2 molal solution of a weak acid HA has a free | vedclass

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(A) The freezing point depression is given by $\Delta T_f = T_f^{\circ} - T_f = 0^{\circ} C - (-3.885^{\circ} C) = 3.885 \ K$.For a weak acid $HA$ und

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) The freezing point depression is given by $\Delta T_f = T_f^{\circ} - T_f = 0^{\circ} C - (-3.885^{\circ} C) = 3.885 \ K$.For a weak acid $HA$ undergoing dissociation,$HA ightleftharpoons H^+ + A^-$,the van't Hoff factor $i = 1 + \alpha$.Using the formula $\Delta T_f = i 	imes K_f 	imes m$:$3.885 = (1 + \alpha) 	imes 1.85 	imes 2$.$3.885 = (1 + \alpha) 	imes 3.7$.$1 + \alpha = \frac{3.885}{3.7} = 1.05$.$\alpha = 1.05 - 1 = 0.05$.Expressing $\alpha$ in terms of $10^{-3}$:$\alpha = 0.05 = 50 	imes 10^{-3}$.Thus,the value is $50$.

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