Relationship between van’t Hoff’s factor (i) and degree of dissociation $\left(\alpha \right)$ is

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