A solution of urea in water has boiling point of ${100.15}^{o} C$ . Calculate the freezing point of the same solution if $K_{f}$ and $K_{b}$ for water are 1.87 K kg ${mol}^{- 1}$ and 0.52 K kg ${mol}^{- 1}$ respectively.

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$- {0.54}^{o} C$

$- {0.44}^{o} C$

$- {0.64}^{o} C$

$- {0.34}^{o} C$

answer is A.

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Detailed Solution

${ΔT}_{b} = (100 .15 - 100) = 0 {.15}^{o} C$

We know that, ${ΔT}_{b} = molality \times K_{b}$

$\begin{array}{l} Molality = \frac{{ΔT}_{b}}{K_{b}} = \frac{0 .15}{0 .52} = 0 .2884 \\ {ΔT}_{f} = molality \times K_{f} \\ = 0 .2884 \times 1 .87 = 0 {.54}^{o} C \end{array}$

Thus, the freezing point of the solution

$= - 0 {.54}^{o} C .$

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