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The volume of oxygen liberated from 0.68...

The volume of oxygen liberated from `0.68g` of `H_(2)O_(2)` is

A

112 m l

B

224 ml

C

56 ml

D

336 ml

Text Solution

Verified by Experts

The correct Answer is:
B
We know that
`{:(2H_(2)O_(2), to, 2H_(2)O + O_(2)),(2xx34 g,,22400 ml):}`
`:'` 2x 34 gm =68 gm of `H_(2)O_(2)`
68 gm of `H_(2)O_(2)` =2240 ml `O_(2)` at STP
`:.` 0.68 gm of `H_(2)O_(2)=(22400xx0.68)/68=224 ml`
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