KCET · Chemistry · Electrochemistry
\(9.65 \mathrm{C}\) 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
- A \(5 \times 10^{-4}\)
- B \(1 \times 10^{-4}\)
- C \(5 \times 10^{-5}\)
- D \(1 \times 10^{-5}\)
Answer & Solution
Correct Answer
(C) \(5 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
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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