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When 20 g of naphthoic acid $(C_{11} H_{8} O_{2})$ is dissolved in $50 g$ of benzene $(K_{f} = 1.72 Kkg {mol}^{- 1})$ , a freezing point depression of $2 K$ is observed. The Van't Hoff factor is

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0.5

answer is A.

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Detailed Solution

$Molaloity = \frac{20 \times 1000}{172 \times 50} = 2.32$

∴ Depression in freezing point $= 2 = 2.32 \times 1.72 \times i$

$2 = i \times 1.72 \times [\frac{\frac{20}{172}}{50}] \times 1000 \Rightarrow i = 0.5$

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