A 1.00- g sample of H_(2)O_(2)solution containing x percent H_(2)O_(2)by weightrequiresx mLof a KMnO_(4)solution for complete oxidation under acidic conditions .Calculate the normality of the KMnO_(4) solution .

A 1.00- g sample of H_(2)O_(2)solution containing x percent H_(2)O_(2)

1.	A 1.00- g sample of H_(2)O_(2)solution containing x percent H_(2)O_(2)by weightrequiresx mLof a KMnO_(4)solution for complete oxidation under acidic conditions .Calculate the normality of the KMnO_(4) solution .
Answer» Solution :`{:(2MnO_(4),+ 5H_(2)O_(2),6H^(+),to 5O_(2),+ 2Mn^(2+),+8H_(2)O),((+7),,,,(+2),):}` FROMTHE 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 eqof `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.eof `KMnO_(4)` `(10 x)/17 = x N ` ( N = normality of `KMnO_(4)`) `:. ` normality of `KMnO_(4)` solution = `10/17eq. " lit"^(-1)`