SolutionHard
Question
An aqueous solution of non-volatile and nonelectrolyte solute boils at 100.78°C. It should freeze at (for water, Kf and Kb are 1.86 and 0.52 K-kg/mol, respectively)
Options
A.2.79°C
B.270.21°C
C.−2.79°C
D.−270.21°C
Solution
$\frac{\Delta T_{f}}{\Delta T_{b}} = \frac{K_{f}}{K_{b}} \Rightarrow \frac{\Delta T_{f}}{0.78} = \frac{1.86}{0.52} \Rightarrow \Delta T_{f} = 2.79^{0}C$
∴ F.P. of solution = –2.79° C
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