A current of $0.5\,\text{A}$ is passed through an aqueous $\mathrm{CuSO_4}$ solution for $32\,\text{minutes}\,10\,\text{seconds}$. The mass of copper deposited at the cathode is approximately ($\mathrm{Cu} = 63.5\,\text{g/mol}$, $F = 96500\,\text{C mol}^{-1}$):

Charge passed $Q = I t = 0.5 \times (32 \times 60 + 10) = 0.5 \times 1930 = 965\,\text{C}$. For $\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}$, $n = 2$, so moles of Cu $= Q/(nF) = 965/(2 \times 96500) = 5 \times 10^{-3}\,\text{mol}$. Mass $= 5 \times 10^{-3} \times 63.5 \approx 0.32\,\text{g}$.