Question

At a given temperature, total vapor pressure (in torr) of a mixture of volatile components A and B is given by ${{\text{P}}_{{\text{Total}}}} = 120 - 75{{\text{X}}_B}$ hence, the vapour pressure of pure A and B respectively (in Torr) are:
(A) 120, 75
(B) 120,195
(C) 120,45
(D) 75,45

Answer 1

To answer this question, you must recall Raoult's law. The Raoult’s law gives an explanation about the change occurring in the vapour pressure of the solvent in a solution. It is applicable only for dilute solutions in case of real gas but is applicable for ideal solutions at all concentrations and temperature.
Formula used: ${{\text{P}}_{\text{A}}} = {\text{P}}_{\text{A}}^{\text{o}}{{{\chi }}_{\text{A}}}$ and ${{\text{P}}_{\text{B}}} = {\text{P}}_{\text{B}}^{\text{o}}{{{\chi }}_{\text{B}}}$
Where ${{\text{P}}_{\text{A}}}$ represents the vapour pressure of A in the mixture
${{\text{P}}_{\text{B}}}$ represents the vapour pressure of A in the mixture
${\text{P}}_{\text{A}}^{\text{o}}$ represents the vapour pressure of pure A
${\text{P}}_{\text{B}}^{\text{o}}$ represents the vapour pressure of pure B
${{{\chi }}_{\text{A}}}$ represents the mole fraction of A in the solution
And, ${{{\chi }}_{\text{B}}}$ represents the mole fraction of B in the solution.

Complete step by step solution
Raoult’s law states that in a mixture of volatile liquids, the partial pressure of each component of the mixture is directly proportional to the mole fraction of the component in the mixture. The sum of the partial pressures of all the components gives us the total pressure of the system.
So the total pressure of the given mixture will be ${{\text{P}}_{{\text{Total}}}} = {{\text{P}}_{\text{A}}} + {{\text{P}}_{\text{B}}}$
Using the Raoult’s law, we can write, ${{\text{P}}_{{\text{total}}}} = {\text{P}}_{\text{A}}^{\text{o}}{{{\chi }}_{\text{A}}} + {\text{P}}_{\text{B}}^{\text{o}}{{{\chi }}_{\text{B}}}$
We know that the mixture consists only of two components, i.e. A and B. So The sum of the mole fractions of A and B will be unity. Thus, we can write,
${{{\chi }}_{{A}}} + {{{\chi }}_{\text{B}}} = 1$
$\Rightarrow {{{\chi }}_{\text{A}}} = 1 - {{{\chi }}_{\text{B}}}$
Substituting this value in the equation of total pressure, we get,
${{\text{P}}_{{\text{total}}}} = {\text{P}}_{\text{A}}^{\text{o}}\left( {1 - {{{\chi }}_{\text{B}}}} \right) + {\text{P}}_{\text{B}}^{\text{o}}{{{\chi }}_{\text{B}}}$
$\Rightarrow {{\text{P}}_{{\text{total}}}} = {\text{P}}_{\text{A}}^{\text{o}} - {\text{P}}_{\text{A}}^{\text{o}}{{{\chi }}_{\text{B}}} + {\text{P}}_{\text{B}}^{\text{o}}{{{\chi }}_{\text{B}}}$
Simplifying:
$\Rightarrow {{\text{P}}_{{\text{total}}}} = {\text{P}}_{\text{A}}^{\text{o}} + \left( {{\text{P}}_{\text{B}}^{\text{o}} - {\text{P}}_{\text{A}}^{\text{o}}} \right){{{\chi }}_{\text{B}}}$
Comparing this equation with the relation given in the question, ${{\text{P}}_{{\text{Total}}}} = 120 - 75{{\text{X}}_B}$ , we get,
${\text{P}}_{\text{A}}^{\text{o}} = 120{\text{ torr}}$ and ${\text{P}}_{\text{B}}^{\text{o}} = 45{\text{ torr}}$
The correct answer is C.

Note
Raoult’s law states that in a mixture of volatile liquids, the partial pressure of each component of the mixture is directly proportional to the mole fraction of the component in the mixture. It is applicable only for dilute solutions in case of real gas but is applicable for ideal solutions at all concentrations and temperature.