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If x' litres of O(2) is released at STP ...

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

A

25 ml

B

`12.5 ml`

C

10 ml

D

`7.5 ml`

Text Solution

Verified by Experts

The correct Answer is:
B
`underset(("2moles")) (2H_(2)O_(2)) to underset((22.4 lt) STP) (2H_(2)O+O_(2))`
`:. M=4N=22.4 vol =6.8 %w//v`
`0.4xx0.5 ( :. 0.1 MkMnO_(4)-=0.5 KMnO_(4))`
`V=12.5 ml`
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