`100 mL `of `0.01 M KMnO_4` oxidises `100 mL H_2O`, in acidic medium. Volume of the same `KMnO_4` required in alkaline medium to oxidise 100 mL of the same `H_2O_2` will be (`MnO_4^(-)` changes to `Mn^(2+)` in acidic medium and to `MnO_2` in alkaline medium)

Options

A. `(100)/(3) mL`
B. `(500)/(3) mL`
C. `(300)/(5) mL`
D. `100/5 mL`

Answer

B

Explanation

`MnO_4^(-) + 5e^(-) to Mn^(2+) ("acidic") MnO_4^(-) + 3e^(-) to MnO_2 ` (basic)
`100 mL H_2O_2 -= 100 xx 5 N MnO_4^(-) -= V xx 3 N MnO_2 , N = 500/3 mL`