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`

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Correct Answer - A
`W = (E xxi xx t)/(96500) = (63.5 xx 2 .6 xx380)/(2 xx 96500) = 0.325` g

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