Question

Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300K, when mixed in the molar ratio 1:1 and a vapour pressure of 350 mm when mixed in the molar ratio of 1:2 at the same temperature. The vapour pressure of the two pure liquid X and Y respectively are:
A.250mm,550mm
B.350mm,450mm
C.350mm,700mm
D.550mm,250mm

Answer 1

Hint: Vapour pressure of mixture is the sum of vapour pressure of volatile constituent. Mole fraction of constituent and its pure vapour pressure can be correlated with vapour pressure of the mixture.
Formula Used:
 ${{\text{P}}_{\text{T}}} = {{\text{P}}_{\text{A}}} \times {\chi _A} + {{\text{P}}_{\text{B}}} \times {\chi _B}$
Where, ${{\text{P}}_{\text{T}}}$ is the total vapour pressure, ${{\text{P}}_{\text{A}}}{\text{ and }}{{\text{P}}_{\text{B}}}$ are vapour pressures of A and B and ${\chi _A}{\text{ and }}{\chi _B}$ are mole fractions of A and B.

Complete step by step answer:
Vapour pressure is pressure exerted by the vapour on the surface of its liquid form in a closed vessel. Only volatile (which can go in vapour state) substances, for example water, benzene and alcohol can have vapour pressure. Glucose, sucrose and urea can't have vapour pressure. If a vessel contains only one volatile substance, the vapour pressure is said to be its pure vapour pressure. But for a solution to have more than one volatile substance, vapour pressure of solution is the sum of vapour pressure of all its volatile constituents. Ideal solution is that solution which follows Raoult's law (observed vapour pressure of mixture is equal to theoretical vapour pressure).
Vapour pressure of a constituent in a mixture is equal to vapour pressure of that component in its pure form multiple by its mole fraction. In the above question, X and Y are in molar ratio 1:1(we can say there are 2 moles in mixture with one mole of X and one mole of Y). So the molar fraction of X and Y will be \[\dfrac{{\text{1}}}{2}\] and \[\dfrac{{\text{1}}}{2}\] respectively.
At this state their solution is having a vapour pressure of 400mm. It can be written as:
\[400 = {\text{ }}{P_x} \times \dfrac{1}{2} + {P_y} \times \dfrac{1}{2}\]
When X and Y are in molar ratio 1:2(we can say there are 3 moles in vessel with one mole of X and 2 mole of Y) giving molar fraction of X as \[\dfrac{{\text{1}}}{3}\] and molar fraction of Y as \[\dfrac{2}{3}\].
At this state their solution is having a vapour pressure of 350mm. It can be written as:
\[350 = {\text{ }}{P_x} \times \dfrac{1}{3} + {P_y} \times \dfrac{2}{3}\]
Multiplying equation 1 by \[\dfrac{2}{3}\] and subtracting the above equation we will get:
\[\dfrac{1}{3}{{\text{P}}_{\text{y}}} = 83.333\]
Or \[{{\text{P}}_{\text{y}}} = 250{\text{mm}}\]
Similarly \[{{\text{P}}_{\text{x}}} = 550{\text{mm}}\]

Thus, option D is correct.

Note:
Mole fraction of a substance is equal to no. of moles of that substance present divided by total no. of moles of all substance present. More the substance is volatile; more will be its vapour pressure. Vapour pressures of non volatile substances are taken as zero. A non ideal solution doesn't ensure that the sum of vapour pressure of all volatile substances is equal to vapour pressure of solution.