A solution of copper sulphate (CuSO$$_4$$) is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at the cathode (at. mass of Cu = 63u) is :

The total charge $$Q$$ passed through the solution is $$Q = I \times t = 1.5 \times (10 \times 60) = 900\text{ C}$$.

Copper is deposited at the cathode via the reaction $$Cu^{2+} + 2e^- \rightarrow Cu(s)$$, requiring $$2F$$ ($$2 \times 96500\text{ C}$$) to deposit one mole ($$63\text{ g}$$) of copper.

The mass $$m$$ of copper deposited is $$m = \frac{M \times Q}{n \times F} = \frac{63 \times 900}{2 \times 96500} \approx 0.2938\text{ g}$$.