Relationship between vant Hoff factor (i) and degree of dissociation $(\alpha)$ is given by
$\alpha=\frac{1-i}{n'-1}$
where, n is the number of ions formed after dissociation
The relationship can be obtained as follows:
$\underset{\text{Initially}}{\text{For the reaction,}} \underset{1\,mole}{A} \rightleftharpoons \underset{0}{n' B}$
After dissociation $(1-\alpha)$ mole n' $\alpha$
Total number of moles present in the solution
$=(1-\alpha)+n' \alpha=1+(n' -1)\alpha$
van't Hoff factor, $i=1+\left(n'-1\right), \alpha>\,1$ if $n' \geq 2$
$\therefore \alpha=\frac{i-1}{n'-1}$