Explanation

(c)
$MgCl_{2}\ \rightarrow \ Mg^{2 +} + \ 2Cl^{-}$
$Mg^{2 +}\ + \ 2e^{-}\ \rightarrow \ Mg$ (at cathode)
2F 1 mol
$\because\ 2F(2\ \times \ 96500\ C)deposits\ \ Mg\ = \ 1\ mol$
$\therefore\ 9.65\ C\ charge\ will\ deposit\ Mg\ = \frac{1\ \times 9.65}{2\ \times \ 96500}$
$= \ 5\ \times \ 10^{- 5}$ mol
$RBr\ \ + \ \ Mg\ \ \ \overset{Dry\ ether\ \ }{\rightarrow}\ \ \ \ RMgBr$
Grignard reagent
In order to prepare Grignard reagent, one mole of Mg is used per mole of reagent obtained. Thus, by $5\ \ \times \ 10^{- 5}$ mol mg , $5\ \times 10^{- 5}$ mole of Grignard reagent is obtained.