The mass of copper deposited when a current of 5.0 A flows for 30 minutes through a solution of CuSO₄ (M_Cu = 63.5 g/mol, F = 96500 C) is approximately
A{'text': '2.96 g', 'label': 'A'}
B{'text': '1.98 g', 'label': 'B'}
C{'text': '5.92 g', 'label': 'C'}
D{'text': '11.85 g', 'label': 'D'}
Answer & Solution
Correct answer: A. {'text': '2.96 g', 'label': 'A'}
1. Charge Q = I × t = 5.0 × 30 × 60 = 9000 C.
2. Moles of electrons = 9000 / 96500 ≈ 0.0933.
3. Cu²⁺ needs 2 electrons per Cu atom, so moles of Cu = 0.0933 / 2 ≈ 0.0466.
4. Mass = 0.0466 × 63.5 ≈ 2.96 g of copper deposited.
_Source: NCERT Class 12 Chemistry, Unit 3 "Electrochemistry", §3.4_
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