Question

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

• A. 31% Mg⁺+69% Mg⁺²
• B. 69% Mg⁺+31% Mg⁺²
• C. 86% Mg⁺+14% Mg⁺²
• D. 14% Mg⁺+86% Mg⁺²

Answer 1

The correct option is B

69% Mg⁺+31% Mg⁺²

Energy absorbed for converting $M{g}_{\left(g\right)}$ $\to$ ${M{g}_{\left(g\right)}}^{+}$ = 750 kJ

Energy left unconsumed = 1200 - 750 = 450 kJ.

This energy will be required to convert ${M{g}_{\left(g\right)}}^{+}$ to ${M{g}_{\left(g\right)}}^{+2}$

Thus % of ${M{g}_{\left(g\right)}}^{+2}$ = $\frac{450}{1450}$ $\times$ 100 = 31%

% of ${M{g}_{\left(g\right)}}^{+}$ = 100 - 31 = 69%