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When 0.04 F of electricity is passed through a solution of CaSO 4 , then the weight of Ca 2 + metal deposited at the cathode is

Options

  1. A0.2 g
  2. B0.4 g
  3. C0.6 g
  4. D0.8 g

Correct answer

D. 0.8 g

Step-by-step solution

Ca 2 + + 2 e - → Ca ​ E Ca = 4 0 2 = 2 0 w Ca = E Ca × No. of faradays = 2 0 × 0.04 = 0.8 g .

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