When electricity is passed through molten AlCl₃, 13.5 g. of Al is deposited. The number of Faradays must be

When AlCl₃ is lectrolysed ,the reaction taking palce at cathode is 

$A{l}^{3+\hspace{0.17em}}+ 3e^{-}\to Al (n=3)\hspace{0.17em}$

$$\begin{gathered} w= \frac{E\times Q}{F} \\ 13.5= \frac{27 \times Q}{3\times F} \\ Q= 1.5 F \end{gathered}$$