A solution of copper sulphate is electrolysed for 10 minutes with a...

The reaction at the cathode is $\text{Cu}^{2+} + 2\text{e}^{-} \rightarrow \text{Cu}$.

The quantity of charge $Q$ passed through the solution is given by $Q = I \times t$.

Substituting the given values:
$Q = 1.5 \text{ A} \times (10 \times 60) \text{ s} = 900 \text{ C}$

From Faraday's first law of electrolysis, the mass of copper deposited is:
$m = \dfrac{M \times Q}{n \times F}$

Here, $M = 63 \text{ g mol}^{-1}$, $n = 2$, and $F = 96487 \text{ C mol}^{-1}$.

$m = \dfrac{63 \times 900}{2 \times 96487}$

$m = \dfrac{56700}{192974} \approx 0.2938 \text{ g}$

Answer: $0.2938$ g