Question

Vapour pressure of a mixture of benzene and toluene is given by $P = 179{X_B} + 92$, where ${X_B}$ is the mole fraction of benzene
If vapours are removed and condensed into liquid then what will be the ratio of mole fraction of benzene and toluene in first condensate
(A) 2.8
(B) 1.5
(C) 3.5
(D) 4.5

Answer 1

Hint: Raoult's law states that the vapour pressure of a component at a given temperature equals the mole fraction of that component times the vapor pressure of that component in the pure state. The mole fraction of benzene will be found out by comparing the given equation with the standard equation that the partial pressure is equal to the pressure of pure component times the mole fraction of that particular component

Complete step by step answer:
We know that by Raoult's law that the vapour pressure of an ideal solution is equal to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
We have been given vapour pressure of the mixture of benzene and toluene which is given as follows
$P = 179{X_B} + 92$
Where ${X_B}$ is the mole fraction of benzene
For pure benzene, the mole fraction of benzene will be one
$ \Rightarrow {X_B} = 1$
$\therefore P_B^ \circ = 179 + 92 = 271$ atm
For pure toluene, the mole fraction will be zero
$ \Rightarrow {X_B} = 0$ mm
Now we know according to Raoult's law that partial pressure of the component is equal to the pressure of pure component times the mole fraction of the component. Then
${P_M} = P_B^ \circ {X_B} + P_T^ \circ {X_T}$
$ \Rightarrow (271 \times \dfrac{{12}}{{12 + 8}}) + (92 \times \dfrac{8}{{12 + 8}})$
$ \Rightarrow {P_M} = 199.4\;mm$
$P_T^ \circ = 179 \times 0 + 92 = 92$
Now the number of molecules of benzene is given as follows
$n = \dfrac{{936}}{{78}} = 12$
Now several moles of toluene is given as follows
$n = \dfrac{{736}}{{92}} = 8$
Now the mole fraction of benzene in the vapor phase of the initial mixture
$X_B^1 = \dfrac{{162.6}}{{199.4}}$
The mole fraction of toluene in vapor phase the of in the initial mixture
$X_T^1 = \dfrac{{36.8}}{{199.4}}$
Hence the ratio of the mole fraction of benzene and toluene in the first condensate would be as follows
$\therefore \dfrac{{X_B^1}}{{X_T^1}} = \dfrac{{162.6}}{{36.8}} = 4.418$
So $\dfrac{{X_B^1}}{{X_T^1}} \simeq 4.5$
So the ratio of the mole fraction of benzene and toluene in the first condensate would be 4.5

So, the correct answer is Option D.

Note: The ideal solution is defined as the one in which no volume exchange takes place. It obeys Raoult's Law. Here no enthalpy takes place and examples are benzene and toluene.