A) \[5\times {{10}^{-4}}\]
B) \[1\times {{10}^{-4}}\]
C) \[5\times {{10}^{-5}}\]
D) \[1\times {{10}^{-5}}\]
Correct Answer: C
Solution :
96500 Coulombs of electric current deposits = 12 g of magnesium. 9.65 Coulombs of electric current deposits \[=\frac{9.65\times 12}{96500}\] \[=1.2\times {{10}^{-3}}g\] of magnesium \[\therefore \]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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