A 0.2 , m aqueous solution of a weak acid (HX | vedclass

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(B) For a weak acid $HX$ undergoing dissociation: $HX \rightleftharpoons H^+ + X^-$. The van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degr

Vedclass · Accepted Answer

$ - 0.45 $

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(B) For a weak acid $HX$ undergoing dissociation: $HX ightleftharpoons H^+ + X^-$. The van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation. Given $\alpha = 20 \% = 0.2$,so $i = 1 + 0.2 = 1.2$. The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Substituting the values: $\Delta T_f = 1.2 	imes 1.86 \, ^o C/m 	imes 0.2 \, m = 0.4464 \, ^o C$. The freezing point of the solution is $T_f = T_f^o - \Delta T_f = 0 \, ^o C - 0.4464 \, ^o C = - 0.4464 \, ^o C \approx - 0.45 \, ^o C$.

$A$ $0.2 \, m$ aqueous solution of a weak acid $(HX)$ is $20 \% $ ionised. The freezing point of this solution is .............. $^o C$ (Given ${K_f} = 1.86 \, ^o C/m$ for water)

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