If 10.8 g of silver is deposited in a silver coulometer volume of oxygen liberated is?
-
0.56 dm3
- 5.60 dm3
- 11.20 dm3
- 22.40 dm3
The correct answer is: A
Explanation
Ag+ + e-H = 108 g Ag
= 0.11 10.8 Ag
4OH → O2 + 2H2O + 4e
4F ↔ 1mole O2
4F = 22.4 dm3 χ 0.1 = 0.56 dm3