A solution of 10ml of  $\frac{\mathrm{M}}{10}\mathrm{Fe}S{O}_4$was titrated with 0.02 M ${\mathrm{KMnO}}_4$ solution in acidic medium, the volume of ${\mathrm{KMnO}}_4$ used will be

${\mathrm{MnO}}_4^{-}+5{\mathrm{Fe}}^{2+}+8{\mathrm{H}}^{+}\to 5{\mathrm{Fe}}^{3+}+{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

Hence, 1 mole of ${\mathrm{KMnO}}_4$ is required to titrate 5 moles of ${\mathrm{FeSO}}_4$.

Moles of ${\mathrm{FeSO}}_4$ titrated = $\mathrm{Molarity}\times \mathrm{volume}=10 \mathrm{ml}\times \frac{\mathrm{M}}{10}=1 \mathrm{m} \mathrm{mole}$

The number of moles of ${\mathrm{KMnO}}_4$ used=$\frac{\mathrm{Moles} \mathrm{of} {\mathrm{FeSO}}_4}{5}=\frac{1 \mathrm{milli} \mathrm{mole}}{5}=0.2 \mathrm{milli} \mathrm{mole}$

So, $0.2 \mathrm{milli} \mathrm{mole}\equiv 10 \mathrm{ml}\times 0.02 \mathrm{M} {{\mathrm{KMnO}}_4}_{ }$

Therefore, 10 ml of 0.02 M ${\mathrm{KMnO}}_4$ is required to titrate the ${\mathrm{FeSO}}_4$solution.