9.65 C of electric current is passed through | vedclass

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(C) The electrolysis reaction is: $MgCl_{2} \longrightarrow Mg^{2+} + 2Cl^{-}$At the cathode: $Mg^{2+} + 2e^{-} \longrightarrow Mg$Since $2 \ F$ $(2 \

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$5 	imes 10^{-5}$

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(C) The electrolysis reaction is: $MgCl_{2} \longrightarrow Mg^{2+} + 2Cl^{-}$At the cathode: $Mg^{2+} + 2e^{-} \longrightarrow Mg$Since $2 \ F$ $(2 	imes 96500 \ C)$ of charge deposits $1 \ mol$ of $Mg$,$9.65 \ C$ of charge will deposit $Mg = \frac{1 	imes 9.65}{2 	imes 96500} \ mol = 5 	imes 10^{-5} \ mol$.To prepare a Grignard reagent $(R-Mg-X)$,$1 \ mol$ of $Mg$ is required per mole of reagent.Therefore,$5 	imes 10^{-5} \ mol$ of $Mg$ will produce $5 	imes 10^{-5} \ mol$ of Grignard reagent.

$9.65 \ C$ 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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