Calculate freezing point of 0.01
mol solution glucose in water kf=1.86Kkgmol
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Answer:
The expression for the depression in the freezing point and the molality is K
f
=
m
ΔT
f
.
The molality is 0.1 m.
The freezing point of 0.1 M glucose solution is 1.86
0
C.
The depression in freezing point is 0−(1.86
0
C)=1.86
0
C.
Substitute values in the above expression.
K
f
=
0.1
1.86
=18.6
When an equal volume of 0.3 M glucose solution is added, the molality of the resulting solution is given by the expression
0.1×∨+0.3∨=M
3
×2∨.
Hence, M
3
=0.2M.
The expression for the freezing point depression will be ΔT
f
=18.6×0.2=3.72
o
C.
The freezing point of the mixture will be T
f
=0−3.72≃−3.72
o
C.
Note: Molarity ≈ molality.
Option C is correct.
Explanation:
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