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On passing 'c' ampere of current for tim...

On passing 'c' ampere of current for time 't' s through 1 L of 2(M) `CuSO_(4)` solution (atomic mass of Cu = 63.5), the amount 'm' of Cu (in g) deposited on cathode will be -

A

`m = ct//(63.5xx96500)`

B

`m=ct//(31.25xx96500)`

C

`m=(cxx96500)//(31.25t)`

D

`m=(31.25xx c xxt)//96500`

Text Solution

Verified by Experts

The correct Answer is:
D
`m=(Ect)/(F)=((63.5)/(2)xxc xxt)/(96500)=(31.75xxc xxt)/(96500)`
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