Given: Current ($I$) = $9.6487\text{ A}$ Time ($t$) = $100\text{ s}$ Molar mass of Cu ($M$) = $63\text{ g mol}^{-1}$ Faraday's constant ($F$) = $96487\text{ C mol}^{-1}$

Total charge ($Q$) passed through the solution is: $Q = I \times t = 9.6487 \times 100 = 964.87\text{ C}$

The reduction reaction for copper from copper sulphate is: $Cu^{2+} + 2e^- \rightarrow Cu(s)$ Here, $n = 2$ moles of electrons are required to deposit $1\text{ mole}$ ($63\text{ g}$) of copper.

According to Faraday's first law of electrolysis, the mass ($m$) deposited is: $m = \frac{M \times Q}{n \times F}$ $m = \frac{63 \times 964.87}{2 \times 96487}$ $m = \frac{63}{2} \times \left(\frac{964.87}{96487}\right)$ $m = 31.5 \times 0.01 = 0.315\text{ g}$