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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

10 it

C

27 lit

D

`4.5` lit

Text Solution

Verified by Experts

The correct Answer is:
B
`6.8%H_(2)O_(2)=22.4 vol :. 30% H_(2)O_(2) =100 vol`
1ml perhydrol gives `100 ml O_(2)`
100 ml pehydrol gives `10 lt O_(2)`
But 90% dcomposed `:.` 9lt is released
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