A 1.00 g sample of `H_(2)O_(2)` solution containing x percent `H_(2)O_(2)` by weight required x ml of a `KMnO_(4)` solution for complete oxidation under acidic condition. Calculate normality of `KMnO_(4)` solution.

Explanation

The complete oxidation under acidic conditions can be represented as follows :
`5H_(2)O_(2) + 2MnO_(4)^(-) + 6H^(+) rarr 5O_(2) + 2Mn^(2+) + 8 H_(2)O`
Since 34 g of `H_(2)O_(2) = 2000` mol of 1`N H_(2)O_(2) {:' "Eq.w.t. or "H_(2)O_(2) = 34/2}`
`:. 34" g of "H_(2)O_(2) = 2000" ml of "1 N KMnO_(4) [N_(1)V_(1) = N_(2)V_(2)]or X/10 g of H_(2)O_(2) = 2000 xx X/100 xx 34` ml of `1 N KMnO_(4)`
Therefore the unknown normality = `2000 xx X//100xx34 = 10/17 or 0.588 N`