Question

First and second ionisation energy of Mg(g) are $720$ k/J mol and $1440$ kJ/mol respectively. Calculate the $\%$ of $M{g}^{+}$ ions if one gram of Mg(s) absorbs $50$ kJ of energy. (Atomic mass of Mg is $24$ amu.)

• A. $31.65\%$
• B. $21.65\%$
• C. $11.65\%$
• D. $41.65\%$

Answer 1

The correct option is A $31.65\%$

Amount of $Mg$ atoms=$\frac{1}{24}=4.167\times {10}^{-2}mol$

Energy absorbed in ionizing $Mg$ to $M{g}^{+}=4.167\times {10}^{-2}\times 740$

$=30.84KJ$

Energy absorbed in ionizing $M{g}^{+}$ to $M{g}^{2+}=(50-30.84)KJ$

$=19.16KJ$

Amount of $M{g}^{+}$ converted to $M{g}^{2+}=\frac{19.16}{1450}$

$=1.321\times {10}^{-2}mol$

Amount of $M{g}^{+}$ remaining as such=$4.167\times {10}^{-2}-1.321\times {10}^{-2}$

$=2.846\times {10}^{-2}mol$

Composition of final mixture would be as follows:

% of $M{g}^{+}=\frac{2.846\times {10}^{-2}}{4.167\times {10}^{-2}}\times 100=68.3$%

% of $M{g}^{2+}=100-68.3=31.7$ %