Answer
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Hint: An ideal solution is that solution in which each component obeys Raoult’s law under all conditions of temperatures and concentrations.
In general, for a solution containing a number of volatile components (liquids), for any component ‘I’,
${{\text{P}}_{\text{i}}}{\text{ = }}{{\text{Y}}_{\text{i}}}{{ \times }}{{\text{P}}_{{\text{Total}}}}$
Here, ${{\text{P}}_{\text{i}}}$ is partial vapor pressure of the component ‘i’, ${{\text{Y}}_{\text{i}}}$ is the mole fraction of the component ‘i’ in the vapor phase and ${{\text{P}}_{{\text{Total}}}}$ is the total pressure of the system.
Complete step by step answer:
According to Raoult’s law for volatile solutes, i.e., for liquid – liquid solutions, the vapour pressure of a component at a given temperature in a solution is equal to the product of the mole fraction of that component in the solution and the vapor pressure of that component in the pure state.
According to the question, A and B are two completely miscible volatile liquids having mole fractions ${{\text{X}}_{\text{A}}}$ and ${{\text{X}}_{\text{B}}}$ respectively in the liquid phase. Their partial vapor pressures are ${{\text{P}}_{\text{A}}}$ and ${{\text{P}}_{\text{B}}}$ respectively. Also, ${{\text{P}}_{\text{A}}}^0$ and ${{\text{P}}_{\text{B}}}^0$ are the vapor pressures of A and B respectively in the pure state. Then, according to Raoult’s law,
${{\text{P}}_{\text{A}}} = {{\text{X}}_{\text{A}}} \times {{\text{P}}_{\text{A}}}^0$
And ${{\text{P}}_{\text{B}}} = {{\text{X}}_{\text{B}}} \times {{\text{P}}_{\text{B}}}^0$
Thus, ${{\text{P}}_{\text{A}}}^0 = \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{X}}_{\text{A}}}}}$
And ${{\text{P}}_{\text{B}}}^0 = \dfrac{{{{\text{P}}_{\text{B}}}}}{{{{\text{X}}_{\text{B}}}}}$
According to the question, ${{\text{P}}_{\text{A}}}^0 > {{\text{P}}_{\text{B}}}^0$, thus,
$\dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{X}}_{\text{A}}}}} > \dfrac{{{{\text{P}}_{\text{B}}}}}{{{{\text{X}}_{\text{B}}}}}$
Rearrange the relation to get:
$
\dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{B}}}}} > \dfrac{{{{\text{X}}_{\text{A}}}}}{{{{\text{X}}_{\text{B}}}}} \\
\Rightarrow \dfrac{{{{\text{X}}_{\text{A}}}}}{{{{\text{X}}_{\text{B}}}}} < \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{B}}}}} \to 1 \\
$
The composition of the vapour phase in equilibrium with the solution can be calculated from the partial pressures of the two components in the vapor phase. By question, ${{\text{Y}}_{\text{A}}}$ is the mole fraction of A in vapour phase and ${{\text{Y}}_{\text{B}}}$ is the mole fraction of B in vapour phase. If ${{\text{P}}_{{\text{Total}}}}$ is the total pressure of the system, then:
${{\text{Y}}_{\text{A}}} = \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{{\text{Total}}}}}}$
And ${{\text{Y}}_{\text{B}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{B}}}}}{{{{\text{P}}_{{\text{Total}}}}}}$
Thus,
$
\dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{B}}}}} = \dfrac{{\dfrac{{{{\text{Y}}_{\text{A}}}}}{{{{\text{P}}_{{\text{Total}}}}}}}}{{\dfrac{{{{\text{Y}}_{\text{B}}}}}{{{{\text{P}}_{{\text{Total}}}}}}}} \\
\Rightarrow \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{B}}}}} = \dfrac{{{{\text{Y}}_{\text{A}}}}}{{{{\text{Y}}_{\text{B}}}}} \to 2 \\
$
Combining the equation 1 and the equation 2, we get:
$\dfrac{{{{\text{X}}_{\text{A}}}}}{{{{\text{X}}_{\text{B}}}}} < \dfrac{{{{\text{Y}}_{\text{A}}}}}{{{{\text{Y}}_{\text{B}}}}}$
Therefore, option C is correct.
Note:
The Raoult’s law is applicable only if the two components form a solution (a homogeneous mixture). It is not applicable if the two liquids are not completely miscible. Liquid pairs that are ideal solutions are benzene plus toluene, hexane plus heptanes etc. Depending upon the vapor pressures of the pure components, total vapor pressure over the solution may increase or decrease with increase in the mole fraction of a component.
In general, for a solution containing a number of volatile components (liquids), for any component ‘I’,
${{\text{P}}_{\text{i}}}{\text{ = }}{{\text{Y}}_{\text{i}}}{{ \times }}{{\text{P}}_{{\text{Total}}}}$
Here, ${{\text{P}}_{\text{i}}}$ is partial vapor pressure of the component ‘i’, ${{\text{Y}}_{\text{i}}}$ is the mole fraction of the component ‘i’ in the vapor phase and ${{\text{P}}_{{\text{Total}}}}$ is the total pressure of the system.
Complete step by step answer:
According to Raoult’s law for volatile solutes, i.e., for liquid – liquid solutions, the vapour pressure of a component at a given temperature in a solution is equal to the product of the mole fraction of that component in the solution and the vapor pressure of that component in the pure state.
According to the question, A and B are two completely miscible volatile liquids having mole fractions ${{\text{X}}_{\text{A}}}$ and ${{\text{X}}_{\text{B}}}$ respectively in the liquid phase. Their partial vapor pressures are ${{\text{P}}_{\text{A}}}$ and ${{\text{P}}_{\text{B}}}$ respectively. Also, ${{\text{P}}_{\text{A}}}^0$ and ${{\text{P}}_{\text{B}}}^0$ are the vapor pressures of A and B respectively in the pure state. Then, according to Raoult’s law,
${{\text{P}}_{\text{A}}} = {{\text{X}}_{\text{A}}} \times {{\text{P}}_{\text{A}}}^0$
And ${{\text{P}}_{\text{B}}} = {{\text{X}}_{\text{B}}} \times {{\text{P}}_{\text{B}}}^0$
Thus, ${{\text{P}}_{\text{A}}}^0 = \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{X}}_{\text{A}}}}}$
And ${{\text{P}}_{\text{B}}}^0 = \dfrac{{{{\text{P}}_{\text{B}}}}}{{{{\text{X}}_{\text{B}}}}}$
According to the question, ${{\text{P}}_{\text{A}}}^0 > {{\text{P}}_{\text{B}}}^0$, thus,
$\dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{X}}_{\text{A}}}}} > \dfrac{{{{\text{P}}_{\text{B}}}}}{{{{\text{X}}_{\text{B}}}}}$
Rearrange the relation to get:
$
\dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{B}}}}} > \dfrac{{{{\text{X}}_{\text{A}}}}}{{{{\text{X}}_{\text{B}}}}} \\
\Rightarrow \dfrac{{{{\text{X}}_{\text{A}}}}}{{{{\text{X}}_{\text{B}}}}} < \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{B}}}}} \to 1 \\
$
The composition of the vapour phase in equilibrium with the solution can be calculated from the partial pressures of the two components in the vapor phase. By question, ${{\text{Y}}_{\text{A}}}$ is the mole fraction of A in vapour phase and ${{\text{Y}}_{\text{B}}}$ is the mole fraction of B in vapour phase. If ${{\text{P}}_{{\text{Total}}}}$ is the total pressure of the system, then:
${{\text{Y}}_{\text{A}}} = \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{{\text{Total}}}}}}$
And ${{\text{Y}}_{\text{B}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{B}}}}}{{{{\text{P}}_{{\text{Total}}}}}}$
Thus,
$
\dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{B}}}}} = \dfrac{{\dfrac{{{{\text{Y}}_{\text{A}}}}}{{{{\text{P}}_{{\text{Total}}}}}}}}{{\dfrac{{{{\text{Y}}_{\text{B}}}}}{{{{\text{P}}_{{\text{Total}}}}}}}} \\
\Rightarrow \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{B}}}}} = \dfrac{{{{\text{Y}}_{\text{A}}}}}{{{{\text{Y}}_{\text{B}}}}} \to 2 \\
$
Combining the equation 1 and the equation 2, we get:
$\dfrac{{{{\text{X}}_{\text{A}}}}}{{{{\text{X}}_{\text{B}}}}} < \dfrac{{{{\text{Y}}_{\text{A}}}}}{{{{\text{Y}}_{\text{B}}}}}$
Therefore, option C is correct.
Note:
The Raoult’s law is applicable only if the two components form a solution (a homogeneous mixture). It is not applicable if the two liquids are not completely miscible. Liquid pairs that are ideal solutions are benzene plus toluene, hexane plus heptanes etc. Depending upon the vapor pressures of the pure components, total vapor pressure over the solution may increase or decrease with increase in the mole fraction of a component.
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