During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 Amperes.

$\text{ During the electrolysis of molten sodium chloride, }$

$\text{ At cathode }:2{\mathrm{Na}}^{+}+2{\mathrm{e}}^{-}⟶2\mathrm{Na}$

$\text{ At anode }:2{\mathrm{Cl}}^{-}⟶{\mathrm{Cl}}_2+2e^{-}$

$\text{ Net reaction : }2{\mathrm{Na}}^{+}+2{\mathrm{Cl}}^{-}⟶2\mathrm{Na}+{\mathrm{Cl}}_2$

$\mathrm{According} \mathrm{to} \mathrm{Faraday}\text{'}\mathrm{s} \mathrm{first} \mathrm{law} \mathrm{of} \mathrm{electrolysis},$

$\begin{array}{l}\mathrm{w}=\mathrm{Z}\times \mathrm{I}\times \mathrm{t}\\ \mathrm{w}=\frac{\mathrm{E}}{96500}\times \mathrm{I}\times \mathrm{t}\end{array}$

$\mathrm{No}.\mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{Cl}}_2 \mathrm{gas} \times \mathrm{Mol}.\mathrm{wt}.\mathrm{of} {\mathrm{Cl}}_2 \mathrm{gas}$

$\begin{array}{l}=\frac{\text{ Eq. wt. of }{\mathrm{Cl}}_2\mathrm{gas}\times \mathrm{I}\times \mathrm{t}}{96500}\\ 0.10\times 71=\frac{35.5\times 3\times \mathrm{t}}{96500}\\ \mathrm{t}=\frac{0.10\times 71\times 96500}{35.5\times 3}=6433.33\sec \end{array}$

$\mathrm{t}=\frac{6433.33}{60} \min =107.22 \min \approx 110 \min$