Question

$1g$ of $Mg$ atoms in the vapour phase absorbs $50.0kJ$ of energy. Find the composition of ${Mg}^{2+}$ formed as a result of absorption of energy. ${IE}_1$ and ${IE}_2$ for $Mg$ are $740$ and $1450kJ$ ${mol}^{-1}$ respectively.

Answer 1

mole of $Mg=\frac{1}{24}$
These mole of $Mg$ will be converted to ${Mg}^{+}$ and ${Mg}^{2+}$. Energy consumed for the conversion of Mg into $M{g}^{+2}$ is 740 + 1450 $=2190$.

Let $a$ mole of ${Mg}^{+}$ are formed, then
$a\times 740+(\frac{1}{24}-a)\times 2190=50$
$\therefore a=0.02845$
$\therefore$ %$of{Mg}^{+}=\frac{0.02845}{1/24}\times 100=68.28$
$\therefore$ %$of{Mg}^{+2}=31.72$