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 27g $$mol^{-1} $$ and 35.5 g $$mol^{-1}$$ respectively. Faraday constant $$=96500C mol^{-1}$$ ]

We need to find the mass of aluminium deposited during electrolysis.

According to Faraday’s law of electrolysis, the mass deposited is given by the relation $$m = \frac{M \times I \times t}{n \times F}$$ where M = molar mass, I = current, t = time, n = number of electrons transferred, and F = Faraday constant.

For the reduction of aluminium ions $$Al^{3+} + 3e^- \rightarrow Al$$ the number of electrons transferred is n = 3. The molar mass M of aluminium is 27 g/mol, the current I is 2 A, the time t is 30 min = 30 × 60 = 1800 s, and the Faraday constant F is 96500 C/mol.

Substituting these values into the formula gives

$$m = \frac{27 \times 2 \times 1800}{3 \times 96500}$$

$$= \frac{97200}{289500}$$

$$= 0.3355 \text{ g}$$

$$\approx 0.336 \text{ g}$$

The correct answer is Option 2: 0.336 g.