When 0.1 kg of 16% copper sulfate solution was electrolyzed using inert electrodes, 17 g of substances were separated at the electrodes. Determine how many farads of current passed through the solution.
- 1.2
- 0.4
- 0.2
- 1
📝 Detailed explanation
0.1 kg of 16% solution contains 16 g CuSO₄ (0.1 mol).Electrolysis of 0.1 mol CuSO₄ yields 6.4 g Cu and 1.6 g O₂, total 8 g.
Remaining mass = 17 - 8 = 9 g, which comes from water electrolysis.
9 g H₂O is 0.5 mol.
Farads for CuSO₄ = 0.1 mol × 2 = 0.2 F.
Farads for H₂O = 0.5 mol × 2 = 1.0 F.
Total = 0.2 + 1.0 = 1.2 F.