To deposit $0.634 \mathrm{~g}$ of copper by electrolysis of aqueous cupric sulphate solution, the amount of electricity required (in coulombs) is

In the electrolysis of cupric sulphate, the reaction that occurs at cathode is $$ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} $$ Thus $2 \mathrm{~F}$ or $2 \times 96500 \mathrm{C}$ of electricity is required to deposit $=1 \mathrm{~mol}$ of $\mathrm{Cu}=63.5 \mathrm{~g}$ of $\mathrm{Cu}$ It means that to deposit $63.5 \mathrm{~g}$ of $\mathrm{Cu}$, the amount of electricity required $=2 \times 96500 \mathrm{C}$ So, to deposit $0.634 \mathrm{~g}$ of $\mathrm{Cu}$, the amount of $$ \begin{aligned} & \text { electricity required }=\frac{2 \times 96500}{63.5} \times 0.634 \\ & \approx 1930 \mathrm{C} \end{aligned} $$