unit-i - physical chemistry - solution(final)

Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124(1)

Section A : Straight Objective Type1. Answer (2)

[2C + O2 2CO] 3

[3CO + Fe2O3 2Fe + 3CO2] 2

6C + 3O2 2Fe2O3 4Fe + 6CO23 moles oxygen gives 2 mole Fe2O33y gm oxygen gives 2z gm Fe2O3Q 2z gm Fe2O3 require 3y gm oxygen

x gm Fe2O3 require = z2y3x

= z2xy3

2. Answer (2)

Number of moles = weightmolecularweight

= 2562.56

Number of moles = 102

Number of molecules = 102 N0Q one molecule contain 16 lone pair electrons 102 N0 molecule will contain = 10

2 N0 16

= 0.16 N03. 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) 2O31000 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 + O2moles of I2 = 0.01

weight of I2 liberated = 0.01 254 = 2.54 g

5. Answer (3)

We know that

N1V1 = N2V2x1y1 = 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 = 3105

22400112 =

moles of IClx = x5.35127625.1

+

moles of Cl2 = 3105

x5.35127625.1

2x =

+1.625x = 10 127 103 + 10 (35.5 103 x)

x (1.625 0.355) = 1.27

x = 127.127.1 =

7. Answer (4)

2KClO3 KCl + KClO4 + O2Molecular mass of KClO3 = (39 + 35.5 + 16 3) gm

= 122.5

moles of KClO3 = 1.05.12225.12 =


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

= 20 103

equivalent mass = 50201000

10201

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 H2O2100 1 5 = milli eq. of H2O2Q In basic medium n factor of KMnO4 = 3milli equivalent of H2O2 = milli equivalents of KMnO4500 = 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 = 1.02442 =

moles of Mg reacted with nitrogen is 0.1 0.02

3 Mg + N2 Mg3N23 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 2 103

X = 1 103 (in 10 mL)

In one litre = 1 101

mass of Na2CO3 = 5.3 gm

methyl orange indicator

Y2X + = 2.5 0.2 2 103

Y = 1 103 0.5 103 = 0.5 103 (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

= 4 104

Q 5 ml contains 4 104 equivalent of oxalate ion (equivalents of KMnO4 = equivalents of oxalate ion)

200 ml contains =5

104200 4

= 16 103

weight of oxalate = 16 103 44

= 704 103

% of oxalate = 1005.110704 3

= 47%



18. Answer (2)

Molecular weight of lewsite

= 2422

24

22

1066.11025.1122

1066.11078.112

++

+

= 208.53 amu.

19. Answer (3)

SO2 + H2O2 H2SO4m. 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 103

= 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)

X = P

(X)2 4h

= 4hX



X V = m4h

m4hV

4h

=

=h

m21V

23. Answer (2)

Angular momentum (mvr) = 2nh

= =h5.1

2h3

24. Answer (3)h = h0 + eVstop

= 0 +

he Vstop

Stopping potential

0

=hetan

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

1434

1910414.2

10626.6106.1

he =

=

25. Answer (1)

= 221

111096781

= 9.1176 106 cm= 911.76

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 Js1

Energy of single photon = J103.310600

103106.6hc 199

834

==

No. of photon ejected per second = 19

19 1006.6103.320 =

44

41910

10101006.6 == NAV

28. Answer (3)Angular momentum mvr = 2

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

x = Ph

From Heisenbergs uncertainty principle

4hPx

4

1PP

hP

4hP

Minimum percentage error in measuring velocity would be

8~96.74

100100PP

VV100 ==

= .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 1025

v2 = 3125

101.91055.42

= 1 106

v = 103 m/s

331

34

10101.910626.6

mv ==

= 7.28 107 m

34. Answer (2)

Let the no. of photons required to be n

1710nhc =

834

91717

10310626.61055010

hc10n

==

= 27.6 = 28 photons



35. Answer (1)

= 2221

2

n1

n1Rz1

2Z1 Since He+ Z = 2 its wavelength is one fourth of atomic hydrogen.

36. Answer (3)

Ionisation energy of He = 13.6 Z2/n2 eV

= 13.6 22

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 = h h0

0hhh43 =

=41

0

16102.341 =

= 8 1015 Hz

40. Answer (2)

Ehc =

=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 ==

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)

= 222

21

11ZR1

= 222

31

11RZ1

89

43 =

3227=

3227= 0.32 = 0.27



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)

n3 = 2r33 = 2 9r = 6r

50. Answer (1)

Orbital angular momentum = + 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. 2.3 103 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 = 60cos2 1121

21 ++

= 21)36.0(2)36.0()36.0( 222 ++

= 336.0 = 0.36 1.732

= 0.62 D

60. Answer (4)

ClCl

= 60cos2 212121 ++

= 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

180 BeCl2120 BCl3

10928 CCl4< 10928 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 p p bonding itself and in CN there is p p bonding between C and N.

67. Answer (2)

P

P

PP60

68. Answer (2)

NH3 hybridisation is sp3, 4

235

2N =+= 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

OO1.66

O = Cl OO

O

1.75

Bond length is inversely proportional to bond order.

73. Answer (3)

Dipole moment = charge (q) distance

1.03 1018 = charge 1.275 108

Charge = 818

10275.11003.1

Percentage ionic character = 81018

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+ + O2

2

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 = 10001.0

12012

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 = 300RR3004

R2736

+

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 + 8O2for 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 2

= 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 2molesYmolesY2

SOH 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)

H = 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

H = E + nRT

E = 48.6 (5 2 103 373) = 44.87 kcal.



99. Answer (4)

At constant temperature T = 0

E = 0, H = 0 and at constant temperature, PV = K(constant) Boyles law

When temperature is constant PV is constant

H = E + (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 .

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 decompositionH > 0 further more entropy increases S > 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

=

1

1

2

1

1

2 2VV

TT

=

=

Since is more for the gas X. The temperature will also be more for it.

105. Answer (1)

For an adiabatic process

PV = constant

log P = log V + constant

Thus slope of log P versus log V graph is . The value of is maximum for helium monoatomic gas. Thuscurve C should respond to helium.



106. Answer (2)

BaCl22H2O + aq Ba2+(aq) + 2Cl(aq) + 2H2O; H = 200 kJ/mol

BaCl2 + 2H2O BaCl22H2O; H = 150 kJ/mol

On adding both equations we get

BaCl2 + aq Ba2+(aq) + 2Cl(aq); H = 50 kJ/mol.

107. Answer (4)

By definition heat of neutralization we have,

21

H2C2O4 + NaOH 21

Na2C2O4 + H2O; H = 53.35 kJ

21

H2C2O4 + OH

21

C2O42 + H2O; H = 53.35 kJ (1)

H+ + OH H2O; H = 57.3 kJ (2)

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

21

H2C2O4 21

C2O42 + H+; H = 3.95

H2C2O4 C2O42 + 2H+; H = 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 T be the change in temperature

PV = mST

607.8 = 180 4.184 T

T = 0.81 K

Tf = Ti + T = 290.8.

109. Answer (1)

N N + 21

(O = O) NN =+ = O(g)

Hf = )418607(49821946 +

+ = 170 kJ mol1

Resonance energy = observed heat of formation calculated heat of formation = 82 170 = 88 kJ/mol.

110. Answer (3)

S(g) + 6F(g) SF6(g); H = 1100 kJ mol1

S(s) S(g); H = +275 kJ/mol

21

F2(g) F(g); H = 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 be

neglected

HCl H+ + Cl

0.1 0.1 0.1

[H+] = 0.1 M

pH = log[H+] = log 0.1

= 1

113. Answer (1)

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

Kp = 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 100C the % of CO2 in air must be greater than 0.95%.

114. Answer (1)

pH = +acid

salta C

ClogpK pH = pKa + log acidsalt

CC

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 = 510

x1.0x)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 ) C2 C2 C2 = [NH4OH] = 0.15 M

NaOH Na+ + OH

C1 C1

Kb = =+=

+1

2

212

4

4 C)1(C

)CC(C]OHNH[

]OH][NH[

= 41

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)

CKfKb

D

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] = 1921

100.105.0

105 =

147

2

2101101

]SH[]S][H[ + =

[H+] = 1921

1011.0101

pH = 1.50.

121. Answer (3)

Its equilibrium constant Keq = wba

KKK

= 1410

101024.3

= 1.8 1.8 104

122. Answer (2)

Solubility of PbSO4 = M102.11044.1K44

sp ==Solubility of PbSO4 = 1.2 10

4 = 1.2 104 303 103.

= 36.36 mg litre1.

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 + 1.5 103

= 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 = 1.27 102 moles/litre

atm38.01

363082.01027.1V

nRTP2

SH2 ===



128. Answer (3)

pH = 11.27 log[H+] = 11.27[H+] = 5.37 1012

[OH] = 31215

10322.11037.5

101.7 =

Kb = 5

2321075.1

1.0)10322.1(

C)C( ==

129. Answer (3)

Since PCO < 2COP

200010500

1PP

K 6CO

COp

2 ===

Since 2.303 RTlogKp = opG

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] = %9110011010100

]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) = 3.98 104 M

M1099.12

1098.3x 44 ==

Ksp = [M2+][OH]2 = 4x3 = 4 (1.99 104)3 = 3.15 1011



133. Answer (4)

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

[Mg2+][OH]2 = 1.2 1011

[OH]2 = 1.2 1010

[OH] = 1.1 105

pOH = log 1.1 105 = 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 ++ ++anode

cathode

cell 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+ +

S1 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)

Ecell = Ecathode Eanode

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

= 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 Ame2AmThus 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 + +

106 M

Cathode 2He2H2 ++

Ecell = 0.118 = 26cathode

2

)10(]H[log

2059.0

+

On solving we get

[H+]cathode = 104 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 > C2146. Answer (2)

Let the formula of mercury ion is +nnHg then the formula of mercury nitrate would be Hgn(NO3)n. The reactionsoccuring at two electrodes are

Cathode ( ) nHgneHg Cnn ++Anode ( ) + + neHgnHg AnnNet reaction ( ) ( )AnnCnn 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 = 5.30591.0492.0699.0 =

149. Answer (1)

H2O H+ + OH ; w1 KlnRTG = o

H+ + e 21 H2 ;

oo2H/H2

EFnG +=

H2O + e

21 H2 + OH ; oo = OH,H/OH3 22EFnG

Since 0GSo0E o2H/H 2 ==+o

Thus o3o1 GG =

RTln Kw = o

OH,H/OH 22EFn

)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; o1G = 2F(0.337)Cu+ + e Cu; o2G = 1F(0.153)

Cu2+ + e Cu+; o3G = 1FEo3G = oo 21 GG

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 =+=+=

Degree of dissociation = 10.024224 =

=

.

155. Answer (2)

5.1901.0

105.191000N

K1000 5 ===

Degree of dissociation = 05.0390

5.19 ==

156. Answer (1)

oooanodeRPcathodeRPcell EEE = = 0.13 (0.34) = 0.47

2Tl + Sn4+ Sn2+ + 2Tl+

Ecell = 0.47 ]Sn[]Sn[]Tl[log

2059.0

4

22

+++

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

157. Answer (2)

)44.0(23.1Ecell =o = 1.23 + 0.44 = 1.67G = ocellnFE = 2 96500 1.67 103 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 > E2161. Answer (4)

Fe2+ + 2e Fe; o1G = 2F(0.44)

Fe3+ + 3e Fe; o2G = 3F(0.036)

Fe3+ + e Fe2+; o3G = 1FEo3G = oo 12 GG

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 beproduced.



164. Answer (2)

AgCl(s) Ag+(aq) + Cl(aq) ; G1 = 2.303 RT log Ksp

Ag+(aq) + e Ag(s) ; G2 = Ag/AgEF1 +

AgCl(s) + e Ag(s) + Cl(aq) ; G3 = Ag/AgCl/ClEF1

From the given equations

G3 = G1 + G2

Ag/AgCl/ClE = Ag/AgE + + 2.303 RT log Ksp0.15 = 0.80 + 2.303 RT log Ksp

log Ksp = 101.16059.095.0 =

Ksp = 7.92 1017.

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 = 90.173. Answer (1)

Total volume of sphere = 33 r3

16r344 =

For fcc unit cell, a = r22

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

Packing fraction = 23r216

r3

16

cubeofvolumesphereofvolume

3

3 =

=

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

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 = .pm61.136732.0100

732.0r ==+

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, = = 90; = 120.



185. Answer (1)

In simple cubic unit-cell the packing fraction = 3

3

)r2(

r34

=

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 = = 6 108

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 = P0At 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 =

N = 21n

0 211

21N

=

N = 7.0414.11

21 ==

Disintegrated amount = 1 0.7 = 0.3 g

PoPo 20682He210

8442

Volume of helium accumulated = g3.0210

22400 = 32 ml.

192. Answer (2)

2252 BAB2BA +

dt]B[d2

dt]AB[d

21

dt]BA[d 2252 +=+=

]BA[k2]BA[k21]BA[k 523522521 ==

So, the relation becomes

2k1 = k2 = 4k3193. Answer (3)

t

0NNlog

t303.2=

303.2t

NNlog

t

0 =

Comparing with y = mx + C then

Slope = 303.2+

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 = H (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)

1n )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 Roults law,

Ps = BBAA XPXPoo +

Ps = )X1(PXP ABAA + oo (Q XA + XB = 1)

Ps = ABBAA XPPXPooo +

Ps = )PP(XP ABABooo (i)

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

oBP = 210 and

oAP = 90

203. Answer (1)

Tf = 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 = 5.55181000 =

xurea = 5.561

5.5511

nnn

waterurea

urea =+=+204. Answer (2)

Freezing will start at 1.86C not 0C because Tf = 1.86 and as the value of Tf increases then the molality alsoincreases due to the freezing of water. Glucose doesnt freeze.

205. Answer (3)

Tf = Kf m (i)

Tb = Kb m (ii)

By adding (i) and (ii)

Tf + Tb = Kf m + Kb m = m(Kf + Kb)

(Q Tf + Tb = 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

Tf for NaCl = i Kf m = 2 1.86 0.1 = 0.372

Tf for Ba(NO3)2 = i Kf m = 3 1.86 0.1 = 0.558

Total depression in freezing point = Tf for NaCl + Tf for Ba(NO3)2 = 0.372 + 0.558 = 0.93

Hence, freezing point of solution = 0 0.93 = 0.93C.



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

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)

Tb = i Kb m = 3 0.52 0.5 = 0.78

Hence, B.P. of solution = 100 + 0.78 = 100.78C

Tf = i Kf m = 3 1.86 0.5 = 2.79

Hence, F.P. of solution = 0 2.79 = 2.79C

B.P. F.P. = 100.78 (2.79) = 103.57C.

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 =

Tf = i Kf m = 1.2 1.86 0.2 = 0.45

F.P. of solution = 0 0.45 = 0.45C

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 Waals 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 = %5210011652424 =+++

%C in C6H12O6 = %4010096127272 =++

%C in CH3COOH = %401001321231224 =++++

%C in C2H5NH2 = %5310021452424 =+++

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)2S2O8n = 2 n = 1equivalents of MnO2 = equivalents of (NH4)2SO41 2 = 1xx = 2

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

5 1 + 20 21 + 30 3

1 = N4(1000)

4N100025 =

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 H2SO4m. 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 103 = 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 = 844/363 = gm.

10. Answer (1, 3, 4)

OH2ClMnClHCl4MnO 20

2

2

2

14

2 +++++

n factor of HCl = 21

42 =

n factor of MnO2 = 2

equivalent mass of MnO2 = 2massmolecular

.

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 = 10033100 =

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 = 298

= 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)

+= 222

m )1m(1

11Rz1

for nth line n2 = (n +1)

+= 222

n )1n(1

11Rz1

++

++=

1)1m(1)1n(

)1n()1m(

2

2

2

2

n

m

16. Answer (1, 2, 3)

Number of scattered -article in Rutherford experiment is inversely proportional to square of kinetic energyof incident -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 =

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 42

1

2n

1

n ==

=

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 0 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

O2 ionic radius 1.40

and

Li+ ionic radius 0.68

Mg2+ ionic radius 0.60

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 BF

F

F

F sp2 sp sp sp2 planar

N SiH3SiH3

SiH3

Lone pair of nitrogen is involve in p-d 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

( 0)

N

N

OO

OO( 0)=

Br

Cl

( 0)~

C

C

HCH3

H C H2 5

( 0)41. Answer (2, 3, 4)

Statement 2 and 3 are facts

Unit of P = Unit of 22

Van

Unit of a = atm L2mol2

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+ + C3

4

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

CaCN2 Ca2+ + CN2

2

[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 HNO348. Answer (1, 2, 3, 4)

KK (2s2) (*2s2) ( 2px2 = 2py

2)

Four electrons are present in 2 molecular orbitals thats why double bond contains both bonds.

49. Answer (1, 3, 4)

(1)

Cl

P

Cl

BrBr

Cl

dipole moment 0

(2)

Cl

P

Br

BrBr

Cl

dipole moment = 0

(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. In

H3PO3, H3PO2 there is PH 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 Waals 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

57. Answer (2, 4)

For spontaneous process

G < 0, H < 0 and E < 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)

E is a state function

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

In isothermal process E = 0.

63. Answer (1, 2, 4)

In isothermal process T = constant

P1V1 = P2V2 (Boyles law at constant T)

U = 0

H1 = H264. Answer (2, 4)

Standard heat of formation of all elements in their standard states is zero

Hf(O) 0 and

Hf (diamond) 0 because these are not standard state.

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 G 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+) = 101.5 = 3.2 102 M

69. Answer (1, 3)

At equilibrium G = 0

G = 2.303 RT Log K

nF cellE = 2.303 RT log K

At 25C

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+ + CO3

2

CO32 + 2H2O H2CO3 + 2OH

(basic)

73. Answer (1, 2)

N2(g) + 3H2(g) 2NH3(g) H = 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 106 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)

=211

2T1

T1

R303.2H

KKlog

Since in the option (3)

H = 0 because (Eaf = Eab)

Therefore only this reaction is independent of temperature and (K2 = K1) on the other hand there is notH = 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.

==

314

2)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 Ecell = 0

Ecell = C

2A

2

]Pb[]Pb[log

2059.0

++

Ksp = [Pb2+][SO4

2] = 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 = Ecell ]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 H1 +2 +1 1

O2

O2

+1 +2 total charge at s = +6

Cr

O2

O

O

O

O

11

11

+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 = 2325

)10()10(log

2059.00

= positive

0E2H/H

= +

Cr Cr3+(aq) + 3e

Cu2+(aq) + 2e Cu

2Cr + 3Cu2+ 2Cr3+ + 3Cu

Ecell = Ecell 3223

)Cu()Cr(log

6059.0

++

= Ecell 32

)2.0()1.0(

6059.0

= Ecell + 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 =

Number of C atoms = 11241 + = 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 =

+ 12414 =

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

Number of C atoms = 21212

414 =

+

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 = 8124 = 3.75

Hence, formula per unit cell is Na4Cl3.75.

IInd Combination : Two face Cl ions are removed


Number of Cl ions = 2214 = 3

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


Number of Cl ions = 1211

814 + = 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 eEa/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)

303.2tk

]R[]R[log 0 =

Comparing with y = mx + C

then we get (1) and (4) t does not depend on concentration for first order reaction.

117. Answer (1, 2, 4)

1n )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 -decay, positron emission & k-electron capture, n/p ratio increases while in -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 Raoults law while ethanol and water shows positivedeviation from Raoults 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 > 105 cm in diameter.

Size of colloidal particles are 107 105 cm in diameter.

Size of true particle < 107 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 (Habers process)

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

2SO3 (Contact process).

135. Answer (1, 3)

Scattering of light cant 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 B10 = 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 100601818 +=

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)

= 4hpx (x = p)

= 4hx



= 4hvmx

xm4hv = h

4m4

h =

m1

4h =

1.910

14.3410626.6 3134

=

= 8 1012 m/s2. Answer (2)

x = vm4h

23

34

1021014.3410626.6

=

= 2.64 1030 m3. Answer (1)

= v (given)

= 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 = 22

nZ6.13

eV46.13

216.13 2

==

= 3.4 eV2. Answer (4)

1n529.0r

2

H == 0.529 1 = 0.529

42529.0r

2

Be3 =+= 0.529

3. 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 H = ionisation energy (1)Na+(g) + e Na H = 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

..

= 0 Addition Subtraction BF3 < NF3 < NH3.

2. Answer (3)

Cl

Cl

Cl = 03. Answer (2)

% ionic character = 16 (EN) + 3.5 (EN)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

OBond order = 33.13

11 =+

3. Answer (3)CO molecule

Total no. of electrons = 14.

Arrangement 1s2 *1s2 2s2 * 2s2

2z

2y

p2p2

2px2.

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)

5.104cos2 2121

21R ++=

2. Answer (3) = q d1.03 3.33 1030 = charge 1.27 1010

Charge = 2.7 1020

Percentage ionic character = %87.16100106.1107.2

19

20 =

~ 17%

3. Answer (1)

O = C = O+ = 0

C14. 1. Answer (4)

22O

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 - 1s2 * 1s2 2s2 *2s2 2px2

2z

2y

p2p2

* 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

68.0x17.3x17.2

unit-i - physical chemistry - solution(final)

Documents

y gm oxygen x gm fe2o3

mole131 gm

answer 4mno2

answer 1k2cr2o7

gm metal oxide

gm 1312x19131 gm of

mole hcl

moles of kclo3