Question

One mole of magnesium in the vapour state absorbed 1200 kJ mol${}^{-1}$ energy . If the first and second ionisation energies of Mg are 750 kJ mol${}^{-1}$ and 1450 kJ mol${}^{-1}$ respectively, the final composition of the mixture is :

• A. $86\%M{g}^{}+14\%M{g}^{2+}$
• B. $69\%M{g}^{+}+31\%M{g}^{2+}$
• C. $14\%M{g}^{+}+86\%M{g}^{2+}$
• D. $31\%M{g}^{+}+69\%M{g}^{2+}$

Answer 1

The correct option is B $69\%M{g}^{+}+31\%M{g}^{2+}$

Energy absorbed in the ionisation of 1 mole of $Mg$ to $M{g}^{+}$(g) $=750$ kJ

Energy left unused $=1200-750=450$ kJ

% of $M{g}^{+}$(g) converted into $M{g}^{2+}$(g)

$=\frac{450}{1450}\times 100=31$%

Hence, the % of $M{g}^{+}=100-31=69$%