Question

The number of electrons required to deposite 1 g-atom of Al (atomic mass $=$ 27) from a solution of$AlC{l}_3$ are:

• A. 1 N
• B. 2 N
• C. 3 N
• D. 4 N

Answer 1

The correct option is C 3 N

1 g-atom means one mole of Al. In $AlC{l}_3$ solution, Al contains $+3$ charge. So one mole of $A{l}^{3+}$ will require three moles of electrons to deposit.
$A{l}^{+3}+3e^{-}\to Al$