`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

Options

A. `100/3 mL`
B. `500/3 mL`
C. `300/5 mL`
D. None

Answer

B

Explanation

`{:(MnO_(4)^(ɵ)+5e^(ɵ)toMn^(2+),("acidic")),(MnO_(4)^(ɵ)+3e^(ɵ)toMnO_(2),("basic")):}`
`100 mL H_(2)O_(2)=100xx5N MnO_(4)^(ɵ)-=Vxx3N MnO_(2)`
`N=500/3 mL`