Question

$I{E}_1$ and $I{E}_2$ of magnesium are $7.6\text{eV}$ and $15.0\text{eV}$ respectively. The amount of energy in $\text{kJ}$ needed to convert all the atoms of magnesium into $M{g}^{2+}$ ions present in $36\text{mg}$ of magnesium vapour will be:
[Given: $1\text{eV}=96.5{\text{kJ mol}}^{–1}$].

• A. $3.27$
• B. $1.62$
• C. $2.18$
• D. $2.51$

Answer 1

The correct option is A $3.27$

Given: $Mg\to M{g}^{+};I{E}_1=7.6\text{eV}$
$M{g}^{+}\to M{g}^{2+};I{E}_2=15.0\text{eV}$
Hence,
$Mg\to M{g}^{2+};IE=22.6\text{eV}/atom$
Given:
$1eV=96.5kJmo{l}^{-1}$

IE in kJ/mol$=22.6\times 96.5=2181kJmo{l}^{-1}$

Moles of $Mg$ in 36 mg of Mg vapours,
$=\frac{36\times {10}^{-3}\text{g}}{24\text{g}}=1.5\times {10}^{-3}mol$

$1$ mol of $Mg$ has an ionisation energy $=2181\text{kJ}$
$\therefore 1.5\times {10}^{-3}$ mol will have an ionisation energy $=1.5\times {10}^{-3}\times 2181\text{kJ}$
$=3271.5\times {10}^{-3}\text{kJ}$