What volume of `01` M `KMnO_4` is required to oxidise 100 " mL of " 0.2 M `FeSO_4` in acidic medium ? The reaction involved is

Explanation

`5e^(-)MnO_4^(ɵ)toMn^(2+)(n=5)`
(reduction or oxidising agent)
`Fe^(2+)toFe^(3+)+e^(-)(n=1)` (oxidation or reducing agent)
`MnO_4^(ɵ)-=Fe^(2+)`
`mEq=mEq`
`N_1V_1(mL)-=N_2V_2(mL)`
`V_1(mL)xx0.1Mxx5` (n-factor)`-=0.2Mxx1` (n-factor) `xx100`
`0.5V_1-=0.2xx100`
`V_1-=40mL`