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When 9.65 Coulombs of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.87 taking as 108.0) the amount of Silver deposited is
  • a)
    10.8 mg
  • b)
    5.4 mg
  • c)
    16.2 mg
  • d)
    21.2 mg
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
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).
Free Test
Community Answer
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.
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When 9.65 Coulombs of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.87 taking as 108.0) the amount of Silver deposited isa)10.8 mgb)5.4 mgc)16.2 mgd)21.2 mgCorrect answer is option 'A'. Can you explain this answer?
Question Description
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