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# Vapour pressure of pure benzene is ${\text{119 torr}}$ and of toluene is ${\text{37}}{\text{.0 torr}}$ at the same temperature mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene ${\text{0}}{\text{.50}}$ , will be:A) $0.137$B) $0.205$C) $0.237$D) $0.435$

Last updated date: 17th Sep 2024
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Hint: The mole fraction of toluene in the vapour phase can be calculated by dividing the total pressure of the solution to the partial vapour pressure of toluene in the solution phase. We also know that the partial vapour pressure of toluene in solution phase is equal to mole fraction of toluene and vapour pressure of pure toluene.
Formula used:
${{ X}_{toluene}}P_{toluene}^ \circ = {P_{total}}{Y_{toluene}}$
Here, ${{\text{{ X}}}_{{\text{toluene}}}}$ = mole fraction of toluene in solution phase
${\text{P}}_{{\text{toluene}}}^{\text{o}}$ = vapour pressure of pure toluene
${{\text{P}}_{{\text{total}}}}$ = total pressure of solution
${{\text{Y}}_{{\text{toluene}}}}$ = mole fraction of toluene in vapour phase

At any temperature, the vapour phase of a component is always greater in the more volatile component as compared to the solution phase. We can also say that mole fraction of the more volatile component is always greater in the vapour phase than in the solution phase.
Here, we know toluene is less volatile than benzene. So, the mole fraction of toluene in the vapour phase will be less than that of mole fraction of toluene in solution phase.
Let us substitute the given values in the equation.
${{\rm X}_{toluene}}P_{toluene}^ \circ = {P_{total}}{Y_{toluene}}$ ……(equation 1)
Here, we are given the value of mole fraction of toluene in the solution phase.
${{\text{{ X}}}_{{\text{toluene}}}} = 0.50$ ……(equation 2)
We are also given the value of vapour pressure of pure toluene.
${\text{P}}_{{\text{toluene}}}^{\text{o}} = 37.0$ ……(equation 3)
We are not given the value of ${{\text{P}}_{{\text{total}}}}$.
We know that the total pressure in the solution is given as:
${P_{total}} = P_{toluene}^ \circ {{\rm X}_{toluene}} + P_{benzene}^ \circ {{\rm X}_{benzene}}$
Now, we know the relation of mole fraction of benzene and toluene in the solution is given as: ${{\rm X}_{benzene}} = 1 - {{\rm X}_{toluene}}$
So, the mole fraction of benzene becomes:
${{\text{{ X}}}_{{\text{benzene}}}}{\text{ = }}1 - 0.50 \\ = 0.5 \\$
By substituting the values we get:
${{\text{P}}_{{\text{total}}}}{\text{ = 37}}{.0 \times 0}{\text{.5 + 119} \times 0}{\text{.5}}$
By solving this equation, we will get the total pressure as:
${{\text{P}}_{{\text{total}}}}{\text{ = 78 torr}}$ ……(equation 4)
Now, by substituting the values of equation 2, equation 3 and equation 4 in equation 1 we get:
${\text{0}}{\text{.50} \times 37}{\text{.0 = 78} \times }{{\text{Y}}_{{\text{toluene}}}}$
By multiplying left side of the above equation we get:
${\text{18}}{\text{.5 = 78} \times }{{\text{Y}}_{{\text{toluene}}}}$
Now by taking numerical values on one side we get
${{\text{Y}}_{{\text{toluene}}}}{\text{ = }}\dfrac{{{\text{18}}{\text{.5}}}}{{{\text{78}}}}$
We get the answer for mole fraction of toluene in the vapour phase as:
${{\text{Y}}_{{\text{toluene}}}}{\text{ = 0}}{\text{.237}}$

Therefore, we can conclude that the correct answer to this question is option C.

Note:
The addition of mole fraction of two components is always equal to unity, either the components are in solution phase or vapour phase.