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What will be the weight of deposited silver on passing 965 coulombs of electricity in solution of AgNO₃?
  • a)
    1.08
  • b)
    2.16
  • c)
    0.54
  • d)
    0.27
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
What will be the weight of deposited silver on passing 965 coulombs of...

Calculation of Weight of Deposited Silver:

To calculate the weight of deposited silver, we can use Faraday's Law of Electrolysis which states that the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the solution.

Given Data:
- Charge passed (Q) = 965 C
- Atomic weight of silver (Ag) = 107.87 g/mol
- 1 Faraday constant (F) = 96500 C/mol

Formula:
Weight of deposited substance = (Q * Atomic weight) / (F * valency)

Calculation:
Weight of deposited silver = (965 * 107.87) / (96500 * 1)
Weight of deposited silver = 104072.55 / 96500
Weight of deposited silver ≈ 1.08 g

Therefore, the weight of deposited silver on passing 965 coulombs of electricity in a solution of AgNO₃ would be approximately 1.08 grams. Option 'A' is the correct answer.
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What will be the weight of deposited silver on passing 965 coulombs of electricity in solution of AgNO₃?a)1.08b)2.16c)0.54d)0.27Correct answer is option 'A'. Can you explain this answer?
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