When a current of 10   A is passes through molten AlCl 3 for 1 . 608 minutes. The mass of Al deposited…

Given data :-

Current = $10 \mathrm{A}$.

Time = $1.608 \min =96.48 \mathrm{s}$

Atomic mass of $\mathrm{Al}$=$27 \mathrm{g}/\mathrm{mol}$

Oxidation state of $\mathrm{Al}=+3$

Mass of aluminium deposited = $\frac{\mathrm{Atomic} \mathrm{mass} \times \mathrm{time}\times \mathrm{current}}{\mathrm{F}\times \mathrm{valency} \mathrm{factor}}$

Mass of aluminium deposited = $\frac{27\times 96.48\times 10}{96480\times 3}=0.09 \mathrm{g}$

Hence, the mass of $\mathrm{Al}$ deposited will be $0.09 \mathrm{g}$.