Number of electrons required to deposit 56 grams of Iron from molten ${\mathrm{FeCl}}_3$  will be

$\mathrm{GEW} \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{of} \mathrm{Iron} \mathrm{in} {\mathrm{FeCl}}_3=\frac{56}{3}\mathrm{g}$

$1 F\mathrm{deposits}\frac{56}{3}\mathrm{grams}\mathrm{of}\mathrm{Fe}$

‘X’ F depositis 56 grams of Fe

X = 3F

$\therefore$Number of electrons required $=3\times 6.023\times {10}^{23}$