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$9.65 \mathrm{C}$ of electric current is passed through fused alhydrous $\mathrm{MgCl}_{2}$. The magnesium metal thus obtained is completely converted into a Grignard reagent. The number of moles of Grignard reagent obtained is
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The correct answer is:
$5 \times 10^{-5}$
96500 Coulombs of electric current deposits $=12$ of magnesium.
$9.65$ Coulombs of electric current deposits
$$
\begin{aligned}
&=\frac{9.65 \times 12}{96500} \\
&=1.2 \times 10^{-3} \mathrm{~g} \text { of magnesium }
\end{aligned}
$$
$\therefore$ The number of moles of Grignard reagent obtained is
$$
=\frac{1.2 \times 10^{-3}}{24}=0.05 \times 10^{-3}=5 \times 10^{-5}
$$
$9.65$ Coulombs of electric current deposits
$$
\begin{aligned}
&=\frac{9.65 \times 12}{96500} \\
&=1.2 \times 10^{-3} \mathrm{~g} \text { of magnesium }
\end{aligned}
$$
$\therefore$ The number of moles of Grignard reagent obtained is
$$
=\frac{1.2 \times 10^{-3}}{24}=0.05 \times 10^{-3}=5 \times 10^{-5}
$$
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