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A current of 2.6 ampere was passed throu...

A current of 2.6 ampere was passed through `CuSO_(4)` solution for 380 sec . The amount of Cu deposited is (atomic mass of Cu 63.5)

A

`0.3250 g`

B

`0.635` g

C

`6.35` g

D

`3.175 g`

Text Solution

Verified by Experts

The correct Answer is:
A
`W = (E xxi xx t)/(96500) = (63.5 xx 2 .6 xx380)/(2 xx 96500) = 0.325` g
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