Addition of 0.643 g of a compound to 50 ml of benzene (density = 0.879 g ${mol}^{- 1}$ ) lowers the freezing point from ${50.51}^{o} C$ to ${50.03}^{o} C$ . If $K_{f}$ for benzene is 5.12, the molecular mass of the compound is

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answer is A.

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Detailed Solution

Weight of benzene $= V \times d = 50 \times 0 .879 g$

$W = 43 .95 g$

Weight of compound = 0.643 gm

$\begin{array}{l} {ΔT}_{f} = 50 .51 - 50 .03 = 0 .48 \\ ∴ {ΔT}_{f} = K_{f} \times \frac{w}{m} \times \frac{1000}{w} \\ 0 .48 = 5 .12 \times \frac{0 .643}{m} \times \frac{1000}{43 .95} \\ ∴ m = 156 .05 gr \end{array}$

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