Question

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

• A. $86 \% Mg ^{+}+14 \% Mg ^{2+}$
• B. $36 \% Mg ^{+}+64 \% Mg ^{2+}$
• C. $69 \% Mg ^{+}+31 \% Mg ^{2+}$
• D. $31 \% Mg ^{+}+69 \% Mg ^{2+}$

Answer 1

Answer: (C)

Number of moles of one $g$ of $ Mg =\frac{1}{24} $
$=0.0417$
$1 \,g$ of $ Mg ( g )$ absorbs $=\frac{1200}{24}= 50\, kJ$
Energy required to convert $Mg ( g )$ to $Mg ^{+}( g )$
$=0.0417 \times 750=31.275 \,kJ$
Remaining energy $=50-31.275 $
$=18.725\, kJ$
Number of moles of $Mg ^{2+}$ formed $=\frac{18.725}{1450}$
$=0.013$
Thus remaining $Mg ^{+}$will be
$=0.0417-0.013=0.0287$
So,$\% Mg ^{2+} =\frac{0.0287}{0.0417} \times 100=68.82 \%$
$\% Mg ^{2+} =100-68.82=31.18 \%$