In a aqueous solution of a weak monoprotic acid , the experimental degree of ionization is . Taking — Solutions and Colligative Properties Chemistry Question
Question
In a $0.2\text{ molal}$ aqueous solution of a weak monoprotic acid $HX$, the experimental degree of ionization is $\alpha = 0.20$. Taking the cryoscopic constant $K_f$ for water as $1.86\text{ K kg mol}^{-1}$, the freezing point of the solution will be:
Answer: A
💡 Solution & Explanation
The weak acid dissociates as $HX \rightleftharpoons H^+ + X^-$, so $n=2$. The van't Hoff factor is $i = 1 + \alpha(n-1) = 1 + 0.20(1) = 1.20$. The depression is $\Delta T_f = i \times K_f \times m = 1.2 \times 1.86 \times 0.2 = 0.4464^\circ\text{C}$. The new freezing point is $-0.45^\circ\text{C}$.
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