Q. One mole of magnesium in the vapour state absorbed 1200 $kJmol^{- 1}$ of energy. If the first and second ionisation energies of Mg are 750 and 1450 $kJmol^{- 1}$ respectively, the final composition of the mixture is
A) $31\% Mg^{+} + 69\% Mg^{2 +}$
B) $69\% Mg^{+} + 31\% Mg^{2 +}$
C) $86\% Mg^{+} + 14\% Mg^{2 +}$
D) $14\% Mg^{+} + 86\% Mg^{2 +}$
Explanation
(b)
Mg$\rightarrow Mg^{+},E = 750kJ$
Remaining energy =1200$-$750=450kJ
Energy needed to convert 1 mole of ${Mg}^{+}$ to ${Mg}^{2 +} = 1450$
Number of moles ${Mg}^{2 +}$produced
$= \frac{1}{1450} \times 450$
=0.31
=31%
Number of moles of $Mg^{+}produced = 1 - 0.31$
=0.69
=69%
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.