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