The freezing point of a solution containing 0 | vedclass

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(C) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$.First,calculate the molality $(m)$:$m = \frac{w_2 	imes 1000

Vedclass · Accepted Answer

$73.5$

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(C) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$.First,calculate the molality $(m)$:$m = \frac{w_2 	imes 1000}{M_2 	imes w_1} = \frac{0.85 	imes 1000}{136.3 	imes 125} = 0.0499 \ mol \ kg^{-1}$.Given $\Delta T_f = 0.23 \ K$,so $0.23 = i 	imes 1.86 	imes 0.0499$.$i = \frac{0.23}{1.86 	imes 0.0499} \approx 2.478$.For $ZnCl_2 ightarrow Zn^{2+} + 2Cl^-$,the number of ions $n = 3$.The degree of dissociation $\alpha = \frac{i - 1}{n - 1} = \frac{2.478 - 1}{3 - 1} = \frac{1.478}{2} = 0.739$.Thus,$\alpha \approx 73.9\%$,which is closest to $73.5\%$.

The freezing point of a solution containing $0.85 \ g$ of $ZnCl_2$ in $125.0 \ g$ of water is $-0.23^\circ C$. The degree of dissociation of this salt is .......... $\%$. (Given: $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$,atomic mass of $Zn = 65.3$,$Cl = 35.5$)

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