Given that the same amount of electricity is passed through both solutions, according to Faraday's second law of electrolysis, the mass of the deposited substance is proportional to its equivalent mass. Let $W_{\text{Na}}$ and $W_{\text{Ca}}$ be the masses of deposited sodium and calcium, $M_{\text{Na}}$ and $M_{\text{Ca}}$ be their atomic masses, and $n_{\text{Na}}$ and $n_{\text{Ca}}$ be the number of electrons involved in the reduction of their respective ions. Here, $W_{\text{Na}} = 2.3 \text{ g}$ $M_{\text{Na}} = 23, n_{\text{Na}} = 1$ (since $\text{Na}^+ + e^- \rightarrow \text{Na}$) $M_{\text{Ca}} = 40, n_{\text{Ca}} = 2$ (since $\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}$) According to the law: $$ \begin{aligned} \frac{W_{\text{Ca}}}{W_{\text{Na}}} &= \frac{M_{\text{Ca}} / n_{\text{Ca}}}{M_{\text{Na}} / n_{\text{Na}}} \\ \Rightarrow W_{\text{Ca}} &= W_{\text{Na}} \times \frac{M_{\text{Ca}}}{n_{\text{Ca}}} \times \frac{n_{\text{Na}}}{M_{\text{Na}}} \\ \Rightarrow W_{\text{Ca}} &= 2.3 \times \frac{40}{2} \times \frac{1}{23} \\ \Rightarrow W_{\text{Ca}} &= 2.3 \times 20 \times \frac{1}{23} \\ \Rightarrow W_{\text{Ca}} &= 2 \text{ g} \end{aligned} $$ Now, to find the amount of electricity ($Q$) passed: Using the formula $W = \frac{M \cdot Q}{n \cdot F}$, where $F = 96500 \text{ C/mol}$: $$ \begin{aligned} Q &= \frac{W_{\text{Na}} \cdot n_{\text{Na}} \cdot F}{M_{\text{Na}}} \\ Q &= \frac{2.3 \times 1 \times 96500}{23} \\ Q &= 9650 \text{ C} \end{aligned} $$ Thus, 2 grams of calcium will be deposited, and the total electricity passed is 9650 Coulombs.