Question

When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.3, approximately 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

The correct option is A 10.8 mg

The amount of silver deposited can be given by,
$W_{A}g=\frac{E_{A}g\times Q}{96500}=\frac{108\times 9.65}{96500}$
$=1.08\times {10}^{-}2gm=10.8mg$
The amount of silver deposited is 10.8 mg.