Unit 2 Solutions (Intext Questions-3)
Intext Question
2.8 The vapour pressure of pure liquids A and B are 450 and $700 {~mm} \hspace{0.5mm} {Hg}$ respectively, at $350 {~K}$. Find out the composition of the liquid mixture if total vapour pressure is $600 {~mm} \hspace{0.5mm} {Hg}$. Also find the composition of the vapour phase.
Show Answer
Answer
It is given that:
$p_{{A}}^{0}=450 {~mm}$ ${Hg}$
$p_{{B}}^{0}=700 {~mm}$ ${Hg}$
$p_{\text {total }}=600 {~mm}$ ${Hg}$
From Raoult’s law, we have : $ p_{{A}}=p_{{A}}^{0} \hspace{0.8mm} \chi_{{A}}$
$p_{{B}}=p_{{B}}^{0} \hspace{0.8mm} \chi_{{B}}=p_{{B}}^{0}\left(1-\chi_{{A}}\right)$
Therefore, total pressure $p_{\text {total }}=p_{{A}}+p_{{B}}$
$\Rightarrow p_{\text {total }}=p_{{A}}^{0} \hspace{0.8mm} \chi_{{A}}+p_{{B}}^{0}\left(1-\chi_{{A}}\right)$
$\Rightarrow p_{\text {total }}=p_{{A}}^{0} \hspace{0.8mm} \chi_{{A}}+p_{{B}}^{0}-p_{{B}}^{0} \chi_{{A}}$
$\Rightarrow p_{\text {total }}=\left(p_{{A}}^{0}-p_{{B}}^{0}\right) \chi_{{A}}+p_{{B}}^{0}$
$\Rightarrow 600=(450-700) \chi_{{A}}+700$
$\Rightarrow-100=-250 \chi_{{A}}$
$\Rightarrow \chi_{{A}}=0.4$
Therefore, $\chi_{{B}}=1-\chi_{{A}}$
$ \hspace{2.5cm}=1-0.4$
$ \hspace{2.5cm} =0.6$
Now, $p_{{A}}=p_{{A}}^{0} \hspace{0.8mm} \chi_{{A}}$
$ \hspace{1.5cm}=450 \times 0.4$
$ \hspace{1.5cm}=180 {~mm}$ ${Hg}$
$p_{{B}}=p_{{B}}^{0} \hspace{0.8mm} \chi_{{B}}$
$ \hspace{0.5cm} =700 \times 0.6$
$ \hspace{0.5cm} =420 {~mm}$ ${Hg}$
Now, in the vapour phase:
Mole fraction of liquid ${A}=\dfrac{p_{{A}}}{p_{{A}}+p_{{B}}}$
$ \hspace{4.2cm} \begin{aligned} & =\frac{180}{180+420} \\ & =\frac{180}{600} \\ & =0.30 \end{aligned} $
And, mole fraction of liquid $B=1-0.30$ $=0.70$
BETA
