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The no. of coulombs required to liberate...

The no. of coulombs required to liberate `0.224 dm^(3)` of chlorine at `0^(@)C` and 1 atm passed are :

A

`2 xx 965`

B

96500

C

96.5

D

965/2

Text Solution

Verified by Experts

The correct Answer is:
A
`22.4 dm^3 of Cl_2=1" mole of "Cl_2=2F`
`therefore 22.4 dm^3 =2F`
Then 0.224 =0.02F
`therefore` Coulombs =`0.02xx96500 or 2xx965`
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