What volume of 0.5M `KMnO_(4)` solution is required to oxiside 2.0g of `FeSO_(4)`to acid medium ?

Explanation

In acid medium , `KMnO_(4)` oxidises `Fe^(2+)`to`Fe^(3+)`and itself gets reduced to manganous salt.
`MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O`
So, in the reaction ,1 mol of of `KMnO_(4)-=5`mol of `Fe^(2+)` ions `therefore`1000mL of 1M `KMnO_(4)` solution -= 5 mol of `Fe^(2+)`ions
Now ,1 mol of `FeSO_(4)`-=mol of `Fe^(2+)` -=55.85g of `Fe^(2+)`
or ,151.85g of `FeSO_(4)-=55.85g of Fe^(2+)`
[Molar mass of `FeSO_(4)=151.85]`
or, 2g of `FeSO_(4)-=0.7356g` of `Fe^(2+)`
Again , 5xx55.85g of `Fe^(2+)-=1000mL` of 1M`KMnO_(4)`solution
or, 0.7356g of `Fe^(2+)-=(1000xx0.7356)/(5xx55.85)mL of 1M KMNO(4)`
Solution -=2.63mL of 1M `KMnO_(4)`solution
`=(2.63)/(0.05)`mL of 0.005 `KMnO_(4)`soution
52.6 mL of 0.05M `KMnO_(4)` solution