When 9.65 Coulombs of electricity is passed through a solution of silv...
Given:
Charge passed (Q) = 9.65 C
Atomic weight of Ag = 108.0 (rounded off from 107.87)
To find: Amount of silver deposited
Formula:
Amount of substance (in moles) = Charge (in Coulombs) / Faraday's Constant
where Faraday's Constant = 96500 C/mol
Amount of substance (in moles) of silver deposited = Q / (Faraday's Constant)
Step-by-step solution:
1. Calculate the amount of substance (in moles) of silver deposited:
Amount of substance (in moles) of silver deposited = Q / (Faraday's Constant)
= 9.65 C / (96500 C/mol)
= 0.0001 mol
2. Calculate the mass of silver deposited:
Mass = amount of substance x molar mass
= 0.0001 mol x 108.0 g/mol
= 0.0108 g
3. Convert mass to milligrams:
Mass in mg = 0.0108 g x (1000 mg/g)
= 10.8 mg
Therefore, the amount of silver deposited is 10.8 mg (option A).
When 9.65 Coulombs of electricity is passed through a solution of silv...
Given:
Charge passed (Q) = 9.65 C
Atomic weight of silver (Ag) = 108.0
We know that the amount of substance deposited (m) is directly proportional to the quantity of electricity passed (Q).
Mathematically,
m = ZIt
where Z = electrochemical equivalent,
I = current in amperes,
t = time in seconds
We need to calculate the electrochemical equivalent (Z) of silver.
Z = atomic weight / 96500
= 108.0 / 96500
= 0.001119 g/C
Now, we can calculate the amount of silver deposited (m) using the formula:
m = ZQ
= 0.001119 x 9.65
= 0.0108 g
Converting grams to milligrams,
m = 10.8 mg
Therefore, option (a) is the correct answer.