Question

A solution containing $0.5g$ of $KCl$ dissolved in $100g$ of water and freezes at ${0.24}^{o}C$. Calculate the degree of dissociation of the salt. ($K_f$ for water $={1.86}^{o}C$. Atomic weights [$K=39,Cl=35.5$].

Answer 1

Given: Wt. of $KCl\left(w\right)=0.5g$
Wt. of water $\left(W\right)=100g$
Molecular wt ($m$) of $KCl=39+35.5=74.5$
Apply the relation of freezing point depression ($\Delta T_f$) with molecular weight of the solute:
$\Delta T_{f_{cal}}=\frac{K_f\times 1000\times w}{W\times m}=\frac{1.86\times 1000\times 0.5}{100\times 74.5}={0.124}^{o}C$
$\Delta T_{f_{obs}}=0^{o}C+{0.24}^{o}C={0.24}^{o}C$
VAn't Hoff factor $\left(i\right)=\frac{\Delta T_{f_{obs}}}{\Delta T_{f_{cal}}}=\frac{0.24}{0.124}=1.93$
The degree of dissociation ($\alpha$) of $KCl$ is:
$\alpha =\frac{i-1}{n-1}$ where $n=$ number of ions
Here $n=2$ for $K^{+}$ and ${Cl}^{-}$
$=\frac{1.93-1}{2-1}=\frac{0.93}{1}$
$\therefore \alpha =0.93$ or $93$%
Thus, the degree of dissociation of the salt is $93$% or $0.93$.