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

Hint: 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.