Number of moles of ${\mathrm{MnO}}_4^{-}$  required to oxidize one mole of ferrous oxalate completely in acidic medium will be

In acidic medium ${\mathrm{MnO}}_4^{-}$ oxidises ferrous oxalate as follows :
$2{\mathrm{MnO}}_4^{-}+5{\mathrm{C}}_2{\mathrm{O}}_4^{-2}+16{\mathrm{H}}^{+}\to 2{\mathrm{Mn}}^{2+}+10{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}$
${\displaystyle \because }$ 5 moles of oxalate ions are oxidised by 2 moles of ${\mathrm{MnO}}_4^{-}$.
${\displaystyle \therefore }$ 1 mole of oxalate ion is oxidised by
$=\frac{2}{5}\mathrm{mole} of{\mathrm{MnO}}_4^{-}=0.4\mathrm{moles}$