Solutions and Colligative PropertiesmediumMCQ SINGLE

A aqueous solution of a weak monoprotic acid () is ionized at a given temperature. The cryoscopic coSolutions and Colligative Properties Chemistry Question

Question

A $0.2\text{ molal}$ aqueous solution of a weak monoprotic acid ($HX$) is $20\%$ ionized at a given temperature. The cryoscopic constant ($K_f$) for water is $1.86\text{ K kg mol}^{-1}$. The expected freezing point of this resulting solution will be:

Answer: A

💡 Solution & Explanation

The weak acid dissociates as $HX \rightleftharpoons H^+ + X^-$, so $n=2$. Given $\alpha = 0.20$, the van't Hoff factor is $i = 1 + \alpha(n-1) = 1 + 0.20(2-1) = 1.20$. The depression in freezing point $\Delta T_f = i \times K_f \times m = 1.20 \times 1.86 \times 0.20 = 0.4464^\circ\text{C}$. Therefore, the new freezing point is $-0.4464^\circ\text{C} \approx -0.45^\circ\text{C}$.

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