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What freezing point depression would be predicted for 0.2 molal solution of benzoic acid in benzene if latent heat of fusion is 40.00 cal $g^{- 1}$ at 280 K (freezing point) for benzene? (assume no change in molecular state)

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${0.52}^{o}$

${0.0052}^{o}$

${0.78}^{o}$

${5.2}^{o}$

answer is D.

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Detailed Solution

T_o=Freezing point of solvent

${ΔH}_{f} =$ Latent heat of fusion

$K_{f} = \frac{{RT}_{0}^{2}}{1000 {ΔH}_{f}} = \frac{2 \times {(280)}^{2}}{1000 \times 40} = \frac{156800}{40000} = 3 .92 K {mol}^{- 1} kg$

${ΔT}_{f} = molality \times K_{f} = 0 .2 \times 3 .92 = 0 {.78}^{o}$

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