Question

First and second ionization energies of Mg are $7.646$ and $15.035$ eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into $M{g}^{2+}$ ions present in 12 mg of magnesium vapours is:

[Given: $1eV$$=96.5$ $kJmo{l}^{-1}$]

• A. $1.5$
• B. $2.0$
• C. $1.1$
• D. $0.5$

Answer 1

Answer: (C)

$1.1$

The energy required to convert one $Mg$ atom to $M{g}^{2+}$ ion is the sum of first and second ionization potentials = $7.646+15.035=22.681eV$

$=96.5\times 22.681=2188.7kJ/mol$

The atomic mass of Mg is 24 g/mol.
12 mg of Mg corresponds to 0.5 mmoles.

The amount of energy required to convert 0.5 mmol of $Mg$ atoms to $M{g}^{2+}$ ions = $0.5\times {10}^{-3}\times 2188.7=1.1kJ$

Hence, option C is correct.