$C_{6} H_{6}$ freezes at 5.5°C . The temperature at which a solution of 10 g of $C_{4} H_{10}$ in 200 g of $C_{6} H_{6}$ freeze is __°C. (The molal freezing point depression constant of $C_{6} H_{6}$ is 5.12°C/m)

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Detailed Solution

$\begin{array}{l} ∆ T_{f} = K_{f} \times m \\ T_{f}^{0} - T_{f} = K_{f} \times m \\ 5.5 - (T_{f}) = 5.12 \times \frac{10}{58} \times \frac{1000}{200} \\ 5.5 - T_{f} = \frac{5.12 \times 5 \times 10}{58} ≅ 1 ° C \end{array}$

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