Mass of copper deposited by passing 9.6487 A of current through copper sulfate solution for 100 seconds is:(Given: Molar mass of Cu = 63 g/mol , 1F = - Chemistry MCQ NEET (UG) 2024 2026 | ExamTest.live

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Mass of copper deposited by passing 9.6487 A of current through copper sulfate solution for 100 seconds is:(Given: Molar mass of Cu = 63 g/mol, 1 F = 96487 C)

Solution & Step-by-step Explanation

According to Faraday's first law: w = \frac{E \times I \times t}{F} E (Equivalent weight of Cu^{2 +}) = \frac{Molar Mass}{n} = \frac{63}{2} = 31.5 g . I = 9.6487 A . t = 100 s . F = 96487 C . w = \frac{31.5 \times 9.6487 \times 100}{96487} = \frac{31.5 \times 964.87}{96487} = 31.5 \times 0.01 = 0.315 g

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Mass of copper deposited by passing 9.6487 A of current through copper sulfate solution for 100 seconds is:(Given: Molar mass of Cu = 63 g/mol, 1 F = 96487 C)

0.315 g

31.5 g

0.0315 g

3.15 g

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