100mL of 1M `KMnO_(4)` oxidised 100mL of `H_(2)O_(2)` in acidic medium ( when `MnO_(4)^(-)` is reduced to `Mn^(2+)` ) , voluem of same `KMnO_(4)` required to oxidise 100mL of `H_(2)O_(2)` in basic medium ( when `MnO_(4)^(-)` is reduced to `MnO_(2)` ) will be `:`

Options

A. `(100)/(3)mL`
B. `(500)/(3)mL`
C. `(300)/(5)mL`
D. none of these

Answer

D

Explanation

In acidic medium: `m" Eq of "KMnO_(4)-=m" Eq of "H_(2)O_(2)`
`(0.01xx5)xx100-=Nxx100`
In basic medium: `m" Eq of "KMnO_(4)-=m" Eq of "H_(2)O_(2)`
`[MnO_(4)^(ɵ)underset(strong)oversetoverset(ɵ)(O)H)toMnO_(4)^(2-)]`
`(0.01xx1)xxV=Nxx100`
From equation (i) and (ii) we get
`(0.01xx1)xxV-=(0.01xx5)xx100impliesV=500mL`