Calculate the amount of electricity required to deposit 1 mole of...

The dissociation of $\text{AlCl}_3$ gives $\text{Al}^{3+}$ ions.

The reduction reaction at the cathode is:

$\text{Al}^{3+} + 3e^{-} \rightarrow \text{Al}$

From the stoichiometry of the reaction, $1$ mole of $\text{Al}$ requires $3$ moles of electrons.

Since $1$ mole of electrons carries a charge of $1$ Faraday, the amount of electricity required is $3$ Faraday.

Answer: $3$ Faraday