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 $27 \mathrm{~g} \mathrm{~mol}^{-1}$ and $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively. Faraday constant $\left.=96500 \mathrm{C} \mathrm{~mol}^{-1}\right]$

<p>$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$</p> <p>The mass of aluminium deposited can be calculated using Faraday’s laws of electrolysis. The steps are as follows:</p> <p><p>Compute the total charge passed:</p> <p>$Q = I \cdot t = 2\,\text{A} \cdot (30 \times 60\,\text{s}) = 3600\,\text{C}$</p></p> <p><p>Determine the moles of aluminium deposited. Since three electrons are required to deposit one mole of aluminium, the number of moles is given by:</p> <p>$$ n(\text{Al}) = \frac{Q}{3F} = \frac{3600}{3 \times 96500} \approx 0.01242\,\text{mol} $$</p></p> <p><p>Finally, calculate the mass of aluminium using its molar mass ($M = 27\,\text{g/mol}$):</p> <p>$m(\text{Al}) = n(\text{Al}) \times M = 0.01242 \times 27 \approx 0.335\,\text{g}$</p></p> <p>Thus, the amount of aluminium deposited at the cathode is approximately</p> <p>$0.336\, \text{g}$</p>