A solution containing 0.85 g of ZnCl_2 in 12 | vedclass

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(B) The molar mass of $ZnCl_2$ is $65.3 + 2 	imes 35.5 = 136.3 \ g \ mol^{-1}$.The observed molar mass $(M_{obs})$ is calculated using the depression

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$73.5 \ \%$

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(B) The molar mass of $ZnCl_2$ is $65.3 + 2 	imes 35.5 = 136.3 \ g \ mol^{-1}$.The observed molar mass $(M_{obs})$ is calculated using the depression in freezing point formula:$\Delta T_f = K_f 	imes \frac{w 	imes 1000}{M_{obs} 	imes W}$$0.23 = 1.86 	imes \frac{0.85 	imes 1000}{M_{obs} 	imes 125}$$M_{obs} = \frac{1.86 	imes 0.85 	imes 1000}{0.23 	imes 125} \approx 55.0 \ g \ mol^{-1}$.The Van't Hoff factor $(i)$ is given by:$i = \frac{M_{normal}}{M_{obs}} = \frac{136.3}{55.0} \approx 2.478$.For the dissociation $ZnCl_2 ightleftharpoons Zn^{2+} + 2Cl^-$,the Van't Hoff factor is $i = 1 + (n-1)\alpha$,where $n=3$.$i = 1 + 2\alpha$.$2.478 = 1 + 2\alpha \implies 2\alpha = 1.478 \implies \alpha = 0.739$.Thus,the degree of dissociation is approximately $73.5 \ \%$.

$A$ solution containing $0.85 \ g$ of $ZnCl_2$ in $125.0 \ g$ of water freezes at $-0.23 \ ^\circ C$. The apparent degree of dissociation of the salt is ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$,atomic mass: $Zn = 65.3$ and $Cl = 35.5$)

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