When of naphthoic acid () is dissolved in of benzene (), a freezing point depression of is observed. — Solutions and Colligative Properties Chemistry Question
Question
When $20\text{ g}$ of naphthoic acid ($C_{11}H_8O_2$) is dissolved in $50\text{ g}$ of benzene ($K_f = 1.72\text{ K kg mol}^{-1}$), a freezing point depression of $2\text{ K}$ is observed. What is the experimental van't Hoff factor ($i$)?
Answer: B
💡 Solution & Explanation
Molar mass of naphthoic acid $= 132 + 8 + 32 = 172\text{ g/mol}$. Molality $m = \frac{20/172}{50/1000} = \frac{20 \times 1000}{172 \times 50} \approx 2.325\text{ m}$. $\Delta T_f = i \times K_f \times m \implies 2 = i \times 1.72 \times 2.325 \implies 2 = i \times 4.0 \implies i = 0.5$.
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