What mass of copper would be formed when a current of 10.0 A is passed through a solution of CuSO4 for 1 hour? [Cu = 63.5; 1F = 96500C]

Analyze the question: We need the mass of copper deposited by electrolysis of CuSO₄ when 10.0 A flows for 1 hour. This is a Faraday's law problem.

Concept:

• Charge: Q = I × t
• Cu²⁺ + 2e⁻ → Cu, so 2 F (2 × 96500 C) deposits 1 mole (63.5 g) of Cu.

Solution:

1. Time in seconds: t = 1 hour = 3600 s
2. Charge: Q = 10.0 × 3600 = 36,000 C
3. Mass of Cu:

mass = (Q × molar mass) / (96500 × 2) mass = (36,000 × 63.5) / 193,000 mass = 2,286,000 / 193,000 mass ≈ 11.84 g

Answer: about 11.8 g, which matches option B.

Tip: First find the number of electrons per ion (here 2 for Cu²⁺). Then use: mass = (I × t × M) / (n × 96500). A quick check: 36,000 C is about 0.373 F, and half of that in moles (≈0.187 mol) times 63.5 gives about 11.8 g.