Question

A $0.2$ molal solution of KCl freezes at $-{0.68}^{o}C$. If $K_f$ for $H_{2}O$ is $1.86$, the degree of dissociation of $KCl$ is:

• A. $75\%$
• B. $83\%$
• C. $65\%$
• D. $92\%$

Answer 1

The correct option is B $83\%$

$△=i\times K_f\times m$
$0.68=i\times 1.86\times 0.2$
$i=1.83$
$\alpha =\frac{i-1}{n-1}$
$=\frac{1.83-1}{2-1}=0.83$
Ionization $=83\%$