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2H(2)O(2)(l) rarr 2H(2)O(l) + O(2)(g) ...

`2H_(2)O_(2)(l) rarr 2H_(2)O(l) + O_(2)(g)`
100 mL of 2.5 molar `H_(2)O_(2)` gives `O_(2)` gas under the condition when 1 mol occupies 24 L. The volume of `O_(2)` evolved in liters is:

A

2.5

B

`1.0`

C

3

D

`0.25`

Text Solution

Verified by Experts

The correct Answer is:
C
`{:(2H_(2)O_(2)(l),rarr 2H_(2)O(l)+,O_(2)(g),,),((1)/(4)"moles",,(1)/(8) "moles",,,"Value of "O_(2) " evolved" = (1)/(8) xx 24 = 3L):}`
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