Question

The number of mole of electrons required to deposit 1g equivalent aluminium (atomic wt $=27$) from a solution of aluminium chloride will be:

• A. $3$
• B. $1$
• C. $4$
• D. $2$

Answer 1

The correct option is B $1$

$1$ eq of $Al=\frac{27}{3}g=9g$

$A{l}^{3+}+3e^{-}⟶Al$

Hence, $Q=3F$

By Faraday's Law,

$3F$ of charge deposits $27g$ of $Al$

$1F$ of charge deposits $9g$ of $Al$ which is $1gram$ eq of $Al$ .

1F means one mole of electrons.