The volume of _{$0.05MKMn{O}_4$}  solution required to oxidize completely 2.70g of oxalic acid _{$\left(H_2{C}_2{O}_4\right)$}  in acidic medium will be

The balanced equation is

_{$2KMn{O}_4+3H_{2}S{O}_4+5H_2{C}_2{O}_4\to$}

                     _{$K_{2}S{O}_4+2MnS{O}_4+8H_{2}O+10C{O}_2$}

5 moles of oxalic acid react with _{$KMn{O}_4=2$} moles

_$2.70g$  oxalic acid _{$=\frac{2.70}{90}$} mole

_{$\therefore \frac{2.70}{90}$}  mole of oxalic acid will react with

                 _{$KMn{O}_4=\frac{2}{5}\times \frac{2.70}{90}=0.012$} mole

_$0.05$  mole of _{$KMn{O}_4$}  sol an present in _$1000c{m}^3$

_{$\therefore 0.012$}  mole of _{$KMn{O}_4$}  will be present in

                   _{$\frac{1000}{0.05}\times 0.012c{m}^3=240c{m}^3$}