Question

The number of electrons required to deposit $1$ g atom of Al (atomic wt. $=27$) from a solution of $AlC{l}_3$ is:

• A. $1N_A$
• B. $2N_A$
• C. $3N_A$
• D. $4N_A$

Answer 1

The correct option is C $3N_A$

1 g atom equals to 1 mole.

$AlC{l}_3$ solution decomposes to $A{l}^{3+}$ and $3C{l}^{-}$.

To obtain $Al$ from $A{l}^{3+}$, 3 mole of electrons are required.

$Moles=\frac{numberofatoms}{N_A}$

Number of electrons required=$3N_A$