0.5 text{ molal} aqueous solution of a weak a | vedclass

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Question

(B) The weak acid $HX$ dissociates as: $HX \rightleftharpoons H^+ + X^-$.Since $n=2$ (number of ions produced) and degree of ionisation $\alpha = 0.2$

Vedclass · Accepted Answer

$1.12$

Vedclass · Answer

(B) The weak acid $HX$ dissociates as: $HX ightleftharpoons H^+ + X^-$.Since $n=2$ (number of ions produced) and degree of ionisation $\alpha = 0.2$,the van't Hoff factor $i$ is calculated as:$i = 1 + (n-1)\alpha = 1 + (2-1) 	imes 0.2 = 1.2$.The depression in freezing point $\Delta T_f$ 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 \ K \approx 1.12 \ K$.

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

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