Question

When electricity is passed through a solution of $AlC{l}_3$ and $13.5g$ of Al is deposited, the number of Faraday required must be:

• A. $0.5$
• B. $1.0$
• C. $1.5$
• D. $2.0$

Answer 1

The correct option is C $1.5$

$A{l}^{3+}+3e^{-}\to Al$

1 mol of Al = 27 g = 3 F

$\therefore 13.5gofAl=\frac{3}{27}\times 13.5\equiv 1.5F$

Hence, option C is correct.