Question

1 g magnesium atoms in vapour phase absorb 50 kJ of energy to convert all Mg into Mg ions. The energy absorbed is needed for the following changes:
${\text{Mg(g)}} \to {\text{M}}{{\text{g}}^{\text{ + }}}{{(g) + e; \Delta H = }}740{\text{ kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
${\text{M}}{{\text{g}}^{\text{ + }}}{\text{(g)}} \to {\text{M}}{{\text{g}}^{{\text{2 + }}}}{{(g) + e; \Delta H = 1450 kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
Find out the % of ${\text{M}}{{\text{g}}^{\text{ + }}}$ and ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ in the final mixture.
(A) ${{\% M}}{{\text{g}}^{\text{ + }}}{\text{ = 68}}{{.28\% , \% of M}}{{\text{g}}^{{\text{2 + }}}}{\text{ = 31}}{{.72\% }}$
(B) ${{\% M}}{{\text{g}}^{\text{ + }}}{\text{ = 58}}{{.28\% , \% of M}}{{\text{g}}^{{\text{2 + }}}}{\text{ = 41}}{{.72\% }}$
(C) ${{\% M}}{{\text{g}}^{\text{ + }}}{\text{ = 78}}{{.28\% , \% of M}}{{\text{g}}^{{\text{2 + }}}}{\text{ = 21}}{{.72\% }}$
(D) None of these

Answer 1

Magnesium is an element with symbol Mg and its atomic number is twelve. It is an element of group 2 that is of alkaline earth metals. It possesses the rank ninth among the most abundant elements in the universe.
The following formula is used to find the number of moles,
${\text{Moles = }}\dfrac{{{\text{Given weight}}}}{{{\text{Molecular weight}}}}$

Complete Step by step solution:
The given weight of magnesium is 1 g. The atomic weight of magnesium is twenty-four. In order to calculate the moles of magnesium, the formula to be used is as follows,
${\text{Moles of magnesium = }}\dfrac{{{\text{Given weight of magnesium}}}}{{{\text{Molecular weight of magnesium}}}}{\text{ = }}\dfrac{1}{{24}}$ moles
$\dfrac{1}{{24}}$ of magnesium gets converted into ${\text{M}}{{\text{g}}^{\text{ + }}}$ and ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ .
Let the moles of magnesium converted into ${\text{M}}{{\text{g}}^{\text{ + }}}$ be x.
Let the moles of magnesium converted into ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ be y.
The total moles including ${\text{M}}{{\text{g}}^{\text{ + }}}$ and ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ is $\dfrac{1}{{24}}$ moles.
Therefore, $x + y = \dfrac{1}{{24}}$ …(i)
Also, as magnesium gets converted into ${\text{M}}{{\text{g}}^{\text{ + }}}$ , 740 kJ of energy is consumed and when it transforms to ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ , 1450 kJ of energy is used. But magnesium atoms absorb only 50 kJ of energy to convert all Mg into Mg ions. Hence the equation obtained is as follows,
$50{\text{ }} = {\text{ }}740x{\text{ }} + {\text{ }}2190y$ …(ii)
From equations (i) and (ii), we get the values of x and y as,
x = $2.845 \times {10^{ - 2}}$
y = $1.322 \times {10^{ - 2}}$
Hence the moles of ${\text{M}}{{\text{g}}^{\text{ + }}}$ is $2.845 \times {10^{ - 2}}$ and the moles of ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ is $1.322 \times {10^{ - 2}}$ .
${\text{The percent of M}}{{\text{g}}^ + }{\text{ = }}\dfrac{{{\text{Moles of M}}{{\text{g}}^ + }}}{{{\text{Total moles}}}}{\text{ }} \times {\text{ 100 = }}\dfrac{{2.845 \times {{10}^{ - 2}}}}{{\dfrac{1}{{24}}}}{\text{ }} \times {\text{ }}100{\text{ = 68}}{{.28\% }}$
Thus, the percent of ${\text{M}}{{\text{g}}^{\text{ + }}}$ is ${\text{68}}{{.28\% }}$ .
${\text{The percent of M}}{{\text{g}}^{2 + }}{\text{ = }}\dfrac{{{\text{Moles of M}}{{\text{g}}^{2 + }}}}{{{\text{Total moles}}}}{\text{ }} \times {\text{ 100 = }}\dfrac{{1.322 \times {{10}^{ - 2}}}}{{\dfrac{1}{{24}}}}{\text{ }} \times {\text{ }}100{\text{ = 31}}{{.72\% }}$
Thus, the percent of ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ is ${\text{31}}{{.72\% }}$ .
Hence, option A is correct.

Note:
After iron and aluminium, magnesium is the most commonly used metal. In super strong, lightweight metals and alloys, magnesium is used. Alloys are mixtures of metals in definite proportion.