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On passing electric current through molten AlCl3, 11.2 litres of Cl2 gas is liberated at the anode at NTP. Find the mass of aluminium deposited at the cathode (in grams). (Atomic mass of Al = 27 g/mol)
- 4.5 g
- 9 g
- 13.5 g
- 27 g
Correct answer: 9 g
Solution
At anode: 2Cl⁻ -> Cl2 + 2e⁻. Moles of Cl2 = 11.2/22.4 = 0.5 mol. Electrons released = 2 * 0.5 = 1 mol of electrons (1 Faraday). At cathode: Al³+ + 3e⁻ -> Al. Moles of Al = (1 mol e⁻) / 3 = 1/3 mol. Mass of Al = (1/3) * 27 = 9 g.
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