Question

When electricity is passed through a solution of $AlC{l}_3$, 13.5 g of Al are deposited. The number of Faradays of electricity passed is: ( write your answer to nearest integer)

Answer 1

$Q=n\times n_f\times Fn=\frac{13.5}{27}{n}_f=3\left(for{Al}^{+3}\right)Q=\frac{13.5}{27}\times 3\times FQ=1.5F$