A solution of a `0.4` g sample of `H_(2)O_(2)` reacted with `0.632 ` g of `KMnO_(4)` in the pressure of sulphuric acid .Calculated the percentage purity of the sample of `H_(2)O_(2)` .

Explanation

`{:(2KMnO_(4),+3H_(2)SO_(4),+5H_(2)O_(2) ,to K_(2)SO_(4),+2MnSO_(4),+8H_(2)O,+5O_(2)),(" +7",,,," +2",,):}`
Eq . Wt . Of `KMnO_(4) = (" mol.wt")/("change in ON per mole ") = 158/5 = 31/6`
Again from the above reaction we see that
2 moles of `KMnO_(4)` combine with 5 moles of `H_(2)O_(2)`
Again from the above reaction we see that
2 moles of `KMnO_(4)` combine with 5 moles of `H_(2)O_(2)`
or 1 equivalent of `KMnO_(4)` combines with `1/2 ` moles of `H_(2)O_(2)`
` :. ` equivalent wt. of `H_(2)O_(2) = (" mol .wt")/2 = 34/2 = 17 `
Now , m.e of `H_(2)O_(2)` = m.e of `KMnO_(4)`
or eq. of `H_(2)O_(2)` = eq. of `KMnO_(4)`
`x/17 =(0.632)/(31.6) " " ` ....(Eqn . 4i)
x being the amount of `H_(2)O_(2)` is grams ,
x = 0.34 g
Percentage of `H_(2)O_(2)` in the sample = `(0.34)/(0.4) xx 100 = 85 % `