In a 0.2 molal aqueous solution of a weak ac | vedclass

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(D) The van't Hoff factor $i$ for the dissociation $HX \rightleftharpoons H^{+} + X^{-}$ is given by $i = 1 + \alpha$,where $\alpha = 0.25$.So,$i = 1

Vedclass · Accepted Answer

$-0.46$

Vedclass · Answer

(D) The van't Hoff factor $i$ for the dissociation $HX ightleftharpoons H^{+} + X^{-}$ is given by $i = 1 + \alpha$,where $\alpha = 0.25$.So,$i = 1 + 0.25 = 1.25$.The depression in freezing point is calculated as $\Delta T_f = i 	imes K_f 	imes m$.Given $K_f = 1.86 \ K \ kg \ mol^{-1}$ and $m = 0.2 \ molal$.$\Delta T_f = 1.25 	imes 1.86 	imes 0.2 = 0.465 \ K$.The freezing point of the solution is $T_f = T_f^{\circ} - \Delta T_f = 0 - 0.465 = -0.465 \ ^oC$.This value is nearest to $-0.46 \ ^oC$.Therefore,the correct option is $D$.

In a $0.2 \ molal$ aqueous solution of a weak acid $HX$,the degree of ionisation is $0.25$. The freezing point of the solution will be nearest to ......... $^oC$. $(K_f = 1.86 \ K \ kg \ mol^{-1})$

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