When a current of $5$ A is passed through a $\mathrm{CuSO_4}$ solution for $193$ seconds, the mass of Cu deposited is: (Atomic mass Cu = 63.5, $F = 96{,}500$ C/mol)

**Charge passed:** $Q = I \cdot t = 5 \times 193 = 965$ C. **Moles of electrons:** $n_e = \dfrac{Q}{F} = \dfrac{965}{96500} = 0.01$ mol. **For Cu deposition** $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$: moles of Cu = $\dfrac{n_e}{2} = \dfrac{0.01}{2} = 0.005$ mol. **Mass:** $0.005 \times 63.5 = 0.3175$ g $\approx 0.318$ g. Option A (0.158 g) is the trap if you forget to divide by 2 (the trap of treating Cu as monovalent). The factor $n = 2$ is the load-bearing detail.