Question

$9.65$ coulombs of electric current is passed through fused anhydrous $MgC{l}_2$ The magnesium metal thus obtained is completely converted into a Gringard regent .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 1

Answer: (C)

$96500$ Coulombs of electric current deposits $=12g$ 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

$=\frac{1.2\times 10^{-3}}{24}=0.05\times 10^{-3}=5\times 10^{-5}$