A 0.5 m aqueous solution of a weak acid (HX) | vedclass

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Question

(A) For the dissociation of a weak acid: $HX \rightleftharpoons H^+ + X^-$,the number of ions produced $n = 2$.The degree of dissociation $\alpha$ is

Vedclass · Accepted Answer

$1.12$

Vedclass · Answer

(A) For the dissociation of a weak acid: $HX ightleftharpoons H^+ + X^-$,the number of ions produced $n = 2$.The degree of dissociation $\alpha$ is given as $20\% = 0.2$.The van't Hoff factor $i$ is calculated as: $i = 1 + \alpha(n - 1) = 1 + 0.2(2 - 1) = 1.2$.The depression in freezing point is given by the formula: $\Delta T_f = i 	imes K_f 	imes m$.Substituting the values: $\Delta T_f = 1.2 	imes 1.86 	imes 0.5$.$\Delta T_f = 1.116 \approx 1.12 \ K$.

$A$ $0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20\%$ ionized. If $K_f$ for the solvent is $1.86 \ K \ kg \ mol^{-1}$,the depression in freezing point of the solution is ..... $K$.

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