Question

Which of the following statements about the composition of the vapor over an ideal 1:1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant ${25^{\text{0}}}{\text{C}}$ , (Given vapor pressure data at ${25^{\text{0}}}{\text{C}}$ Benzene = ${\text{12}}{\text{.8}}\,{\text{K}}\,{\text{Pa}}$, toluene= ${\text{3}}{\text{.85}}\,{\text{K}}\,{\text{Pa}}$)
A.The vapor will contain a higher percentage of benzene
B.The vapor will contain a higher percentage of toluene
C.The vapor will contain equal amount of benzene and toluene
D.Not enough information is given to make prediction

Answer 1

Hint:An ideal mixture is one which obeys Raoult's law. But the fact is that there is no such thing as an ideal mixture , however some liquid mixtures are fairly close to being ideal. Liquid mixtures of two very similar substances can be take as an example here
Hexane and heptane
Benzene and toluene
Formula used: Total vapor pressure of ideal mixture containing liquid A and B is given by
\[{{\text{P}}_{{\text{total}}}}\,{\text{ = }}\,{{\text{P}}_{\text{A}}}\,{\text{ + }}\,{{\text{P}}_{\text{B}}}\]
$
  {{\text{P}}_{\text{A}}}\, - \,{\text{vapor pressure of A}} \\
  {{\text{P}}_{\text{B}}}\, - \,{\text{vapor pressure of B}} \\
 $

Complete answer:
Raoult's law is defined as an ideal law for vapor-liquid phases in equilibrium, where a solution with volatile liquids ,the partial vapor pressure of each component in the solution is directly proportional to its mole fraction ( ${\text{x}}$) at given temperature.
$
  {\text{P}}\, \propto \,{\text{x}}\,\, \Rightarrow \,{\text{P}}\,{\text{ = }}\,{{\text{P}}^0}{\text{x}} \\
  {\text{P - partial vapor pressure}} \\
  {{\text{p}}^0} - {\text{partial vapor pressure of component }}\,{\text{at its}}\,\,{\text{pure}}\,{\text{state}}{\text{.}} \\
 $
For ideal solution (A,B)
$
  {{\text{P}}_{\text{A}}}\,{\text{ = }}\,{{\text{P}}_{\text{A}}}^{\text{0}}\,{{\text{x}}_{\text{A}}} \\
  {{\text{P}}_{\text{B}}}\,{\text{ = }}\,{{\text{P}}_{\text{B}}}^{\text{0}}\,{{\text{x}}_{\text{B}}} \\
 $
Hence the total vapor pressure of ideal mixture containing liquid A and B is given by
               \[{{\text{P}}_{{\text{total}}}}\,{\text{ = }}\,{{\text{P}}_{\text{A}}}\,{\text{ + }}\,{{\text{P}}_{\text{B}}}\]
Composition of liquid A and B in vapor is given by, ${\text{1}}\,{\text{ = }}\,{{\text{y}}_{\text{A}}}\,{\text{ + }}\,{{\text{y}}_{\text{B}}}$
$
  {\text{For A ,}} \\
  {{\text{y}}_{\text{A}}}\,{\text{ = }}\,\dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{{\text{total}}}}}} \\
  {\text{For }}\,{\text{B}} \\
  {{\text{y}}_{\text{B}}}\,{\text{ = }}\,\dfrac{{{{\text{P}}_{\text{B}}}}}{{{{\text{P}}_{{\text{total}}}}}} \\
 $

 Here it's given $1:1$ molar ideal mixture of benzene and toluene. So here we can take \[{\text{A}}\, \to {\text{Benzene ,}}\,{\text{B}}\, \to {\text{Toluene}}\]
$1:1$ molar mixture of A and B means
${{\text{x}}_{\text{A}}}\,{\text{ = }}\,\dfrac{1}{2}$ and ${{\text{x}}_{\text{B}}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{2}}}$
It is given that vapor pressure of benzene at its pure state is ${\text{12}}{\text{.8}}\,{\text{K}}\,{\text{Pa}}$, and for toluene it is ${\text{3}}{\text{.85}}\,{\text{K}}\,{\text{Pa}}$.
Hence total vapor pressure of ideal mixture is given by, \[{{\text{P}}_{{\text{total}}}}\,{\text{ = }}\,{{\text{P}}_{\text{A}}}\,{\text{ + }}\,{{\text{P}}_{\text{B}}}\]
${{\text{P}}_{{\text{total}}}}\, = \,12.8 \times \dfrac{1}{2}\, + \,3.85 \times \dfrac{1}{2}\, = \,8.325\,{\text{KPa}}$
Hence the composition of liquid A in vapor phase is given by ${{\text{y}}_{\text{A}}}\,{\text{ = }}\,\dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{{\text{total}}}}}}$
$\therefore \,\,{{\text{y}}_{\text{A}}}\, = \,\dfrac{{12.8 \times \dfrac{1}{2}}}{{8.325}}\, = 0.768$
For a ideal mixture the composition in vapor is given by ${\text{1}}\,{\text{ = }}\,{{\text{y}}_{\text{A}}}\,{\text{ + }}\,{{\text{y}}_{\text{B}}}$
So here we get the composition of toluene in vapor phase. It is given by
$
  {{\text{y}}_{\text{B}}}\, = \,1\, - \,{{\text{y}}_{\text{A}}} \\
  \therefore \,\,{{\text{y}}_{\text{B}}}\, = \,1\, - \,0.768\, = \,0.232 \\
 $
So the composition of benzene and toluene in vapor phase is $0.768$ and $0.232$ respectively. From the values it is clear that the vapor contains a higher percentage of benzene than toluene.

Hence, the correct answer is option A.

Note:
The main limitations of Raoult's law are,
-Raoult's law is only applicable to very dilute solution
-It is applicable to solution containing non- volatile solute only
-It doesn’t consider the attractive forces in solution
-It is not applicable to the solutes which associate or dissociate in a particular solution.