Use Faraday’s law of electrolysis

The mass deposited is given by $m = \frac{Q}{F}\cdot \frac{M}{n}$ where:

• $Q = 9650\ \text{C}$
• $F = 96500\ \text{C mol}^{-1}$
• $M(\text{Ag}) = 108\ \text{g mol}^{-1}$
• $n = 1$ for $\text{Ag}^+ + e^- \to \text{Ag}$

Calculate moles of electrons passed

$\frac{Q}{F} = \frac{9650}{96500} = 0.1$

So, $0.1$ mole of electrons passes through the solution.

Relate electrons to silver deposited

Since $\text{Ag}^+ + e^- \to \text{Ag}$ 1 mole of electrons deposits 1 mole of silver.

Therefore, moles of silver deposited $= 0.1$ mole.

Calculate mass of silver deposited

$m = 0.1 \times 108 = 10.8\ \text{g}$

Match with options

$\boxed{10.8\ \text{g}}$

Hence, the correct option is A.