When electricity is passed through molten $AlC{l}_3$, 13.5 g. of Al is deposited. The number of Faradays must be

${\mathrm{AlCl}}_3\to {\mathrm{Al}}^{3+}+3{\mathrm{Cl}}^{-}$

Using electrolysis's first law
$w=\frac{\text{ E.i.t }}{F}$

putting the values in the equation-
$\begin{array}{l}13.5=\frac{\frac{27}{3}}{ \mathrm{F}}·\mathrm{t}\\ =1.5 \mathrm{F}\end{array}$

So, charge passed=1.5F

Therefore, the correct option is C 1.5