What volume of 0.2 N `KMnO_4` is required to oxidise 10 " mg of " ferrous oxalate in acidic medium? (Molecular Weight of `FeC_2O_4` is 144 g)

Explanation

n-factor of `FeCO_2O_4=3`
n-factor of `MnO_4^(ɵ)=5`
(Since normality of `MnO_4^(ɵ)` is given so n-factor of `MnO_4^(ɵ)` is not required).
`MnO_4^(ɵ)-=FeC_2O_4`
`1mEq-=1mEq`
`N_1V_1(mL)=("Weight")/("Equivalent weight")xx10^(3)`(mEq)
`0.2NxxV_1(mL)=(10xx10^(-3)gxx10^(3))/((144)/(3))`
`V_1=(10xx3)/(144xx0.2)=(300)/(288)=1.04mL`
Volume of `KMnO_4=1.04` mL