Question

A current of 5 amperes is passed through $\text{AlC}{\text{l}}_{\text{3}}$ solution for 15 minutes using Pt electrodes, when 0.504 g of Al is produced. What is the molar mass of Al?

Answer 1

$\omega =\frac{\text{Mol}\text{.wt}\times \text{I}\times t}{n\times 96500}$

$0.504=\frac{M\times 5\times 15\times 60}{3\times 96500}$

$M=\frac{0.504\times 3\times 96500}{5\times 15\times 60}=32.484$