When of naphthoic acid (, ) is dissolved in of non-polar benzene (), a freezing point depression of — Solutions and Colligative Properties Chemistry Question
Question
When $20\text{ g}$ of naphthoic acid ($C_{11}H_8O_2$, $M_w = 172$) is dissolved in $50\text{ g}$ of non-polar benzene ($K_f = 1.72\text{ K kg mol}^{-1}$), a freezing point depression of exactly $2\text{ K}$ is observed. What is the experimental van't Hoff factor ($i$)?
Answer: B
💡 Solution & Explanation
Theoretical molality $m = \frac{20/172}{50/1000} \approx 2.325\text{ m}$. Using $\Delta T_f = i \times K_f \times m$, we get $2 = i \times 1.72 \times 2.325 \implies 2 = i \times 4.0 \implies i = 0.5$. This indicates $100\%$ dimerization in the non-polar solvent.
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