What quantity of aluminium is deposited when a current of 10A is passed through a solution of an aluminium salt for 1930s?

[Al = 27, F = 96500 C mol-1]

Got it — I’ll answer in English.

Correct answer: 1.8 g (Option B)

Explanation:

• Cathode reaction: Al³⁺ + 3e⁻ → Al(s)
• Charge passed: Q = I·t = 10 A × 1930 s = 19,300 C
• Moles of electrons: n_e− = Q/F = 19,300 / 96,500 = 0.20 mol e⁻
• Moles of Al: n(Al) = n_e− / 3 = 0.20 / 3 ≈ 0.0667 mol
• Mass of Al: m = nM = 0.0667 × 27 ≈ 1.8 g

Common pitfall: Forgetting to divide by 3 because Al³⁺ requires three electrons per atom deposited.