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26. (b) : According to Gay-Lussac’s law,
P1
P
= 2
(n, V constant)
T1
T2
Given that, P 1 = 250 kPa; T 1 = 300 K,
P 2 = 1 × 10 6 Pa; T 2 = ?
3 6
250×
10 1 10
= × ⇒ T 2 = 1200 K
300 T2
Thus, cylinder will burst at 1200 K before it attains
its melting point (1800 K).
27. (d) : Ag is extracted from argentiferrous lead by
Parke’s process where Zn and Pb in molten state
are immiscible and form separate layers, zinc
being lighter forms upper layer (X). Ag is soluble
in both but more soluble in upper layer. So, all the
statements are correct.
28. (b) : Sandmeyer reaction :
C 6 H 5 N + 2Cl – CuCl
C 6 H 5 Cl + N 2
29. (a)
30. (a) :
HCl
Propanoic acid
31. (a) : For the depression in freezing point,
∆T f = 1000× Kf
× w
W×
Mexp.
−3
1000 × 512 . × 20×
10
∴ 069 . =
M exp. × 1
∴ M exp = 148.41 (M normal of phenol = 94)
M
van’t Hoff factor
nor.
() i = = 1− α +
α
Mexp.
2
Mnor.
94
= = − +
Mexp. 148.
41 1 α α ⇒ α = 0.734 or 73.4%
2
32. (a)
33. (d) : Since the amine (C 5 H 13 N) on treatment
with aq. NaNO 2 /HCl evolves N 2 gas, it must be a
1° amine. Since, the amine is optically active, the
–NH 2 group cannot be attached to a chiral centre
because it will rapidly undergo racemisation due to
nitrogen inversion. Therefore, the carbon skeleton
must contain a chiral centre. In other words, the
amine is 2-methylbutanamine. The reaction looks
like,
34. (c)
35. (a) : ∆S p = 2.303 n × C p × log T 2
T1
Entropy change for heating water from 27°C to
100°C ;
∆S p = 2.303 × 1000 4180 × 18 373
× log = 910.55 J
18 1000 300
Entropy change for heating 1 kg H 2 O to 1 kg steam
at 100°C ;
∆S = ∆Hv 5
23 × 10
= = 6166.21 J
T 373
Entropy change for heating 1 kg steam from 373 to
473 K ;
473nCp dT
T
∆S =
⋅ 473 ( 1670 + 049 . )
∫ = m
dT
T
∫
T
373
373
= 396.73 + 49 = 445.73 J, where m = mass in kg
Total entropy change = 910.55 + 6166.21 + 445.73
= 7522.50 J
36. (c) : Magnetic moment (µ) = nn ( + 2)
BM
(n = number of unpaired electrons)
Given that, µ = 1.73 BM.
∴ 1.73 = nn ( + 2)
⇒ n 2 + 2n –(1.73) 2 = 0
On solving this equation we get, n = 1
So, vanadium atom must have one unpaired electron
and thus its configuration is
23V 4+ : 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 1
37. (c) : Buna-S is an elastomer, thus has weakest
intermolecular forces. Nylon-6,6, is a example of
fibres, thus has strong intermolecular forces like
H-bonding. Polythene is a thermoplastic polymer,
thus the intermolecular forces present in polythene
are inbetween elastomer and fibres. Thus, the order
of intermolecular forces of these polymers is
Buna-S < Polythene < Nylon-6,6 i.e., B < C < A
46 CHEMISTRY TODAY | APRIL ‘17