Question

First and second ionization energies of magnesium are $7$ and $15\text{eV}$ respectively. The amount of energy in ${\text{kJ mol}}^{-1}$ needed to convert all the atoms of Magnesium into $M{g}^{2+}$ ions present in $12\text{mg}$ of magnesium vapours is: (Report your answer upto two decimal)
[Given: $1\text{eV}=96.5{\text{kJ mol}}^{-1}]$

Answer 1

$\text{Number of moles of}Mg=\frac{12\text{mg}}{24\text{g}}=0.5\times {10}^{-3}\text{mol}$
$\text{Energy required per atom}=7+15=22\text{eV}$
$\therefore \text{energy required per mole}=(22\times 96.5){\text{kJ mol}}^{-1}=21.23\times {10}^2{\text{kJ mol}}^{-1}$
$E_{needed}=21.23\times {10}^2\times 0.5\times {10}^{-3}$
$=1.06{\text{kJ mol}}^{-1}$