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 $750$ and $1450 \,kJ\, mol ^{-1}$ respectively, the final composition of the mixture is

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

Answer 1

Answer: (B)

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