If x' litres of `O_(2)` is released at STP from one litre of `H_(2)O_(2)` solution due to decomposition of `H_(2)O_(2)` then we lable such solution as 'x volume `H_(2)O_(2)`
`2H_(2)O_(2) to 2H_(2)O+O_(2)`
30% (w/v) `H_(2)O_(2)` is called perhydrol.
How much volume of 224 volumes `H_(2)O_(2)` can completely reduce 10 ml of 0.1M KMnO, Into

Answer

B

Explanation

`2H_2O_2 to 2H_2O + O_2`
(2moles ) (22.4lt) STP
`:. 2M = 4N = 22.4 vol = 6.8% w//v`
`22. 4 vol =4 N rArr 2.24 vol = 0.4 N`
`0.4 xx 0.5 ( :. 0.1 MkMnO_4-=0.5N KMnO_4)`
`V = 12.5ml`