A current of 3.0A is passed through 750 ml of 0.45 M solution of CuSO$_4$ for 2 hours with a current efficiency of 90%. If the volume of the solution is assumed to remain constant, what would be the final molarity of CuSO$_4$ solution?

The reaction for the electrolysis of CuSO$_{4}$ is Cu$^{2+}$ + 2e$^{-} \rightarrow$ Cu(s).

The total charge passed through the solution is $Q = I \times t \times \text{efficiency}$.

$Q = 3.0 \text{ A} \times (2 \times 3600 \text{ s}) \times 0.90 = 19440 \text{ C}$.

The number of moles of electrons passed is $n_{e} = \dfrac{Q}{F} = \dfrac{19440}{96500} \approx 0.20145 \text{ mol}$.

Since 2 moles of electrons are required to reduce 1 mole of Cu$^{2+}$, the moles of Cu$^{2+}$ reduced is $n_{\text{reduced}} = \dfrac{n_{e}}{2} = \dfrac{0.20145}{2} = 0.100725 \text{ mol}$.

The initial moles of CuSO$_{4}$ is $n_{\text{initial}} = M \times V = 0.45 \text{ M} \times 0.750 \text{ L} = 0.3375 \text{ mol}$.

The final moles of CuSO$_{4}$ is $n_{\text{final}} = n_{\text{initial}} - n_{\text{reduced}} = 0.3375 - 0.100725 = 0.236775 \text{ mol}$.

The final molarity is $M_{\text{final}} = \dfrac{n_{\text{final}}}{V} = \dfrac{0.236775}{0.750} = 0.3157 \text{ M} \approx 0.316 \text{ M}$.

Answer: 0.316