Question

One mole of magnesium in the vapour state absorbed $ 1200\text{ }kJ\text{ }mo{{l}^{-1}} $ of energy. If the first and second ionization energies of Mg are 750 and $ 1450\text{ }kJ\text{ }mo{{l}^{-1}} $ respectively, the final composition of the mixture is:

• A. 31 $ \text{ }M{{g}^{+}}+69 $ % $ \text{ }M{{g}^{2+}} $
• B. 69% $ \text{ }M{{g}^{+}}+31 $ % $ \text{ }M{{g}^{2+}} $
• C. 86% $ \text{ }M{{g}^{+}}+14 $ % $ \text{ }M{{g}^{2+}} $
• D. 14% $ \text{ }M{{g}^{+}}+86 $ % $ \text{ }M{{g}^{2+}} $
• E. 13% $ \text{ }M{{g}^{+}}+87 $ % $ \text{ }M{{g}^{2+}} $

Answer 1

Answer: (B)

$ Mg\xrightarrow[{}]{{}}M{{g}^{+}}, $
$ E=750\,kJ $
Remaining energy $ =1200-750=450\text{ }kJ $
Energy needed to convert 1 mol of
$ M{{g}^{+}} $ to $ M{{g}^{2+}}=1450 $
Number of moles of $ M{{g}^{2+}} $ produced
$ =\frac{1}{1450}\times 450 $
$ =0.31 $ $ =31% $
Number of moles of $ M{{g}^{+}} $ produced
$ =1-0.31 $
$ =0.69 $
$ =69% $