For the oxidation of 0 . 2   M   FeSO 4 solution 0 . 965 amperes current is passed through it for…

Given that,

$0.2 \mathrm{M} {\mathrm{FeSO}}_4$ and current = $0.965 \mathrm{amp}-\mathrm{hr}=\left(0.965\times 3600\right) \mathrm{coulomb}$

Dissociation of ${\mathrm{FeSO}}_4={\mathrm{Fe}}^{2+}+{\mathrm{SO}}_4^{2-}$

$\mathrm{n}-$factor = $2$

We know that, 

Normality = Molarity $\times \mathrm{n}-$factor

$\Rightarrow$Normality = $2\times 0.2=0.4 \mathrm{N}$

Here, $96500 \mathrm{coulomb}$ = $1 \mathrm{eq}. {\mathrm{Fe}}^{2+}$.

$\Rightarrow 0.965\times 3600$ $\mathrm{coulomb}$ = $\frac{0.965\times 3600}{96500} \mathrm{eq}. \mathrm{of} {\mathrm{Fe}}^{2+}$ = $0.036 \mathrm{eq}.$

Also, Normality $=\frac{\mathrm{No}. \mathrm{of} \mathrm{gram} \mathrm{eq}.}{\mathrm{Volume}\left(\mathrm{L}\right)}$ $\Rightarrow 0.4=\frac{0.036}{\mathrm{Vol}.\left(\mathrm{L}\right)}\Rightarrow \mathrm{Vol}.\left(\mathrm{L}\right)=\frac{0.036}{0.4}=0.09 \mathrm{L} =90 \mathrm{mL}$

Hence, the volume of the solution that is oxidised in $\mathrm{mL}=90 \mathrm{mL}$