Question

${(IE)_1}$ and ${\left( {IE} \right)_2}$ of $M{g_{(g)}}$ are 740,1450 $kJ/mole$. Calculate the percentage of $M{g^ + }_{(g)}$ and $M{g^{2 + }}_{(g)}$. If 1 gram of $M{g_{(g)}}$absorbs 50 KJ of energy.
A.$\% M{g^ + } = 70\% \;\;\;\;\;\;\;\;\% M{g^{2 + }} = 30\%$
B.$\% {\text{ }}M{g^ + } = 68.35\% \;\;\;\% M{g^{2 + }} = 31.65\%$
C.$\% M{g^ + } = 72\% \;\;\;\;\;\;\;\;\;\;\% M{g^{2 + }} = 28\%$
D.$\% M{g^ + } = 60\% \;\;\;\;\;\;\;\;\;\;\;\% M{g^{2 + }} = 40\%$

Answer 1

Since, Energy of Mg ${(IE)_1}$ and ${\left( {IE} \right)_2}$ is given by740 $kJ/mole$ 1450 $kJ/mole$ respectively. We know that If one gram of Mg absorbs 50 kJ of energy​. Then, we need to calculate the number of moles of Mg Using formula of number of moles, whose equation is
Number of moles of $Mg = \dfrac{1}{{atomic\:mass\:of\:Mg}}$

Complete step by step answer:
Given that,
Energy of Mg ${\left( {IE} \right)_1} = {\text{ }}740{\text{ }}KJ/mole$
Energy of Mg ${\left( {IE} \right)_2} = 1450KJ/mol$
If one gram of Mg absorbs 50 kJ of energy​.
Then, we need to calculate the number of moles of Mg
Using formula of number of moles:
Number of moles of $Mg = \dfrac{1}{{atomic\:mass\:of\:Mg}}$
Now, we have to substitute the value into the given formula, we get:
Number of moles of ${M_g} = \dfrac{1}{{24}}$
Number of moles of $Mg = 0.0417$
Energy absorbed in the ionization of Mg to Mg+
${E_{obs}} = 0.0417 \times 740$
${E_{obs}} = 30.83$
Energy unused is
${\mathbf{E}} = {\mathbf{50}}-{\mathbf{30}}.{\mathbf{83}} = {\mathbf{19}}.{\mathbf{17KJ}}$
19.17 kJ will be used in the ionization of $M{g^ + }_{(g)}$to $M{g^{2 + }}_{(g)}$.
We need to calculate the number of moles of $M{g^ + }_{(g)}$converted to $M{g^{2 + }}_{(g)}$
Number of moles = 19.17 / 1450
Number of moles = 0.01322
We need to calculate the number of moles of $M{g^ + }_{(g)}$left $M{g^{2 + }}_{(g)}$.
Number of moles = 0.0417 - 0.01322
Number of moles = 0.02848
We need to calculate the percentage of $M{g^ + }_{(g)}$
Using formula for percentage
$Mg + (g) = \dfrac{{0.02848}}{{0.0417}} \times 100$
$M{g^ + }_{(g)}\; = 68.297\% \approx 68.35{\text{ }}\%$
We need to calculate the percentage of $M{g^{2 + }}_{(g)}$
Using formula for percentage
$Mg + 2(g) = \dfrac{{0.01322}}{{0.0417}} \times 100$
$M{g^{ + 2}}_{(g)}\; = 31.70\% \approx 31.65\%$
Hence, the percentage of $M{g^ + }_{(g)}$is $70.0\%$.
The percentage of $M{g^{2 + }}_{(g)}$. is $29.9\%$.

Therefore, the correct answer is option (B).

Note: The ionization energy gives us an idea of the reactivity of chemical compounds. It can also be used to determine the strength of chemical bonds. It is measured either in units of electron volts or kJ/mol.