To solve this problem, we will use Faraday's laws of electrolysis.$\\$According to Faraday's first law, the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.$\\$The formula is given by:$\\ m = \frac{Z \cdot Q}{F} \\$where:$\\ m$ is the mass of the substance deposited (in grams),$\\ Z$ is the electrochemical equivalent of the substance,$\\ Q$ is the total charge passed through the electrolyte (in coulombs),$\\ F$ is Faraday's constant (96487 C/mol).$\\$The electrochemical equivalent $Z$ is given by:$\\ Z = \frac{M}{n \cdot F} \\$where:$\\ M$ is the molar mass of the substance (for Cu, $M = 63$ g/mol),$\\ n$ is the number of electrons involved in the reduction (for Cu, $n = 2). \\$Substituting the values, we get:$\\ Z = \frac{63}{2 \cdot 96487} \\$The total charge $Q$ is given by:$\\ Q = I \cdot t \\$where:$\\ I$ is the current (9.6487 A),$\\ t$ is the time (100 seconds).$\\$Substituting the values, we get:$\\ Q = 9.6487 \cdot 100 = 964.87$ C$\\$Now, substitute $Z$ and $Q$ into the formula for $m: \\ m = \frac{\left(\frac{63}{2 \cdot 96487}\right) \cdot 964.87}{1} \\$Simplify the expression:$\\ m = \frac{63 \cdot 964.87}{2 \cdot 96487} \\ m = \frac{63 \cdot 964.87}{192974} \\ m = \frac{60746.81}{192974} \\ m \approx 0.315$ g$\\$Therefore, the mass of copper deposited is approximately $0.315$ g.$\\$This corresponds to Option 4.