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$9.65 \mathrm{C}$ of electric current is passed through fused anhydrous magnesium chloride. The magnesium metal thus, obtained is completely converted into a Grignard reagent. The number of moles of the Grignard reagent obtained is
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$5 \times 10^{-5}$
$\mathrm{MgCl}_{2} \longrightarrow \mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-}$
$\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{1 \mathrm{~mol}}{\mathrm{Mg}}$ (at cathode)
$\because 2 \mathrm{~F}(2 \times 96500 \mathrm{C}) \text { deposits } \mathrm{Mg}=1 \mathrm{~mol}$
$\therefore 9.65 \mathrm{C}(2 \times 96500 \mathrm{C}$ ) deposits $\mathrm{Mg}=1 \mathrm{~mol}$
$=\frac{1 \times 9.65}{2 \times 96500}$
$=5 \times 10^{-5} \mathrm{~mol}$
In order to prepare Grignard reagent, one mole of $\mathrm{Mg}$ is used per mole of reagent obtained. Thus, by $5 \times 10^{-5} \mathrm{~mol} \mathrm{Mg}, 5 \times 10^{-5}$ mole of Grignard reagent are obtained.
$\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{1 \mathrm{~mol}}{\mathrm{Mg}}$ (at cathode)
$\because 2 \mathrm{~F}(2 \times 96500 \mathrm{C}) \text { deposits } \mathrm{Mg}=1 \mathrm{~mol}$
$\therefore 9.65 \mathrm{C}(2 \times 96500 \mathrm{C}$ ) deposits $\mathrm{Mg}=1 \mathrm{~mol}$
$=\frac{1 \times 9.65}{2 \times 96500}$
$=5 \times 10^{-5} \mathrm{~mol}$
In order to prepare Grignard reagent, one mole of $\mathrm{Mg}$ is used per mole of reagent obtained. Thus, by $5 \times 10^{-5} \mathrm{~mol} \mathrm{Mg}, 5 \times 10^{-5}$ mole of Grignard reagent are obtained.
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