Question

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

• A. $5.0$
• B. $1.0$
• C. $1.5$
• D. $3.0$

Answer 1

The correct option is C $1.5$

In $Al{Cl}_3$, $Al$ is in $+3$ Oxidation state.

Thus, the no. of Faraday required for 1 mole of Aluminium is $=3F$

$3F=27g$

$1g=\frac{3F}{27g}$

$13.5g=\frac{3F}{27g}\times 13.5g=1.5F$

Thus $1.5$ Faraday is required for deposition of $13.5g$ of $Al$

Hence, the correct option is $\text{C}$