Question

the amount of electricity that should be passed through CuSO4 solution with Cu electrodes to deposit 0.1 gram atom of Cu is
1.9650 C
2.96500 C
3.19300 C
4.193000 C​

Answer 1

Answer: 3.  19300 C

Explanation:

1 electron carry charge=$1.6\times 10^{-19}C$

1 mole of electrons contain=$6.023\times 10^{23}$ electrons

Thus  1 mole of electrons carry charge=$\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C=1 Faraday$

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

At cathode: $Cu^{2+}+2e^-\rightarrow Cu$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole of copper

1  mole of copper is deposited by = 193000 C

0.1 mole of copper is deposited by =$\frac{193000}{1}\times 0.1=19300C$

Thus 19300 C of of electricity that should be passed through CuSO4 solution with Cu electrodes to deposit 0.1 gram atom of Cu.

Answer 2

1 mol of Cu is deposited by 193000 C of electricity

0.1 mol of Cu is deposited by=(193000×0.1)÷1=19300 C of electricity

Final answer is 3. "19300 C"

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