The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500(mol-1)
-
54.0g
- 27.0g
- 13.5g
- 108.0g
The correct answer is: A
Explanation
Recall that
mass deposited = \(\frac{MmIt}{96500n}\)
Mm =108, t = 4830s
I = 10A, n = 1
m = 108 × 10 × (\(\frac{4830}{96500}\)) × 1
m = 54.0g
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