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If same quantity of electricity is passed through `CuCI` and `CuSO_(4)` the ratio of the weights of `Cu` deposited from `CuSO_(4)` and `CuCI` is:-A. `2:1`B. `1:2`C. `1:1`D. `4:1` |
Answer» Correct Answer - B `W prop E` i.e., `(W_(1))/(W_(2))=(E_(1))/(E_(2))` `(W_(CuCl))/(W_(CuSO_(4)))=(E_(CuSO_(4)))/(E_(CuCl))=(M)/(2M)=(1)/(2)` |
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