Question

1 g of Mg atom in the vapour phase absorbs 500 kJ of energy. Find the composition of Mg formed as a result of absorption of energy. $I{E}_1$ and $I{E}_2$ for Mg are 740 and 1450 kJ /mol respectively.

Answer 1

$1g$ of $Mg=\frac{1}{24}=4.167\times {10}^{-2}mol$

energy used in ionizing $Mg$ to $M{g}^{+}$=$4.167\times {10}^{-2}mol\times 740=30.84kJ$

Energy used in ionizing $M{g}^{+}$ to $M{g}^{2+}=(50-30.84)=19.16kJ$

Amount of $M{g}^{+}$ converted to $M{g}^{2+}=\frac{19.16}{1450}=1.321\times {10}^{-2}mol$

Amount of $M{g}^{+}$ remaining as such= $4.167\times {10}^{-2}-1.321\times {10}^{-2}$

$=2.846\times {10}^{-2}mol$

Composition of final mixture,

% of $M{g}^{+}=\frac{2.846\times {10}^{-2}}{4.167\times {10}^{-2}}\times 100$

$=68.3$

% of $M{g}^{2+}=100-68.3=31.7$