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A current of 0.75 A is passed through an...

A current of `0.75 A` is passed through an acidic solution of `CuSO_4` for `10` minutes. The vlume of oxygen liberated at anode (at STP) will be.

A

` 0.261 sm^(3)`

B

` 0.261 cm^3`

C

` 0.261 xx10^2 mL`

D

` 0. 2661 m^3`

Text Solution

Verified by Experts

The correct Answer is:
C
`2O^(2-) rarr O_2 +4e`
Mole of `E= (0.75 xx 10 xx 60)/(96500)`
Mole of `O_2 = (4.66 xx 10^(-3))/4 = 0. 026 1L` .
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