A) \[5\times {{10}^{-4}}\]
B) \[1\times {{10}^{-4}}\]
C) \[5\times {{10}^{-5}}\]
D) \[1\times {{10}^{-5}}\]
Correct Answer: C
Solution :
\[MgC{{l}_{2}}\xrightarrow{{}}M{{g}^{2+}}+2C{{l}^{-}}\] \[M{{g}^{2+}}+\underset{2\,\,F}{\mathop{2{{e}^{-}}}}\,\xrightarrow{{}}\underset{1\,\,mol}{\mathop{Mg}}\,\] (at cathode) \[\because \,\,2F(2\times 96500\,C)\] deposits \[Mg=1\,\,mol\] \[\therefore \,\,9.65\,C\] charge will deposit \[Mg=\frac{1\times 9.65}{2\times 96500}\] 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 are obtained.
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