Question

The freezing point of 0.2 molal $K_{2}S{O}_4$ is $-1.1^{\circ }C$ Calculate percentage degree of dissociation of $K_{2}S{O}_4$ $K_f$ for water is $1.86^{\circ }$

• A. $97.5$
• B. $90.75$
• C. $105.5$
• D. $85.75$

Answer 1

Answer: (A)

$\Delta T_f$ = freezing point of water - freezing point of solution
$=0^{\circ }C-(-1.1^{\circ }C)=1.1^{\circ }$
We know that,
$\Delta T_f=i\times K_f\times m$
$1.1=i\times 1.86\times 0.2$
$\therefore i=\frac{1.1}{1.86\times 0.2}=2.95$
But we know $i=1+(n-1)\alpha$
$2.95=1+(3-1)\alpha =1+2\alpha$
$\alpha =0.975$
Percentage degree of dissociation = 97.5