JEE Main 2025 · 22 January · morning · Single correct · +4 / −1

A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A . The amount of the aluminium deposited at the cathode is __________ .
[Given : molar mass of aluminium and chlorine are 27 g mol⁻¹ and 35.5 g mol⁻¹ respectively. Faraday constant =96500 C mol⁻¹]

  1. (a) 1.660 g
  2. (b) 1.007 g
  3. (c) 0.336 g
  4. (d) 0.441 g
Answer and solution

Answer (c) 0.336 g

Solution

Chemistry ma’am · 4 steps

Find the charge, turn it into moles of electrons, then remember each Al³⁺ needs three of them.

  1. Charge passed. Convert the time to seconds: 30 min = 1800 s. Then Q = It = 2 × 1800 = 3600 C.
  2. Cathode reaction. At the cathode, Al³⁺ + 3e⁻ → Al. So 3 moles of electrons deposit 1 mole (27 g) of aluminium, and n = 3.

    m = (M I t)/nF = (27 × 3600)/(3 × 96500) g

  3. Put in numbers. Moles of electrons = 3600/96500 = 0.0373. Moles of Al = 0.0373/3 = 0.01244. Mass = 0.01244 × 27 = 0.336 g.
  4. So. About 0.336 g of aluminium is deposited, which is option (c). The molar mass of chlorine is given only as a distractor.

The idea to keep. Mass deposited is M/n × It/F. Always read n from the charge on the ion being discharged, and use the molar mass of the element, not the salt.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 170 times in JEE Main since 2020. This question is from 2025.