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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` gC. `6.35` gD. `3.175 g` |
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Answer» 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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