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Question: 1.0g magnesium atoms in vapour phase absorbs \(50.0kJ\) of energy to convert all \(Mg\) into \(Mg\) ...

1.0g magnesium atoms in vapour phase absorbs 50.0kJ50.0kJ of energy to convert all MgMg into MgMg ions. The energy absorbed is needed for the following changes:
Mg(g)→Mg+(g)+e;ΔH=740kJmol−1Mg\left( g \right) \to M{g^ + }\left( g \right) + e;\Delta H = 740kJmo{l^{ - 1}}
Mg+(g)→Mg2+(g)+e;ΔH=1450kJmol−1M{g^ + }\left( g \right) \to M{g^{2 + }}\left( g \right) + e;\Delta H = 1450kJmo{l^{ - 1}}
Find out the % of Mg+M{g^ + } and Mg2+M{g^{2 + }} in the final mixture.
A. %Mg+=68.28%,%of  Mg2+=31.72%\% M{g^ + } = 68.28\% ,\% of\;M{g^{2 + }} = 31.72\%
B. %Mg+=58.28%,%of  Mg2+=41.72%\% M{g^ + } = 58.28\% ,\% of\;M{g^{2 + }} = 41.72\%
C. %Mg+=78.28%,%of  Mg2+=21.72%\% M{g^ + } = 78.28\% ,\% of\;M{g^{2 + }} = 21.72\%
D. None of these

Explanation

Solution

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 MgMgis 24g24g and hence 24g24g of magnesium is 1mole1mole.
Therefore, the number of moles in 1g1g is 124moles\dfrac{1}{{24}}moles
Therefore, these moles of magnesium are converted to Mg+M{g^ + }and Mg2+M{g^{2 + }} . The energy gets used for Mg→Mg2+Mg \to M{g^{2 + }} can be written as: (740+1450)=2190kJ\left( {740 + 1450} \right) = 2190kJ
If we consider that aais the grams of Mg+M{g^ + } ions and bbis the grams of Mg2+M{g^{2 + }} ions.
Since the total mass of magnesium taken is 1.0g1.0g hence: a+b=1a + b = 1
Moles of Mg+M{g^ + } ions is a24\dfrac{a}{{24}} and the moles of Mg2+M{g^{2 + }} ions is b24\dfrac{b}{{24}} .
Therefore, the energy absorbed for generating the Mg+M{g^ + }ions =a24×740 = \dfrac{a}{{24}} \times 740
The energy absorbed for forming the Mg2+M{g^{2 + }}ions =b24×2190 = \dfrac{b}{{24}} \times 2190
The total energy absorbed is =a24×740+b24×2190 = \dfrac{a}{{24}} \times 740 + \dfrac{b}{{24}} \times 2190
As given here the total energy absorbed is 50kJ50kJ and hence
50=a24×740+b24×219050 = \dfrac{a}{{24}} \times 740 + \dfrac{b}{{24}} \times 2190
1200=740a+2190b1200 = 740a + 2190b
120=74a+219b120 = 74a + 219b
As we all know that from the previous equation: a+b=1a + b = 1
Hence if we put the value in terms of a=1−ba = 1 - b
Putting the value of the aa in the equation:
120=74(1−b)+219b120 = 74\left( {1 - b} \right) + 219b
120=145b+74120 = 145b + 74
b=0.3172b = 0.3172
Therefore bb is the mass of Mg2+M{g^{2 + }}ions is 0.3172g0.3172g which when converted to percentage produces 31.72%31.72\% . The rest is that of the percentage of Mg+M{g^ + } ions =(100−31.72)=68.28%= \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.