First and second ionization energy of Mg(g) are 720 $\mathrm{kJ}/\mathrm{mol}$ and 1440 $\mathrm{kJ}/\mathrm{mol}$ respectively. Calculate the % of $m{g}^{+}$ ions if one gram of Mg(g) absorbs 50 kJ of energy. (Atomic mass of Mg is 24 amu.)

The number of Mg atoms$=\frac{1}{24}=4.167\times {10}^{-2} mol$

absorbed power while ionizing magnesium to $M{g}^{+}$ $M{g}^{+}=4.167\times {10}^{-2}\times 740$

the power used to create ions $M{g}^{+ }to M{g}^{2+}$ $=(50-30.84)KJ =19.16KJ$

Amount of ${\mathrm{Mg}}^{+}$$\text{ converted to }{\mathrm{Mg}}^{2+}=\frac{19.10}{1450}$

$=1.321\times {10}^{-2}\mathrm{mol}$

$M{g}^{+}$ staying in that form in that amount 

$=4.167\times {10}^{-2}-1.321\times {10}^{-2}$

$=2.846\times {10}^{-2}\mathrm{mol}$

The following would make up the final mixture:

$\mathrm{\%}\text{ of }{\mathrm{Mg}}^{+}=\frac{2.846\times {10}^{-2}}{4.167\times {10}^{-2}}\times 100=68.3\mathrm{\%}$

$\mathrm{\%}\text{ of }{\mathrm{Mg}}^{2+}=100-68.3=31.7\mathrm{\%}$