On passing $C$ ampere of current for time $t$ seconds through $1 \ L$ of $2 \ M$ $CuSO_4$ solution (atomic weight of $Cu = 63.5$),the amount $m$ of $Cu$ (in grams) deposited on the cathode will be:
- A$m = Ct / (63.5 \times 96500)$
- B$m = Ct / (31.25 \times 96500)$
- C$m = (C \times 96500) / (31.25 \times t)$
- D$m = (31.75 \times C \times t) / 96500$
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