Question

Given, $1g$ of $Mg$ atoms in the vapour phase absorbs $50.0kJ$ of energy. Find the composition of $M{g^{2 + }}$ formed as a result of an absorption of energy. $I{E_1}$ and $I{E_2}$ for $Mg$ are $740$ and $1450kJ$ mol respectively.

Answer 1

We have to know that the energy required to remove the electron from the isolated gaseous atom is known as ionization energy. And the chemical composition of the element indicates the number, percentage or ratio of the atoms or ions of a chemical substance. And the composition may change according to adding new substances. Here, the magnesium is changed to magnesium ion by the absorption of energy.

Complete answer:
According to the question, the number of magnesium is given as $1g$. Therefore, the number of moles can be found by dividing the given mass with the molecular weight of magnesium, which is equal to $24$.
Hence, number of moles$= \dfrac{1}{{24}}$
And the magnesium is converted $M{g^ + }$ and $M{g^{2 + }}$. The ionization energy for the conversion of $M{g^ + }$ and $M{g^{2 + }}$ is equal to $740$ and $1450kJ$ mol.
Therefore, total energy $= 740 + 1450 = 2190$
Let the number of moles of $M{g^ + }$ formed is considered as ‘a’. Then,
$a \times 740 + \left( {\dfrac{1}{{24}} - a} \right) \times 2190 = 50$
By simplification, $a = 0.02845$
Hence, the percentage of $M{g^ + }$ is calculated by dividing the value of ‘a’ with the number of moles of $Mg$.
Percentage composition of M{g^ + }$$$$ = \dfrac{{0.02845}}{{1/24}} \times 100 = 68.28
Thus, the composition of M{g^{2 + }}$$$$ = 100 - 68.28 = 31.72

Note:
We need to know that the atom will be excited to higher energy levels from the ground state when the atom absorbs energy from its surroundings and this process is known as absorption. The required energy of each atom is different. The energy needed to remove the outer electron is called ionization energy. And it depends upon some factors. The ionization increases with the size of the atom. Because, the effective nuclear charge increases.