Question

The first and second ionization energies of magnesium are $740{\text{ , 1540 kJ mo}}{{\text{l}}^{ - 1}}$ respectively. Calculate the percentage of $M{g_{(s)}}^{ + 1}$ and $M{g_{(s)}}^{ + 2}$, if one gram of $M{g_{(s)}}$ absorbs $50{\text{ kJ}}$ of energy.
A. $\% {\text{M}}{{\text{g}}^ + }{\text{ = 50}}$ and $\% {\text{M}}{{\text{g}}^{ + 2}}{\text{ = 50}}$
B. $\% {\text{M}}{{\text{g}}^ + }{\text{ = 70}}{\text{.13}}$ and $\% {\text{M}}{{\text{g}}^{ + 2}}{\text{ = 29}}{\text{.87}}$
C. $\% {\text{M}}{{\text{g}}^ + }{\text{ = 75}}$ and $\% {\text{M}}{{\text{g}}^{ + 2}}{\text{ = 25}}$
D. $\% {\text{M}}{{\text{g}}^ + }{\text{ = 68}}{\text{.65}}$ and $\% {\text{M}}{{\text{g}}^{ + 2}}{\text{ = 31}}{\text{.65}}$

Answer 1

The first and second ionisation of energy is given for magnesium metal atoms. The total amount of energy it absorbs is $50{\text{ kJ}}$ , this energy will be used for both ionization of magnesium metal. We will find the amount of energy used by a given mole of magnesium for first ionisation and the remaining energy will be used for second ionization. In this way, we can also find the percentage of each ionised metal ion.
Formula used:
Number of moles $= {\text{ }}\dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$

Complete answer: We will find the amount of energy consumed by the first ionisation state of magnesium for one gram of magnesium. The given ionisation of magnesium metals is for one mole. Therefore we will find the number of moles of magnesium metal is given. The number of moles of magnesium can be find as,
Number of moles $= {\text{ }}\dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$
Here, given mass of magnesium is one gram and molar mass or molecular mass of magnesium is $24{\text{ g}}$ , thus we can find the number of moles as,
Number of moles $= {\text{ }}\dfrac{1}{{24}}$
Number of moles $= {\text{ 0}}{\text{.0417}}$
The first ionisation energy of magnesium metal is $740{\text{ kJ mo}}{{\text{l}}^{ - 1}}$ , which means one mole of magnesium requires this amount of energy for first ionisation to become $M{g_{(s)}}^{ + 1}$. But we are provided with only${\text{ 0}}{\text{.0417}}$ moles of magnesium. Thus the amount of energy required by ${\text{ 0}}{\text{.0417}}$ moles of magnesium to become $M{g_{(s)}}^{ + 1}$ will be equal to,
${\text{ 0}}{\text{.0417 }} \times {\text{ 740 = 30}}{\text{.83 kJ}}$
Since according to the question, $50{\text{ kJ}}$ energy is supplied to magnesium metal atoms. After first ionisation the unused energy will be equal to $50{\text{ - 30}}{\text{.83 = 19}}{\text{.17 kJ}}$. Therefore ${\text{19}}{\text{.17 kJ}}$ energy will be supplied to $M{g_{(s)}}^{ + 1}$ become$M{g_{(s)}}^{ + 2}$.
Thus the number of moles of $M{g_{(s)}}^{ + 1}$ which is converted into $M{g_{(s)}}^{ + 2}$ equal to $\dfrac{{19.17}}{{1450}}{\text{ = 0}}{\text{.0132}}$.
The number of moles of $M{g_{(s)}}^{ + 1}$ will be equal to $= {\text{ 0}}{\text{.0417 - 0}}{\text{.0132 = 0}}{\text{.0285}}$ moles. The percentage of each ion can be found as,
$\%$ Of $M{g_{(s)}}^{ + 1}{\text{ = }}\left( {\dfrac{{0.0285}}{{0.0417}}{\text{ }} \times {\text{ 100}}} \right)\% {\text{ = 68}}{{.65\% }}$
$\%$ Of $M{g_{(s)}}^{ + 2}{\text{ = }}\left( {\dfrac{{0.0132}}{{0.0417}}{\text{ }} \times {\text{ 100}}} \right)\% {\text{ = 31}}{{.65\% }}$
Hence the correct option is D. $\% {\text{M}}{{\text{g}}^ + }{\text{ = 68}}{\text{.65}}$ and $\% {\text{M}}{{\text{g}}^{ + 2}}{\text{ = 31}}{\text{.65}}$

Note:
Since the amount of energy left after first ionisation of a magnesium metal atom is much less than required for second ionisation energy, therefore the percentage of $M{g_{(s)}}^{ + 2}$ is very less. The number of moles is just a number, therefore it has no units.