Thank you for reporting, we will resolve it shortly
Q.
The freezing point of 0.2 molal $K_{2}SO_{4}$ is $-1.1^{\circ}C$ Calculate percentage degree of dissociation of $K _{2}SO_{4}$ $K_{f}$ for water is $1.86^{\circ}$
Solutions
Solution:
$\Delta\,T_{f}$ = 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