Question

When 9.65 Coulombs of electricity is passed through a solution of silver nitrate (At. wt. of Ag =107.87 taking as 108) the amount of silver deposited is

• A. 10.8 mg
• B. 5.4 mg
• C. 16.2 mg
• D. 21.2 mg

Answer 1

Answer: (A)

10.8 mg

Answer 2

$W = \frac{\text{EIt}}{\text{96500}} = \frac{108 \times 9.65}{96500} = 0.0108g = 10.8mg.$