Question

One mole of magnesium in the vapour state absorbed $1200kJmo{l}^{-1}$ of energy. If the first and second ionization energies of Mg are 750 and $1450kJmo{l}^{-1}$ respectively, the final composition of the mixture is:

• A. 31 $M{g}^{+}+69$ % $M{g}^{2+}$
• B. 69% $M{g}^{+}+31$ % $M{g}^{2+}$
• C. 86% $M{g}^{+}+14$ % $M{g}^{2+}$
• D. 14% $M{g}^{+}+86$ % $M{g}^{2+}$
• E. 13% $M{g}^{+}+87$ % $M{g}^{2+}$

Answer 1

Answer: (B)

$Mg\to M{g}^{+},$
$E=750kJ$
Remaining energy $=1200-750=450kJ$
Energy needed to convert 1 mol of
$M{g}^{+}$ to $M{g}^{2+}=1450$
Number of moles of $M{g}^{2+}$ produced
$=\frac{1}{1450}\times 450$
$=0.31$ $=31$
Number of moles of $M{g}^{+}$ produced
$=1-0.31$
$=0.69$
$=69$