Question

(ii) The freezing point of a solution of $3.42g$ of cane sugar (molecular weight=342) in $100g$ water is $-{0.185}^{o}C$ . The freezing point of a solution of $12.5g$ $N{H}_{4}Cl$ molecular weight = 53.5 in $100g$ water is $-{6.2}^{o}C$ . Calculate the degree of dissociation of the salt.

Answer 1

$\Delta T_f=i{K}_{f}m$

$\Rightarrow K_f=\frac{\Delta T_f}{im}$

$=\frac{0.185}{1\times \frac{3.42}{342}\times 10}$

$\Rightarrow K_f=1.85$

Far $N{H}_{4}Cl$

$i=\frac{\Delta T_f}{K_{f}m}=\frac{6.2}{1.85\times \frac{\frac{12.5}{53.5}}{100}\times 1000}$

$\Rightarrow i=1.43$

$\Rightarrow 1+\alpha =1.43$

$\Rightarrow \alpha =0.43$