Question

A mixture of two miscible liquids A and B is distilled under equilibrium conditions at \[1{\text{atm}}\] pressure. The mole fraction of A in solution and vapor phase are $0.30$ and $0.60$ respectively. Assuming ideal behavior of solution and vapor, calculate the ratio of vapor pressure of pure A to that of pure B.
A. $4.0$
B. $3.5$
C. $2.5$
D. $1.85$

Answer 1

Hint: Vapor pressure of a liquid is much different in a solution than it is in pure liquid. The dissolved non-volatile solute lowers the vapor pressure. Dalton’s law and Raoult’s law can be used to solve this question.
Given data:
Mole fraction of A in solution, ${{\text{x}}_{\text{A}}} = 0.30$
Mole fraction of A in vapor phase, ${{\text{y}}_{\text{A}}} = 0.60$
Pressure, ${\text{P = 1atm}}$

Complete step by step answer:
It is given that, mole fraction of A in solution, ${{\text{x}}_{\text{A}}} = 0.30$
Therefore mole fraction of B in solution, ${{\text{x}}_{\text{B}}} = 1 - 0.30 = 0.70$
Mole fraction of A in vapor phase, ${{\text{y}}_{\text{A}}} = 0.60$
Similarly mole fraction of B in vapor phase, ${{\text{y}}_{\text{B}}} = 1 - 0.60 = 0.40$
Raoult’s law can be mathematically expressed as:
${{\text{P}}_{{\text{soln}}}} = {\text{x}}{{\text{P}}_{{\text{solvent}}}}$, where ${{\text{P}}_{{\text{soln}}}}$ is the observed vapor pressure.
${{\text{P}}_{{\text{solvent}}}}$is the vapor pressure of solvent,
${\text{x}}$is the mole fraction.
Therefore mole fraction can be written as:
${\text{x}} = \dfrac{{{{\text{P}}_{{\text{soln}}}}}}{{{{\text{P}}_{{\text{solvent}}}}}}$
This is a liquid-liquid solution.
Therefore Raoult’s law is modified as:
${{\text{P}}_{{\text{total}}}} = {{\text{x}}_{\text{A}}}{{\text{P}}_{\text{A}}}^ \circ + {{\text{x}}_{\text{B}}}{{\text{P}}_{\text{B}}}^ \circ $
Using Raoult’s law, it can be expressed as:
${{\text{y}}_{\text{A}}} = \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{{\text{total}}}}}}$
Combining both equations, we get
${{\text{y}}_{\text{A}}} = \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{A}}} + {{\text{P}}_{\text{B}}}}} = \dfrac{{0.30{P_A}^ \circ }}{{0.30{{\text{P}}_{\text{A}}}^ \circ + 0.70{{\text{P}}_{\text{B}}}^ \circ }} = 0.60$
Similarly, ${{\text{y}}_{\text{B}}} = \dfrac{{0.70{{\text{P}}_{\text{B}}}^ \circ }}{{0.70{{\text{P}}_{\text{B}}}^ \circ + 0.30{{\text{P}}_{\text{A}}}^ \circ }} = 0.40$
We have to find the ratio of the vapor pressure of pure A to pure B.
For that, we can divide ${{\text{y}}_{\text{A}}}$by ${{\text{y}}_{\text{B}}}$.
i.e. $
  \dfrac{{{{\text{y}}_{\text{A}}}}}{{{{\text{y}}_{\text{B}}}}} = \dfrac{{0.60}}{{0.40}} = \dfrac{{0.30{{\text{P}}_{\text{A}}}^ \circ }}{{0.30{{\text{P}}_{\text{A}}}^ \circ + 0.70{{\text{P}}_{\text{B}}}^ \circ }} \div \dfrac{{0.70{{\text{P}}_{\text{B}}}^ \circ }}{{0.70{{\text{P}}_{\text{B}}}^ \circ + 0.30{{\text{P}}_{\text{A}}}^ \circ }} = \dfrac{{0.30{{\text{P}}_{\text{A}}}^ \circ }}{{0.70{{\text{P}}_{\text{B}}}^ \circ }} \\
  1.5 = \dfrac{{0.30{{\text{P}}_{\text{A}}}^ \circ }}{{0.70{{\text{P}}_{\text{B}}}^ \circ }} \\
 $
Thus, $\dfrac{{{{\text{P}}_{\text{A}}}^ \circ }}{{{{\text{P}}_{\text{B}}}^ \circ }} = \dfrac{{1.5 \times 0.70}}{{0.30}} = 3.5$
Hence the option B is correct.

Additional information:
Dalton’s law of partial pressure explains that the total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container. For example, air is a mixture of non-reacting gases. The total pressure exerted by air is equal to the sum of partial pressure of each gas.

Note:
Raoult’s law states that the amount of change in the vapor pressure is dependent on the amount of non-volatile solute added to the solution (mole fraction), not the quality of the solute. The number of solvent molecules per unit volume can be decreased by the solute by lowering the tendency for the molecules to escape into vapour.