Considering the reaction $\mathrm{Mg}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{MgSO}}_4+{\mathrm{H}}_2$ quantitative, the volume of $0.4{\mathrm{MH}}_2{\mathrm{SO}}_4$ used to react

with $2.4 \mathrm{g} \mathrm{Mg}$ is:

$\mathrm{Mg}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶\mathrm{Mg}{\mathrm{SO}}_4+{\mathrm{H}}_2$

1 mol                       1 mol

$0.1 \mathrm{mol}$                    $0.1 \mathrm{mol}$

   Moles of  $\mathrm{Mg}=(\mathrm{wt}/\mathrm{g}\text{ at. }\mathrm{wt}.)=2.4/24=0.1$

$1 \mathrm{mol}$                            $1 \mathrm{mol}$

VOl of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in litre = Mol /molarity

  $= 0.1/0.4 = 0.25 \mathrm{L}$.