`100mL` of `H_2O_2` is oxidised by `100mL` of `0.01M KMnO_4` in acidic
Channel: Doubtnut
Date: 26/07/2020

`100mL` of `H_2O_2` is oxidised by `100mL` of `0.01M KMnO_4` in acidic medium `(MnO_4^(ɵ)` reduced to `Mn^(2+)`). `100mL` of the same `H_2O_2` is oxidised by `VmL` of `0.01M KMnO_4` in basic medium. Hence `V` is