When electricity is passed through a solution of AICI₃ 13.5 g of Al is deposited. The number of faradays must be

Reaction took place at cathode was :

Al⁺³ + 3e^-→ Al

One mole of electron is deposited by one faraday, similarly 3 mole of electron is deposited by 3F.

1 mole of Al ⇒ 27 g

So, 13.5 g ⇒ 0.5 mole
If one mole of Al requires 3 F, then the amount of F requires for 0.5 mole of Al is:

$\mathrm{Number} \mathrm{of} \mathrm{Faraday}=0.5 \mathrm{mol}\times \frac{3 \mathrm{F}}{1 \mathrm{mol}}=1.5 \mathrm{F}$