100 ml of a mixture of NaOH and ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is neutralized by 10 ml of 0.5 M ${\mathrm{H}}_2{\mathrm{SO}}_4$. Hence NaOH in 100 mL solution is

As  ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is already neutral so only NaOH will be neutralized by ${\mathrm{H}}_2{\mathrm{SO}}_4$

$\therefore \mathrm{Eq}. \mathrm{of} \mathrm{NaOH} = \mathrm{Eq}. \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4$

            ${\left(\frac{\mathrm{W}}{\mathrm{E}}\right)}_{\mathrm{NaOH}} = {\left(\frac{\mathrm{NV}\left(\mathrm{ml}\right)}{1000}\right)}_{{\mathrm{H}}_2{\mathrm{SO}}_4}$

            $$\begin{gathered} \frac{\mathrm{W}}{40}=\frac{0.5\times 2\times 10}{1000} \\ \mathrm{W} = 40 \times {10}^{-2} = 0.4 \mathrm{g} \end{gathered}$$