A sample of `H_(2)O_(2)` is x% by mass x ml of `KMnO_(4)` are required to oxidize one gram of this `H_(2)O_(2)` sample. Calculate the normality of `KMnO_(4)` solution.

Explanation

Suppose, Mass of `H_(2)O_(2)` solution `= 100g`
Mass of `H_(2)O_(2)` present =x gram
Mass of `H_(2)O_(2)` solution taken = 1 gram
Mass of `H_(2)O_(2)` present in 1 gram solution `=(x)/(100)`
Equivalents of `H_(2)O_(2) = (w)/(E) = (X)/(100 xx 17)` ....(1)
(E for `H_(2)O_(2)) = 17`
equivalents of `KMnO_(4) = N xx V` (litre) `= N xx X xx 10^(-3)`
Putting equivalents of `H_(2)O_(2)` and `KMnO_(4)` equal `(X)/(100 xx 17) = N xx x xx 10^(-3)`
`N = 0.59` (Normality of `KMnO_(4))`