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Pure Benzene freezes at 5.4 °C. A solution of0.223 g of phenyl acetic acid $(C_{6} H_{5} {CH}_{2} COOH)$ in 4.4 g of benzene $(K_{f} = 5.12 K kg {mol}^{- 1})$ freezes at 4.47 °C. From this observation, one can conclude that

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phenyl acetic acid exists as such in benzene

phenyl acetic acid undergoes partial ionization in benzene

phenyl acetic acid undergoes complete ionization in benzene

phenyl acetic acid dimerizes in benzene.

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detailed solution

Correct option is D

$m = \frac{(- Δ T_{f})}{K_{f}} = \frac{(5.4 - 4.47)}{5.12} mol {kg}^{- 1}$ i.e. $\frac{0.223 g / M}{(4.4 / 1000) kg} = \frac{0.93}{5.12} mol {kg}^{- 1}$

Solving for M, we get $M = 279 g {mol}^{- 1}$

Number of molecules of phenyl acetic acid $= \frac{279}{136} = 2$

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Pure Benzene freezes at 5.4 °C. A solution of0.223 g of phenyl acetic acid C6H5CH2COOH in 4.4 g of benzene Kf=5.12 K kg mol-1 freezes at 4.47 °C. From this observation, one can conclude that