One $\text{g}$ equivalent of $Na$ metal is formed from electrolysis of fused $NaCl$. Number of mole of $Al$ from the fused $N{a}_{3}Al{F}_6$ with the same current passed is:

The correct option is C $\frac{1}{3}$
$\text{Gram equivalent}=\text{mole}\times n-factor\text{Gram equivalent of}Na\text{metal}=\text{mole}\times n-factor$
$1=mole\times 1$
$\Rightarrow mole=1$
$$\begin{array}{cccc}& N{a}^{+}& +e^{-}\to & Na\left(s\right)\\ & 1\text{mole}& 1\text{Faraday}& \end{array}$$
According of question, same current is passed
So, $$\begin{array}{cccc}& A{l}^{3+}+& 3e^{-}\to & Al\left(s\right)\\ & & 1\text{faraday}& \end{array}$$
$\text{Hence, number of mole of}Al=\frac{1}{3}\text{mol}$