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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 `O_(2)` is released from 100 ml of Phydrol" due to decomposition of `H_(2)O_(2)` at STP?

A

18 lit

B

9 lit

C

27 lit

D

4.5lit

Text Solution

Verified by Experts

The correct Answer is:
B
6.8 % `H_2O_2 =22.4 vol :. 30% H_2O_2 = 100vol`
1 ml perhydrol gives 100 ml `O_2`
100 ml perhydrol gives 10 lt `O_2`
But 90% decomposed` :. `9lt is released
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