SolutionHard
Question
A 5% (by mass) solution of cane sugar in water has freezing point of 272.85 K. Calculate the freezing point of 5% (by mass) solution of glucose in water, if the freezing point of pure water is 273.15 K.
Options
A.0.57 K
B.272.58 K
C.272.85 K
D.273.72 K
Solution
$\Delta T_{f} = K_{f}.m$
For cane sugar solution:
$(273.15 - 272.85) = K_{f}.\frac{5/342}{95/1000}(1)$
For glucose solution:
$\left( 273.15 - T_{f} \right) = K_{f}.\frac{5/180}{95/1000}(2)$
From (1) and (2), Tf = 272.58 K
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