Question

1.0g magnesium atoms in vapour phase absorbs $50.0kJ$ of energy to convert all $Mg$ into $Mg$ ions. The energy absorbed is needed for the following changes:
$Mg\left( g \right) \to M{g^ + }\left( g \right) + e;\Delta H = 740kJmo{l^{ - 1}}$
$M{g^ + }\left( g \right) \to M{g^{2 + }}\left( g \right) + e;\Delta H = 1450kJmo{l^{ - 1}}$
Find out the % of $M{g^ + }$ and $M{g^{2 + }}$ in the final mixture.
A. $\% M{g^ + } = 68.28\% ,\% of\;M{g^{2 + }} = 31.72\% $
B. $\% M{g^ + } = 58.28\% ,\% of\;M{g^{2 + }} = 41.72\% $
C. $\% M{g^ + } = 78.28\% ,\% of\;M{g^{2 + }} = 21.72\% $
D. None of these

Answer 1

Hint: Total moles for each of the ions are to be tested after the ionization enthalpy when the removal of electrons takes place. The formation of two ions is possible due to two ionization processes. Process of energy absorption is assessed using the given enthalpy changes.

Complete step by step answer:
The molecular weight of $Mg$is $24g$ and hence $24g$ of magnesium is $1mole$.
Therefore, the number of moles in $1g$ is $\dfrac{1}{{24}}moles$
Therefore, these moles of magnesium are converted to $M{g^ + }$and $M{g^{2 + }}$ . The energy gets used for $Mg \to M{g^{2 + }}$ can be written as: $\left( {740 + 1450} \right) = 2190kJ$
If we consider that $a$is the grams of $M{g^ + }$ ions and $b$is the grams of $M{g^{2 + }}$ ions.
Since the total mass of magnesium taken is $1.0g$ hence: $a + b = 1$
Moles of $M{g^ + }$ ions is $\dfrac{a}{{24}}$ and the moles of $M{g^{2 + }}$ ions is $\dfrac{b}{{24}}$ .
Therefore, the energy absorbed for generating the $M{g^ + }$ions $ = \dfrac{a}{{24}} \times 740$
The energy absorbed for forming the $M{g^{2 + }}$ions $ = \dfrac{b}{{24}} \times 2190$
The total energy absorbed is $ = \dfrac{a}{{24}} \times 740 + \dfrac{b}{{24}} \times 2190$
As given here the total energy absorbed is $50kJ$ and hence
$50 = \dfrac{a}{{24}} \times 740 + \dfrac{b}{{24}} \times 2190$
$1200 = 740a + 2190b$
$120 = 74a + 219b$
As we all know that from the previous equation: $a + b = 1$
Hence if we put the value in terms of $a = 1 - b$
Putting the value of the $a$ in the equation:
$120 = 74\left( {1 - b} \right) + 219b$
$120 = 145b + 74$
$b = 0.3172$
Therefore $b$ is the mass of $M{g^{2 + }}$ions is $0.3172g$ which when converted to percentage produces $31.72\% $. The rest is that of the percentage of $M{g^ + }$ ions $ = \left( {100 - 31.72} \right) = 68.28\% $ .

So, the correct answer is Option A.

Note: The conversion of magnesium to the second ionisation state is to be known from the given enthalpy. There are two reactions provided for each ionization and for determining the changes to the second state from the ground state is important before determining the percentage of each in the solution.