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 ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ ions present in $12$ mg of magnesium vapour is:
(Given $1$ eV = $96.5$kJ/mol)
A. $1.5$
B. $2.0$
C. $1.1$
D. $0.5$

Answer 1

Hint:We have to convert the in $12$ mg of magnesium into $Mg^{2+}$ ions so, first we will determine the energy required to convert one mole of Mg into ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ ions. Then we will calculate the mole of Mg present in $12$ mg of magnesium. Then by comparing the calculated moles with energy of one mole we will determine the energy required to convert the $12$ mg of magnesium.

Complete step-by-step solution:It is given that first and second ionization energies of Mg are $7.646$and $15.035$eV respectively. So, after two ionization Mg converts into ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ so, the total energy required to convert the Mg into ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ will be the sum of first and second ionization energy.
So,
$7.646\, + 15.035\, = \,22.681$eV
Now, we will convert the above energy from eV into kJ as follows:
Given, $1$ eV = $96.5$kJ/mol
$22.681$eV = $2188.7$kJ/mol
So, the energy required to convert one mole Mg atom into ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ ions is $96.5$kJ/mol.
Now, we will determine the mole of Mg present in $12$ mg of magnesium vapour as follows:
First we will convert the amount of magnesium vapour form mg to gram as follows:
$1$ mg = ${10^{ - 3}}$ gram
$12$ mg = $12 \times {10^{ - 3}}$ gram
The mole formula is as follows:
\[{\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}\]
Molar mass of Mg is $24$ g/mol.
On substituting $12 \times {10^{ - 3}}$ gram for mass and $24$ g/mol for molar mass,
\[{\text{mole}}\,{\text{ = }}\,\dfrac{{12 \times {{10}^{ - 3}}}}{{24}}\]
\[{\text{mole}}\,{\text{ = }}\,0.5 \times {10^{ - 3}}\]
So, we know that the energy required to convert one mole Mg atom into ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ ions is $96.5$kJ/mol so, the energy will be required to convert the \[0.5 \times {10^{ - 3}}\] mole is,
One mole Mg = $96.5$kJ
\[0.5 \times {10^{ - 3}}\] mol = \[1.1\]kJ
 So, the amount of energy in kJ needed to convert all the atoms of magnesium into ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ ions present in $12$ mg of magnesium vapour is \[1.1\]kJ.

Therefore, option (C) \[1.1\]kJ is correct.

Note:The energy required to remove an electron from an isolated gaseous atom is known as ionization energy. To remove the first electron the energy required is known as first ionization energy and energy required for the second electron is known as second ionization energy. Ionization energy is additive. If we do not convert the mg amount of magnesium vapour into gram then we will get our answer of moles in mmol.