The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500($mol^{-1}$)

Analyze the question: We need the mass of silver deposited by electrolysis, given the current (10 A), the time (4830 s), Ag = 108 and F = 96500 C/mol.

Concept: Faraday's laws of electrolysis

• Charge: Q = I × t
• Ag⁺ + e⁻ → Ag, so 1 mole of electrons (96500 C) deposits 1 mole of Ag (108 g).
• Mass = (Q / 96500) × (molar mass / number of electrons)

Solution:

1. Calculate the charge: Q = I × t = 10 × 4830 = 48,300 C

2. Calculate the moles of electrons: 48,300 / 96,500 = 0.5 mol

3. Silver ions need 1 electron each, so moles of Ag = 0.5 mol

4. Mass of Ag = 0.5 × 108 = 54.0 g

Answer: The mass of silver deposited is 54.0 g, which is option A.

Tip: Notice that 4830 is exactly half of 9660, so 10 × 4830 = 48,300 is exactly half of 96,500. JAMB often chooses numbers like this so the answer comes out as a simple fraction. Whenever you see Q close to 96,500 or a simple fraction of it, you can quickly spot that the answer is that fraction of the equivalent mass (here ½ × 108 = 54 g).