One mole of magnesium in the vapour state absorbed 1200 kJ mole−1 of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ mole−1 respectively, the final composition of the mixture is
69% Mg++31% Mg+2
Energy absorbed for converting Mg(g) → Mg(g)+ = 750 kJ
Energy left unconsumed = 1200 - 750 = 450 kJ.
This energy will be required to convert Mg(g)+ to Mg(g)+2
Thus % of Mg(g)+2 = 4501450 × 100 = 31%
% of Mg(g)+ = 100 - 31 = 69%