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100mL of H2O2 is oxidised by 100mL of 0....

`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

A

500

B

100

C

`(100)/(3)`

D

`(500)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`100xx5N(MnO_(4)^(ɵ))-=Vxx3N(MnO_(4)^(ɵ))`
`V=(500)/(3)mL`
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