Question

One mole of magnesium in the vapour state absorbed 1200 kJ of energy. If the first and second ionization energies of magnesium are 750 and 1450 $kJ mol^{-1}$ 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

Answer: (B)

Energy absorbed in the ionisation of 1 mole of
Mg (g) to $Mg^+4 (g) = 750 kJ$
Energy left unused = 1200 - 750 = 450 kJ
% of $Mg^+$ (g) converted into $Mg^{2+}$ (g)
=$\frac{450}{1450} \times 100$ = 31%
. Thus the percentage of $Mg^+$ (g)
= 100 - 31 = 69%