If $\alpha$ is the degree of dissociation of $N{a}_{2}S{O}_4$ , the van't Hoff factor used tor the calculation of molecular mass is

${\mathrm{Na}}_2{\mathrm{SO}}_4$ ionizes as

$\underset{n(1-\alpha )}{{\mathrm{Na}}_2{\mathrm{SO}}_4\to }\underset{n\left(2\alpha \right)}{2{\mathrm{Na}}^{+}}+\underset{4\alpha }{{\mathrm{SO}}_4^{2-}}$

Total amount of species in solution will be $n^{\mathrm{\prime }}=n(1-a)+2n\alpha +n\alpha =n(1+2\alpha )$

The van't Hoff factor is $i=\frac{n^{\mathrm{\prime }}}{n}=1+2\alpha$