Question

$9.65$ coulombs of electric current is passed through fused anhydrous $MgCl_2$ The magnesium metal thus obtained is completely converted into a Gringard regent .The number of moles of Grignard reagent obtained is _______

• A. $5\times10^{-4}$
• B. $1\times10^{-4}$
• C. $5\times10^{-5}$
• D. $1\times10^{-5}$

Answer 1

Answer: (C)

$96500$ 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

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