Question

1.0 g magnesium atoms in vapour phase absorbs 50.0 kJ of energy to convert all Mg into Mg ions. The energy absorbed is needed for the following changes :

$Mg\left(g\right)⟶M{g}^{+}\left(g\right)+e;\Delta H=740kJmo{l}^{-1}$
$M{g}^{+}\left(g\right)⟶M{g}^{2+}\left(g\right)+e;\Delta H=1450kJmo{l}^{-1}$

Find out the % of $M{g}^{+}$ and $M{g}^{2+}$ in the final mixture.

• A. % $M{g}^{+}=68.28$% , % $ofM{g}^{2+}=31.72$%
• B. % $M{g}^{+}=58.28$% , % $ofM{g}^{2+}=41.72$%
• C. % $M{g}^{+}=78.28$% , % $ofM{g}^{2+}=21.72$%
• D. None of these

Answer 1

The correct option is A % $M{g}^{+}=68.28$% , % $ofM{g}^{2+}=31.72$%

$MolesofMg=\frac{1}{24}$

Let $a$ mole of $M{g}^{+}$ and $b$ mole of $M{g}^{2+}$ are present, then
$a+b=\frac{1}{24}$ ... (i)

Also, $50=a\times 740+b\times 2190$ ... (ii)

$\therefore$ From eqs. (i) and (ii),
$a=2.845\times {10}^{-2}$

$b=1.322\times {10}^{-2}$

% $ofM{g}^{+}=68.28\%$

% $ofM{g}^{2+}=31.72\%$