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On passing C ampere of current for time ...

On passing C ampere of current for time ‘t' sec through 1L of 2 (M) `CuSO_4` solution (atomic weight of Cu 63.5), the amount ‘m’ of Cu (in g) deposited on cathode will be cathode will be

A

`m = Ct/(63.5 xx 96500)`

B

`m = Ct/(31.75 xx 96500)`

C

`m = (Cxx96500)/(31.75xxt)`

D

`m = 31.75 xx C xx t)/(96500))

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The correct Answer is:
D
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