`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^(-) rarr Mn^(2+) ` (acidic ) `MnO_(4)^(-) + 3e^(-) rarr MnO_(2)` (basic )
100 mL `H_(2) O_(2) = 100 xx 5 N MnO_(4)^(-) = V xx 3 N Mn O_(2) , N= (500)/(3)` mL