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Question: Two miscible liquids \[A\] and \[B\] having vapour pressure in pure state \[P_A^o\] and \[P_B^o\] ar...

Two miscible liquids AA and BB having vapour pressure in pure state PAoP_A^o and PBoP_B^o are mixed in mole fraction χA{\chi _A} and χB{\chi _B} to get a mixture having total vapour pressure of mixture PM{P_M}. Which of the following reactions are correct?
A. χA=PM−PBoPAo−PBo{\chi _A} = \dfrac{{{P_M} - P_B^o}}{{P_A^o - P_B^o}}
B. χA(l)χA(V)=PMPAo\dfrac{{{\chi _{A(l)}}}}{{{\chi _{A(V)}}}} = \dfrac{{{P_M}}}{{P_A^o}}
C. χA(l)χA(V)=PMPBo\dfrac{{{\chi _{A(l)}}}}{{{\chi _{A(V)}}}} = \dfrac{{{P_M}}}{{P_B^o}}
D. All of these

Explanation

Solution

The partial pressure of a gas is the pressure exerted by the gas when it solely occupies the whole volume of the space. In a mixture of A and B the new pressure (partial pressure) is dependent on the concentration of individual components in the liquid phase.

Complete step by step answer:
The given question depends on Raoult’s law of gas mixtures. According to Raoult’s law a solvent’s partial vapour pressure in a mixture is equal to the vapour pressure of the pure solvent multiplied by the mole fraction of the solvent present in the solution.
The Raoult’s law is expressed as,
Psoln=Psolvent∘×χsolvent{P_{soln}} = P_{solvent}^\circ \times {\chi _{solvent}}
Where Psoln{P_{soln}} = vapour pressure of the solution, χsolvent{\chi _{solvent}} = mole fraction of the solvent, Psolvent∘P_{solvent}^\circ = vapour pressure of the pure solvent.
In a mixture of two miscible and volatile liquids A and B in a container, both the particles of A and B are present in the vapour phase. So the total pressure of the solution is equal to the sum of partial vapour pressure of particles of both A and B.
ThusPM=PA+PB{P_M} = {P_A} + {P_B} ,
At equilibrium the partial pressure of A and B are equal to
PA=PA∘χA{P_A} = P_A^\circ {\chi _A} and PB=PB∘χB{P_B} = P_B^\circ {\chi _B}
So the total vapour pressure of PM{P_M} for binary mixtures,
PM=PA∘⋅χA+PB∘⋅χB{P_{M}} = P_A^\circ \cdot {\chi _{A}} + P_B^\circ \cdot {\chi _B} ​
Further the total vapour pressure of the mixture is related to the vapour pressure of the pure component in liquid phase as below:
PMχA(V)=PA∘χA(l){P_M}{\chi _{A\left( V \right)}} = P_A^\circ {\chi _{A\left( l \right)}}
So, χA(l)χA(V)′=PMPAo\dfrac{{{\chi _{A(l)}}}}{{\chi _{A(V)}'}} = \dfrac{{{P_M}}}{{P_A^o}}
The sum of the mole fraction of A and B is χA+χB=1{\chi _A} + {\chi _B} = 1
Or χB=1−χA{\chi _B} = 1 - {\chi _A}
Using the relation PM=PA∘⋅χA+PB∘⋅χB{P_{M}} = P_A^\circ \cdot {\chi _{A}} + P_B^\circ \cdot {\chi _B} ​
PM=PA∘χA+PB∘⋅(1−χA){P_M} = P_A^\circ {\chi _{A}} + P_B^\circ \cdot (1 - {\chi _A})
or χA=PM−PBoPAo−PBo{\chi _A} = \dfrac{{{P_M} - P_B^o}}{{P_A^o - P_B^o}}
Hence the options A and B are correct.

Note:
It is evident from Raoult's law that the partial pressure changes with the vapour phase pressure and the mole fraction of the component. Raoult’s law is applicable only in cases when the solute and the solvent have the equal tendency to escape from the pure state.