CF SM XI Chemistry Unit-1 Section-A

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Section A : Straight Objective Type1. Answer (2)

KMnO4 + H2O2 ⎯→ Product

n = 5 n = 2

milli equivalents of KMnO4 = milli equivalents of H2O2

100 × 1 × 5 = milli eq. of H2O2

∵ 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

2. 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.

3. Answer (1)

2IClx ⎯→ I2 + 2x

Cl2

moles of Cl2 = 310522400112 −×=

moles of IClx = x5.35127625.1+

∴ moles of Cl2 = 3105x5.35127

625.12x −×=⎟

⎠⎞⎜

⎝⎛

+

Physical Chemistry UNIT 1


Success Magnet -Solutions (Part-I) Physical Chemistry

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

x (1.625 – 0.355) = 1.27

x = 127.127.1 =

4. Answer (4)

Fe2O3 + 3CO ⎯→ 2Fe + 3CO2

233C O 3CO2

+ → at NTP

Moles of 2 34 1Fe O

160 40= =

⇒ Moles of 1CO 3

40= ×

⇒ Moles of 21 3 3O2 40 80

= × =

⇒ Volume of O2 at NTP = 3 22.4 L80

×

5. Answer (2)

Number of moles = weightmolecularweight

= 2562.56

Number of moles = 10–2

Number of molecules = 10–2 N0

∵ one molecule contain 16 lone pair electrons

∴ 10–2 N0 molecule will contain = 10–2 N0 × 16

= 0.16 N0

6. Answer (2)

Moles of CaO = 5662.1 = moles of CaCl2

Mass of CaCl2 = 5662.1 × 111 = 3.21 gm

% = 1021.3 × 100 = 32.1%

7. Answer (2)

3O2 ⎯→ 2O3

O3 + 2KI + H2O ⎯→ I2 + 2KOH + O2

1344 ml O2 = 21344 mole O22400


Physical Chemistry Success Magnet Solutions (Part-I)

⇒ Moles of 32 1344O3 22400

= ×

⇒ Moles of I2 liberated = 2 13443 22400×

8. Answer (1)

x1 ml solution having normality y1

⇒ milli eq. of solution = x1y1

On dilution normality becomes y2

volume after dilution 1 1

2

x yv

y=

volume required for dilution = 1 1 1

1 12 2

x y yx x 1y y

⎛ ⎞− = −⎜ ⎟

⎝ ⎠

9. Answer (3)

12.25 g of KClO3 sample. (impure)

3 4 22KClO KCl KClO OΔ⎯⎯→ + +

Mole. of 2

1.2 3O32 80

= =

⇒ Mole. of 32 3 3KClO80 40×

= =

mass of KClO3 = 3 122.540

×

% of pure 3

3 122.5KClO 10040 12.25×

= ××

10. 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 ==× −



11. Answer (4)

Z2O3 + 3H2 ⎯→ 2Z + 3H2O(2x + 48) gm 6 gm

∵ (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

12. Answer (1)

K2Cr2O7 + 14 HCl ⎯→ 2KCl + 2 CrCl3 + 7H2O + 3Cl2∵ 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

∵ 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 + eVs

γ

Vs

Slope = eh

25. Answer (1)

⎟⎠⎞

⎜⎝⎛

∞−=

λ 221

111096781

λ = 9.1176 × 10–6 cm

= 911.76 Å

26. Answer (1)

λ = 400 Å

34 8 1

10

hc 6.62 10 J.s. 3 10 msE400 10 m

− −

−

× × ×= =

λ ×

s2K.E. E eV3

= × =

⇒s

2 EV3 e

= ×

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 = π2nh

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 (2)

3d6

If we starts filling from L.H.S in anti clockwise direction

–2 –1 0 +1 +2

n = 3, = 2, m = –2, 1s

2−

=

If we starts filling of electrons from clockwise, R.H.S. in then m = +2, s = 12

+



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)

22 21 2

1 1 1RZn n

2Z1∝

λ Since He+ Z = 2

∴ Its wavelength is one fourth of atomic hydrogen.

36. Answer (3)

Ionisation energy of He = 13.6 Z2/n2 eV

= 13.6 × 2

2

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 22

2

31

11RZ1

89

43×=

λ

λ

α

β

3227=

3227

=λβ × 0.32 Å = 0.27Å

46. Answer (4)

2

1 2

13.6 ZE (He) 13.6 4n

−= = − ×

413.6 4 13.6E

16 4− × −

= =

E4 – E1 = 13.6 × 4 –

13.64

eV

15 hcE 13.64

⎡ ⎤Δ = =⎢ ⎥ λ⎣ ⎦

⇒34 8

–19

6.62 10 3 10 413.6 15 1.6 10

−× × × ×λ =

× × ×

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(

For s-orbital = 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 (1)

Size decreases with increase in the magnitude of +ve charge or oxidation state

Ze ratio of Cl =

17 117

=

17Cl 0.94418

− = =

and 26Fe 1.0824

++ = =

53. Answer (2)

Ionisation potential 1

Size of the atom∝

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.

∵ 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 (1)

Due to metallic oxide

61. Answer (1)

Bond angle Molecules

180º BeCl2120º BCl3

109º28′ CCl4< 109º28′ PCl3

Therefore BeCl2 > BCl3 > CCl4 > PCl3.

62. Answer (1)

63. Answer (4)

Nitrogen is chemically inert due to absence of bond polarity.

64. Answer (3)

52

282N =+=

Attached atoms = 3

∴ T-shaped.

64(a). Answer (4) (IIT-JEE 2012)

Xe

F

F

O

O

⇒ See saw.

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 (3)

Larger anions have greater polarizability, due to expose of its valency shells

70. Answer (4)

4N 6 4SF , 5,2 2

lone pairs = 5 – 4 = 1

4,N 4 4

CF 4,2 2


4N 8 4

XeF , 6,2 2


71. Answer (3)

He2 as bond order is zero

72. Answer (2)

Species Bond orderCl – 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 (3)

75. Answer (2)

52

1362N =++= , attached atoms = 3

∴ T-shaped.

76. Answer (1)

Bond energy ∝ bond order

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)

Equal volume contains equal no. of moles or molecules

n1H2 = n2He = n3O2 = n4O3

⇒ Ratio of atoms → 2 : 1 : 2 : 3

82. Answer (1)

r.m.s.H2 = r.m.s. N2

1 23RT 3 RT2 28

=

⇒ 1T 3002 28=

⇒ 1300T 21.42 K14

= =

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 (1)

r.m.s. is 3RT

M

Depends on temperature and molar mass of gas

92. Answer (1)

d = 0.4 g/L

PM = dRT

dRTPM

=

∴ PCalculate = 0.4 0.082 773 0.8451

30× ×

= atm

Pobserved = 1 atm

⇒ Repulsive force



93. 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

94. 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=

95. 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

==−



96. Answer (2)

6.2nnn

nP

42

22

CHHeH

HH ×

++=

⇒ 1.6 atm

97. Answer (2)

1:2224

4121

mm

nn

rr

2

22

H

He

He

H

He

H ===

98. 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.

99. 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.

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 (1)

V

P

2

1

10 N/m2

BC

A

Work done C to A → Area under the curve

= 1 10 1 5J2× × =

ΔU = q + (–w)

or q = w (ΔU = 0)

5 = (wAB+ wBC + wCA)

5 = (10 + 0 + wCA)

⇒ wCA = –5J

106(a). Answer (1, 3) (IIT-JEE 2012)

Entropy is a state function. In this diagram initial state is X and final state is Z.

Therefore, ΔSX → Z = ΔSX → Y + ΔSY → Z

Work is a path function. Therefore WX → Y → Z = WX → Y



107. Answer (1)

Hexane = 8.6 0.1 mole86

=

Heptane = 10 0.1mole

100=

108. Answer (3)

109. Answer (3)

ΔG = ΔH – TΔS

ΔS = +ve always

110. 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.

111. 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.

112. 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.



113. 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.

114. 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(=

+.

115. 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

116. 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%.



117. 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 =

118. 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 −−

×=× .

119. 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 −×=

120. 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.



121. 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 −

122. 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

123. Answer (2)

[S2–] = 19–21

100.105.0

105 ×=× −

147

2

2101101

]SH[]S][H[ −−

−+

×××=

[H+] = 19

21

1011.0101

−

−

×××

pH = 1.50.

124. Answer (3)

Its equilibrium constant Keq =w

ba

KKK ×

= 14

10

101024.3−

−×

= 1.8 × 1.8 × 104

125. 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.



126. 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

=⎟⎠⎞⎜

⎝⎛=⎟

⎠⎞⎜

⎝⎛

−+

127. 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.

128. 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

129. Answer (1)

50.0XX ClCl2 ==

Kp = )15.0()15.0(

PX)PX(

PP 2

TCl

2TCl

Cl

2Cl

22××

=××

=

= 0.5 M



130. 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=

×××==

−

131. 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 −

132. Answer (3)

Since PCO < 2COP

200010500

1PP

K 6CO

COp

2 =×

==−

Since 2.303 RTlogKp = pGΔ−

2.303 × 8.314 × T log2000 = 20700 + 12T

T = 404.3 K

133. 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=

134. Answer (1)

% of [In–] = %91100110

10100]HIn[]In[

]In[=×

+=×

+−

−



135. Answer (1)

2 4 2N O 2 NO

Molar mass calculated = 28 + 64 = 92

Molar mass observed = 30 × 2 = 60

⇒calculated

observed

M1 2 921 M 60

− α + α= =

⇒92 1 0.53360

α = − =

⇒ % = 53.33 %

136. Answer (1)

CH3COONa CH3COO– + Na+

2H O3 3CH COOH CH COO H− +⎯⎯⎯→ +

It is a buffer solution

a[Salt]pH pK log[Acid]

= +

137. Answer (1)

CH3COOH

1N10

=

α = 1.2 %

log 1.2 = 0.079

31 1.2[H ] C 1.2 1010 100

+ −= ∝= × = ×

pH = –log[H+] = –log(1.2) + 3 = 3 – 0.079 = 2.921

138. Answer (1)

% of h = ?

2H OCN HCN OH− −+

[OH–] = Ch Kh = Ch2

KhKa = Kw

wh

a

KK

K

2 w

a

KCh

K

2w

C

Kh 2.92 10

Ka

% of h = 2.92



139. 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

140. Answer (1)

3 2 2I IO H I H O− − ++ + ⎯⎯→ +

3 2 210I 2IO 12H 6I 6H O− − ++ + ⎯⎯→ +

10 : 2 : 12

or 5 : 1 : 6

141. Answer (1)

142. Answer (3)

In HNO2, N is in +3 oxidation state

So it can oxidise to +5 or reduce to –3

143. Answer (3)

143a. Answer (5) (IIT-JEE 2011)

–5 –2

2Br 2Br 5 Br 2Br 10e

2 e+⎡ ⎤→ ⎡ ⎤× → +⎢ ⎥ ⎣ ⎦+⎢ ⎥⎣ ⎦

Br2 + 5Br2 → 10NaBr + 2NaBrO3 + 6CO2 + 6Na2CO3

3Br2 + 3Na2CO3 → 5NaBr + NaBrO3 + 3CO2

CF SM XI Chemistry Unit-1 Section-A

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Transcript of CF SM XI Chemistry Unit-1 Section-A