Question

The number of moles of electrons required to deposit $36g$ of $Al$ from an aqueous solution of $Al{\left(N{O}_3\right)}_3$ is:

$[At.wt.ofAl=27]$ :

• A. $4$
• B. $2$
• C. $3$
• D. $1$

Answer 1

Answer: (A)

$4$

The mass of Al in 1 mol of $Al(N{O}_3{)}_3$ is the $27g$.

And, for 1 mol of Al ( 27 g ), 3 mol of electrons are required to deposit 27 g of Al.

As 3 mole of electrons required to deposit $⟶27g$ of Al

so for 36 g $⟶36\times \frac{3}{27}=\frac{36}{9}=4moles$

Hence, option A is correct.