The electricity required to produce 20.0 g of calcium from molten ${\mathrm{CaCl}}_2$ is (Given: Molar mass of $\left.\mathrm{Ca}=40 \mathrm{g} {\mathrm{mol}}^{-1}\right)$.

Amount of Ca to be deposited, $n=\frac{m}{M}=\frac{20.0\mathrm{g}}{40\mathrm{g}{\mathrm{mol}}^{-1}}=0.5\mathrm{mol}.$

Since each ${\mathrm{Ca}}^{2+}$ 10n requires $2{\mathrm{e}}^{-}$ to produce Ca, the quantity of electricity required will be $(2\times 0.5\mathrm{mol})F\text{, i.e. }\left(1.0\mathrm{mol}\right)F$.