Question

A $0.5$ percent solution of potassium chloride was found to freeze at $-0.24^{\circ} C$. The percentage dissociation of potassium chloride is _________(Nearest integer)
(Molal depression constant for water is $1.80 \,K \,kg \,mol ^{-1}$ and molar mass of $KCl$ is $74.6\, g\, mol ^{-1}$ )

Answer 1

$0.5 \%$ solution of $KCl$
So $m =\frac{0.5}{74.6} \times \frac{1}{0.1}$
$\Delta T _{ f }= i \times m \times K _{ f }$
$0.24= i \times \frac{0.5}{74.6} \times \frac{1.80}{0.1}$
$i =\frac{0.24 \times 74.6}{0.5 \times 1.80} \times 0.1$
$=1.989$
$1.989=1+\alpha( n -1)$
$1.989=1+\alpha$
$\alpha=.989$
$\% \alpha=98.9 \%$
Ans $99 \%$
If mass of $H _{2} O =99.5$
$ m =\frac{0.5}{74.5} \times \frac{1}{.0995}$
$ i =\frac{0.24 \times 74.6 \times .0995}{.5 \times 1.80} $
$=1.979 $
$1.979=1+\alpha( n -1) $
$1.979=1+\alpha $
$\alpha=.979$
$\% \alpha=97.9 \%$
Ans $98 \%$