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JEE Main Chemistry Class 11 Solutions Part 2 Questions
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© examsnet.com
Question : 19
Total: 80
The solution from the following with highest depression in freezing point/lowest freezing point is
[30-Jan-2024 Shift 2]
180
g
of acetic acid dissolved in water
180
g
of acetic acid dissolved in benzene
180
g
of benzoic acid dissolved in benzene
180
g
of glucose dissolved in water
Validate
Solution:
∆
T
f
is maximum when
i
×
m
is maximum.
1)
m
1
=
180
60
=
3
,
i
=
1
+
α
Hence
∆
T
f
=
(
1
+
α
)
⋅
k
f
=
3
×
1.86
=
5.58
∘
C
(
α
<
<
1
)
2)
m
2
=
180
60
=
3
,
i
=
0.5
,
∆
T
f
=
3
2
×
k
f
′
=
7.68
∘
C
3)
m
3
=
180
122
=
1.48
,
i
=
0.5
,
∆
T
f
=
1.48
2
×
k
f
′
=
3.8
∘
C
4)
m
4
=
180
180
=
1
,
i
=
1
,
∆
T
f
=
1
⋅
k
f
′
=
1.86
∘
C
As per NCERT,
k
f
′
(
H
2
O
)
=
1.86
k
⋅
kg
mol
−
1
k
f
′
(
Benzene
)
=
5.12
k
⋅
kg
mol
−
1
© examsnet.com
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