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

`{:(2MnO_(4),+ 5H_(2)O_(2),6H^(+),to 5O_(2),+ 2Mn^(2+),+8H_(2)O),((+7),,,,(+2),):}`
From the equation , we see that
change in oxidation number of Mn = `7-2 = 5`
` :. ` 1 mole of `KMnO_(4)` = 5 eq. og `KMnO_(4)`
` :. ` 10 eq. of ` KMnO_(4)` = 5 eq. of `KMnO_(4)`
` :. ` 10 eq of `KMnO_(4)` combines with 5 moles of `H_(2)O_(2)` .
` :. ` 1 eq . of `KMnO_(4)` combines with `1/2 ` mole of `H_(2)O_(2)`
` :. ` eq.wt of `H_(2)O_(2) = 34/2 = 17`
Now ,
` :. 100 ` g of `H_(2)O_(2)` solution contains x g of `H_(2)O_(2)`
` :. 100` g of `H_(2)O_(2)` contains `x/17` equivalent of `H_(2)O_(2)`
` :. 1 ` g of `H_(2)O_(2)` contains `x/(17 xx100) ` eq. of `H_(2)O_(2)`
` :. ` number of m.e of `H_(2)O_(2)` in 1 g solution = `x/(17 xx100) xx 1000`
` = (10 x)/17`
m.e of `H_(2)O_(2)` = m.e of `KMnO_(4)`
`(10 x)/17 = x N `
( N = normality of `KMnO_(4)`)
` :. ` normality of `KMnO_(4)` solution = `10/17 eq. " lit"^(-1)`