How many faradays of electricity are required for the deposition of 9 grams of aluminium?

The correct option is A 1 faraday
$1$ faraday of electricity is required for the deposition of $9$ grams of aluminium.
The atomic mass of $Al$ is $27g/mol$. $9g$ of $Al$ corresponds to $\frac{9}{27}=0.333$ moles.
$A{l}^{3+}+3e^{-}\to Al$
$3$ moles of electrons are required for the deposition of $27$ grams (1 mole) of aluminium.
$1$ mole of electrons is required for the deposition of $9$ grams of aluminium.
$1$ mole of electrons $=$ $1$ faraday of electricity.
Hence, $1$ faraday of electricity is required for the deposition of 9 grams of aluminium.