`20 mL` of a solution containing `0.2g` of impure sample of `H_(2)O_(2)` reacts with `0.316 g` of `KMnO_(4)` (acidic). Calculate:
(a) Purity of `H_(2)O_(2)`,
(b) Volume of dry `O_(2)` evolved at `27^(@)C` and `750 mm P`.

Explanation

`2KMnO_4+5H_2O_2+3H_2SO_4to2MnSO_4+K_2SO_4+5O_2+8H_2O`
`MnO_4^(ɵ)+8H^(o+)+5e^(-)toMn^(2+)+4H_2O]xx2`
`Ew(KMnO_4)=31.6`
" Eq of "`KMnO_4=(0.316)/(31.6)=10^(-2)Eq`
`therefore " Eq of "H_2O_2=10^(-2)`
Amount of `H_2O_2=10^(-2)xx17=0.17g`
Purity of `H_2O_2=(0.17)/(0.2)xx100=85%`
" Eq of "`H_2O_2=" Eq of "KMnO_4=" Eq of "O_2=10^(-2)Eq`
Since `O_2` is obtained from both `H_2O_2` and `KMnO_4`,
" Eq of "`O_2=2xx10^(-2)`
`4 " Eq of "O_2=1 mol =22400mL` at STP
`2xx10^(-2)" Eq of "O_2=(22400)/(4)xx2xx10^(-2)`
`=11200xx10^(-2)=112mL` at STP
Volume of `O_2` at `27^@C` and 750 mm Hg pressure
`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`
`(760xx112)/(273)=(750xxV_2)/(300)`
`V_2=(760xx112xx300)/(750xx273)=124.7mL`