = ?

⎥⎥⎦

⎤

⎢⎢⎣

⎡−××=

× − 22

27

9 n1

11410097.1

104.301

n2 = 2

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

⎥⎥⎦

⎤

⎢⎢⎣

⎡−××=

× − 22

27

9 n1

21410097.1

105.1081

∴ n2 = 5.

8. According to de Broglie

matter wavelength ph

mvh

==λ

We have,

Kinetic energy E = 2mv21

= )m2(h

m2p

2

22

λ=

∴mE2h

=λ

The rate of change of de Broglie wavelength λ with the kinetic energy E



dE)E(d·

m2h

dEd 2

1−

=λ

= 2

3

hm

EmE2h

21 λ

−=×

×−

16234

31310

Å1atmJ1007.2

)10626.6(101.9)101(

dEd −

−

−−

=λ

×−=×

×××−=⎟

⎠

⎞⎜⎝

⎛ λ.

9. 98R1

H=λ

λ = 105.14 nm

Ethreshold = 9

834

1023010310626.6hc

−

−

×

×××=

λ

Eincident = 9

834

1014.10510310626.6

−

−

×

×××

K.E. = ⎥⎦⎤

⎢⎣⎡ −

×××−

−

2301

14.1051

1010310626.6

9

834

= 1.026 × 10–18 J.

10. Energy of photon = 9

834

1060010310626.6hc

−

−

×

×××=

λ = 3.313 × 10–19 J

E (required for melting) = kJ67.166618500

=×

Number of photons = 19

3

10313.31067.166

−×

× = 5.03 × 1023

Average energy required/molecule = J1096.910023.6

6000N6000 21

23AV

−×=×

=

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

XH ~ XF = 0.208 [ ]21FFHHFH ]EEE −−− ×−

XH ~ XF = 0.208 21

21

)6.362.104(6.134⎥⎥⎦

⎤

⎢⎢⎣

⎡×−

XH ~ XF = 1.78 and XH < XF

XF = 1.78 + XH = 1.78 + 2.1 = 3.88.



12. Moles of Mg = 241

These mole of Mg will be converted to Mg+ and Mg2+ assume a mole of Mg+ are formed then

502190a241740a =×⎟

⎠⎞

⎜⎝⎛ −+×

a = 0.02845

% of Mg+ = 28.68100

241

02845.0=×

% 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–.

15. (i) O Cl

FO

FF

Trigonal bipyramidalwith hybridisation of Clsp d3

(ii) Xe

OF

Square pyramidalwith hybridisation of Xesp d3 2

F

F

F

(iii) I

O

Square pyramidalwith hybridisation of Isp d3 2

Cl

Cl

Cl

Cl

⎡⎢⎣

⎤⎥⎦

–

(iv) IClCl

⎡ ⎤⎢ ⎥⎣ ⎦

+

V-shapedwith hybridisation of I sp3

16. (i) NO2 is paramagnetic.(ii) 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 bondsof the three.

17. The molecule BrF5 has bromine atom in centre surrounded by six electron pair, five of which are used to formbonds to fluorine atoms. Figure shows the structure of BrF5 molecule. It is better to imagine six electrons pairaround the bromine atom directed towards the corner of octahedron with five of these corners occupied byfluorine atoms. As we know that the non-bonded pair of electrons occupied more space than bonding electronpairs so geometry of BrF5 depart from that of a regular octahedron.

Br long

shortF

F

90°85°

FF

F



18. IV = II < III < I.

19. Effective molar mass of the gas mixture mol/g43.131

273082.06.0p

dRT=

××==

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

NxHy ⇒ )g(H2y)g(N

2x

22 +

initial a 0 0

02

ax2

ay

Thus a22

ay2ax

=+

x + y = 4 …(i)

when O2 is added

)l(OH)g(O21)g(H 222 →+

Since the gaseous mixture was needed to pass through alkaline pyrogallol solution. The oxygen was left inexcess and hydrogen would have been consumed completely. After passing through pyrogallol solution volumeleft would be corresponding to nitrogen only

Thus2a

2ax

= (x = 1)

y = 4 – 1

Formula of nitrogen hydride is NH3.

21. Let the volume of each vessel be V L

moles of Cl2 = R3001V ×

moles of Cl2O = R3005.1V ×

Cl2O → Cl2 + 21

O2

After reaction

total moles of Cl2 = R600V5

R300V5.2

R300V5.1

R300V

==+

moles of O2 = R600V5.1

R3002V5.1

=×



Total moles of gases in two vessel after the reaction = R600V5.6

R600V5.1

R600V5

=+

When the two vessel are kept in water bath at different temperatures diffusion will take place till the pressureof the gases in two vessel becomes equal. Let the moles of gases left in one of vessel maintained at 27°C

be x. Thus moles of gases in another vessel maintained at 52°C would be ⎟⎟⎠

⎞⎜⎜⎝

⎛− x

R600V5.6

.

V325Rx

R600V5.6

V300Rx ×

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

××

x = RV1065.5 3−×

P = atm7.1~695.1V300R

RV1065.5 3 =

××× − .

22. For one mole of a real gas

RT)bV(VaP 2 =−⎟

⎠⎞

⎜⎝⎛ +

If volume correction is ignored

RTvVaP 2 =⎟

⎠

⎞⎜⎝

⎛ +

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 <<< V

∴ RTVVaP 2 =⎟

⎠⎞

⎜⎝⎛ +

RTVaPV =+

VaRTPV −=

RTVa1

RTPV

−=

∴ Z < 1.

∴ At high temperature motion of molecules is so fast that P >>> 2V

a



P(V – b) = RT

PV – Pb = RT

PV = RT + Pb

RTPb1

RTPV

+=

∴ Z > 1.

23. (i) 33.1M32

rr

2O

v ==

M = mol/g09.18)33.1(

322 =

(ii) Molar volume 33 m1025.5036.009.18V −×==

(iii) Compression factor (Z) = 500314.8

10013.11025.50RT

VP 53

××××

=−

= 1.224.

(iv) Repulsive forces dominate since the actual density is less than the density if it were ideal.

24. Work done for irreversible process

W = –P(V2 – V1) = –ΔnRT = 37320181

××⎟⎠⎞

⎜⎝⎛ −− = –41.44 cal/gm

Now q = 540 cal/gm

ΔE = Δq + W = 540 – 41.44 = 498.56 cal/gm

For isothermal process

ΔG = 2.303 nRT2

1

VVlog = 2.303 × 4 × 8.314 × 400 10

1log = –30635.4 J.

25. Since T is constant ΔU = 0

W = 2.303nRT1

2

PPlog = 2.303 × 3 × 8.314 × ⎟

⎠⎞

⎜⎝⎛

15log400 = 1.61 × 104 J

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 mustflow from the gas to surrounding of constant temperature maintain the temperature of the gas at 400 K whenit 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 standardstate for elemental mercury is the pure liquid ΔGf° = 0 for Hg(l) and ΔG° for the vapourisation reaction is simplyequal to ΔGf° for 1 mol of Hg(g)

log Kp = 58.5298314.8303.2

1085.31RT303.2

G 3−=

×××−

=°Δ−

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.



27. The enthalpy change for the given reaction is calculated as

ClO2(g) + 21

O(3(g) → 21

Cl2O7(g) ; ΔH1 = –121.85 kJ

21

Cl2O7(g) → ClO3 + 21

O ; ΔH2 = 143.5 kJ

21

O2(g) + 21

O(g) → 21

O3(g) ; ΔH3 = –53.35 kJ

O(g) → 21

O2(g) ; ΔH4 = –249.17 kJ

ClO2(g) + O(g) → ClO3(g) ; ΔH = –280.87 kJ

Using the enthalpy of formation of ClO2(g) we calculate the bond enthalpy of Cl = O

21

Cl2(g) + O2(g) → ClO2(g)

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 + 1

1

]acid[]salt[log

4 = pKa + 5.0]salt[log 1

4 = pKa + log[salt]1 + log2 …(i)

pH2 = 6 = pKa + log[salt]2 + log2 …(ii)

Equation (ii) – (i) gives

log[salt]2 – log[salt]1 = 2

100]salt[]salt[

1

2 =

After mixing equal volume of 2 buffers

1f ]acid[M5.02

5.05.0]Acid[ =+

=

2]salt[]salt[]Salt[ 21

f+

=

2]salt[101]Salt[ 1

f =

1

1af ]acid[2

]salt[101logpK)pH( +=



2101log)pH(

2101log

]acid[]salt[logpK)pH( 11

1af +=++=

(pH)f = 5.703.

29. V.D. = 12

4000821.03.4p2

dRT×

××= = 70.6

(CF3COOH)2(g) 2CF3COOH(g)

1 2

1 – X 2X

molesinitialmolesmequilibriu

densityvapourmequilibriudensityvapourinitial

=

1x1

6.70114 +

=

X = 0.61

Kp = atm4.21

61.139.0

161.1

61.02

PP

23

3

)COOHCF(

COOHCF =×

⎟⎠⎞

⎜⎝⎛ ×

×

=

We know,

Kp = Kc(RT)Δn, Here Δn = 1

Kc = litre/mol103.74000821.0

4.2RTK 2p −×=

×= .

30. Initially the constant pressure of 275 mm of Hg is obtained when gaseous NH3 and HI are in equilibrium withsolid NH4I

NH4I(s) NH3(g) + HI(g)

Total pressure in atm = 760275

= 0.362 atm

atm181.02362.0PP HINH3

===

(Kp)1 = 22HINH atm033.0~)181.0(Pp

3=×

But when HI starts dissociating the pressure will increase further when both the equilibria are establishedsimultaneously

NH4I(s) NH3(g) + HI(g)

x x – y

2HI(g) H2(g) + I2(g)

x – y2y

2y

(Kp)1 = x(x – y) = 0.033 …(i)



(Kp)2 = 015.0)yx(2y

2y

2 =−

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

015.0)yx(2

y=

− …(ii)

On solving equation (i) and (ii) we get x = 0.202 and y = 0.04 atm

Ptotal = 204.0

204.0)04.0202.0(202.0PPPP

223 IHHINH ++−+=+++

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

CH3COOAg CH3COO– + Ag+

(x – y) x

CH3COO– + H+ CH3COOH

x 10–3 0

x – y 10–3 y

(buffer)

Solubility of CH3COOAg in buffer x = M37.0167

8.61=

Ksp = (x – y)x

(0.37 – y)0.37 = 25 × 10–4

∴ y = 0.3632

and 3a 10)yx(

yK1

−−=

∴ Ka = 1.87 × 10–5 M.

32. Ag2S 2Ag+ + S2– ; Ksp = 1.6 × 10–49

H2S H+ + HS– ; K1 = 7

2101

]SH[]HS[]H[ −

−+

×=

HS– H+ + S2– ; K2 = 142

101]HS[

]S[]H[ −−

−+

×=

Neglecting second ionisation let x is the amount of HS– produced then

72

1011.0

x −×=



x = 1 × 10–4

K2 = 14

2101

]HS[]S[]H[ −

−

−+

×=

[S2–] = 1 × 10–14 Q ([H+) ~ [HS–])

[Ag2+]2 = 14

49

2sp

101106.1

]S[

K−

−

− ×

×=

[Ag+] = 14

49

101106.1

−

−

××

= 4 × 10–18 M.

33. Mole fraction of oxygen in air = 0.21

So partial pressure of oxygen in air

mm72021.0P2O ×= of Hg

= 1.99 × 10–1 atm

Kp = 57

2O

3O 103.1

)P(

)P(

3

2 ×=

= 572

O

31103.1

)P()1099.1(

3

×=× −

atm1046.2P 30O3

−×=

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

= mole101298082.0

101046.2 211030

−−

×=×

××

Number of ozone molecules = 1 × 10–21 × 6.023 × 1023 = 6.02 × 102.

34. (i) The given cell representation can be reduced to

Pt)atm1(H)M10(H)OH(

KCd|Cd 2

122

sp2 −+−

+⎟⎟⎠

⎞⎜⎜⎝

⎛

The cell reaction is

Cd + 2H+ → Cd2+ + H2

Ecell = 2H

2

cell ]H[

P]Cd[log

20591.0E 2

+

+

−°

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−−−=

−+ 22Hsp

]OH[]H[

PKlog

20591.0)4.0(00 2



⎟⎟⎠

⎞⎜⎜⎝

⎛ ×−= 2

w

sp

K

1Klog

20591.04.00

536.130591.0

24.0K

Klog 2

w

sp =×

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Ksp = 3.44 × 1013 × 10–28 = 3.44 × 10–15 M3.(ii) ΔG = –nFE = 0

ΔS = 1

PJK386002.0965002

TEnF −=××=⎟

⎠⎞

⎜⎝⎛

∂∂

ΔH = ΔG + TΔS = 0 + 298 × 386 = 115028 J = 115.028 kJ.

35. Total amount of charge passed = F373.096500

606025=

×××

Sodium nitrite will be formed as2H+ + NO3

– + 2e → NO2– + H2O

0.386 g of NaNO2 = 69386.0

mole NaNO2 = 69386.0

mole NO2–

∴ Equivalent of NO2– reduced from NO3

– = 269386.0

× [n-factor of NO2– = 2]

Amount of charge used to obtain 0.386 g of NaNO2 = F269386.0

×

∴ Current efficiency for NaNO2 = %95.2100373.0011.0

=×

Ammonia will be formed by the reduction of nitrate following the equation9H+ + NO3

– + 8e– → NH3 + 3H2O

0.095 g of ammonia = 17095.0

mole NH3

Equivalent of NH3 obtained = 817095.0

× [n-factor of NH3 is 8]

Charge used for this purpose = F817095.0

× = 0.045 F

Current efficiency for NH3 = %06.12100373.0045.0

=×

Hydrogen gas will evolve through reaction2H+ + 2e → H2

3.55 litre of hydrogen = 4.22

55.3

Equivalent of H2 evolved = 24.22

55.3× [n-factor for H2 = 2]

Charge used to liberate hydrogen = 0.3169 F

Current efficiency for H2 = %96.84100373.03169.0

=× .



35.(a). Answer (2) IIT-JEE 2008

Fit

EW

=

96500t101020.01

EW 3 ××

=×=−

t = 19.3 × 104 s

36. The degree of dissociation of acetic acid is given by

013.07.390

2.5c ==λλ

=α∞

CH3COOH CH3COO– + H+

initially C 0 0

At equilibrium C(1 – α) Cα Cα

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 arerequired to know the [H+] present in this compartment. The reaction occuring in the cell is

Zn(s) + 2H+ → Zn2+ + H2(g)

∴ Ecell = Zn/ZnH/H 2

2EE ++ °−° – 2

2H

]H[

]Zn[Plog

2059.0 2

+

+

0.701 = 2]H[1.01log

2059.0)76.0(

+

×−−−

[H+] = 0.0316 M

Since sufficient NaOH is to be added to cathode compartment to consume entire H+ the equivalent of HCl mustbe equal to the equivalent of NaOH. Let the weight of NaOH added be x gm

10316.0140x

×=×

x = 1.264 g.

After the addition of sufficient NaOH the solution becomes neutral and we have [H+] = 10–7 M

∴ Ecell = 27 )10(1.0log

2059.076.0

−− = 0.3765 volt.

38. H2(g) + 21

O2(g) → H2O(l); ΔG1 = –237.23 kJ

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

Cell reaction H2O(l) + 21

O2(g) + H2(g) → 2H+ (aq) + 2OH–(aq)



The given cell reaction can be obtained from equation (i) are (ii) as

H2(g) + 21

O2(g) → H2O(l); ΔG1 = –237.23 kJ

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

H2O(l) + 21

O2(g) + H2(g) → 2H+(aq) + 2OH–(aq); ΔG3 = –237.23 + (2 × 79.71) = –77.81 kJ

ΔG3 = –nFEcell = –2 ×96500 Ecell

∴ Ecell = 0.403 V

H2(g) + 21

O2(g) → H2O(l); ΔH1 = –285.85 kJ

H2O(l) → H+(aq) + OH–(aq); ΔH2 = 56.9 kJ

ΔH3 of cell reaction = –285.58 + 2(56.9) = –172.05 kJ.

133

pVK00164.0

29896500281.7705.172

nFTGH

TE −−=

××+−

=Δ−Δ

=⎟⎠

⎞⎜⎝

⎛∂∂

39. Cell reaction is Zn + Cu2+ → Zn2+ + Cu

Ecell = E°cell – ]Cu[]Zn[log

F2RT303.2

2

2

+

+×

]Cu[]Zn[log

2059.0)763.0(34.01 2

2

+

+

−−−=

3101]Cu[]Zn[

2

2=

+

+

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

∴ x1x1

]Cu[]Zn[

2

2

−+

=+

+

Hence 3101x1x1

=−+

, x = 0.9993 mole

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 = [Hg2

2+][Br–]2

[Hg22+]anode = 2

sp2

sp

)1.0(

K

]Br[

K=

−

reduction reaction of mercury half cell is

Hg22+(aq) + 2e → 2Hg(l)



sp

2

Hg/Hg K)1.0(log

2059.080.0E 2

2−=+

reduction reaction of manganese half cell

MnO4– + 8H+ + 5e → Mn2+ + 4H2O

V42.1)1.0)(1.0(

)1.0(log5059.051.1E 8Mn/H,MnO 2–

4=−=++

Hg/HgMn/H,MnOcell 22

24

EEE +++− −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

sp

2

K)1.0(log

2059.080.042.121.1

20K

)1.0(logsp

2=

32220

2

sp M10110

)1.0(K −×== .

41. (i) A = 8 × 81

+ 6 × 21

= 4

B = 12 × 41

+ 4 = 3 + 4 = 7

Formula = A4B7

(ii) oh voids occupied = 3 (by edge centre)

Fraction occupied =43

= 0.75

(iii) td void occupied = 4

Fraction occupied = 84

= 0.5

(iv) Total fraction occupied = 58.0127

=

42. Edge length of unit cell = 290 × 2

= 580 pm

V of unit cell = 195.112 × 10–24 cm³

Density =NAaMZ

3 ××

= 2324 10023.610112.1952394

××××−

= 8.135 g/cm3



43.

22a:a

43:

2a

r:r:r fccbcccubicsimple

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

∴rrA = 0.414 (for maximum)

rrA = 0.225 (for minimum)

(r is radius of molecules constituting fcc)

In fcc

4r = a2

r = a42 = pm7.152432

42

=×

∴ 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

46. log K = log A – RT303.2Ea

log K = log(1.26 × 1013) – 800987.1303.2

105.58 3

×××

log K = 13.1003 – 15.9799

log K = – 2.8796

K = 1.3194 × 10–3 sec–1

525103194.1

693.0K693.0t 32/1 =

×==

−seconds.



47. For the (i) equation

]Br][NO[]NOBr[K

2

2eq =

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 = 21

N = 21n

0 211

21N ⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

N = gm7.021

=

Disintegrated amount = 1 – 0.7 = 0.3 gm

PbPo 20682

21084 ⎯⎯ →⎯ α−

So, volume of helium = 3.0210

22400× = 32 cm2 at STP.

49. 2N2O5(g) ⎯→⎯ 4NO2(g) + O2(g)

At t = 0 a 0 0

At t = 30 min (a – x) 2x2x

At t = ∞ 0 2a2a

Number of moles ∝ pressure

At t = ∞ 5.5842aa2 =+

a = 8.2335

25.584=

×

At t = 30 minutes 5.2842x3a =+

x = 8.3323

8.2335.284=×

−

K = )xa(alog

t303.2

−

K = 2008.233log

30303.2

K = 5.206 × 10–3 minute–1.



50. For AB2,

WTwk1000m f

AB2 ×Δ××

=

203.211.51000m

2AB ×××

=

87.110m2AB =

For AB4,

203.111.51000m

4AB ×××

=

15.196m4AB =

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

51.800

1000602

100080060

10102

m

33

×=

××

=

−

0417.0m =

)1(1

COOHCH3

α− α+

α

−

0H

0COOCH3

Given, α = 0.2

2.11

2.011

1i =+

=α+

=

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



This show that KCN is completely dissociated.

−+ nCN)CN(Hg 2 −+

n2n)CN(Hg

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 completedissociation.

So, total molarity = −+

−+ ++ n2n)CN(HgnCNK

= 0.1892 + (0.1892 – 0.095 n) + 0.095

m = f

fkTΔ

0.1892 + (0.1892 – 0.095 n) + 0.095 = 86.153.0

0.4734 – 0.095 n = 0.2849

n ≈ 2

So, the complex becomes −24)CN(Hg

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 andis given by

D4Z πημ

=

Where η is called coefficient of viscosity, D is dielectric constant of the medium and μ is the velocity of thecolloidal 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 sameconditions, the more strongly adsorbable adsorbate is adsorbed to a greater extent irrespective of its amountpresent.

e.g. air besides moisture contains a number of gases such as N2, O2 etc., yet moisture is adsorbed morestrongly 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 goldsol.

Gold number of starch = 0.025 × 1000

= 25 (mg)

unit-i - physical chemistry - solution(final)

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y gm oxygen x gm fe2o3

mole131 gm

answer 4mno2

answer 1k2cr2o7

gm metal oxide

gm 1312x19131 gm of

mole hcl

moles of kclo3