A non-volatile solute (A) is dissolved in a volatile solvent (B). The vapor pressure of the solution is ${P_s}$. The vapor pressure of pure solvent is $P_B^0$.
If X is the mole fraction then which of the following is correct.
A. ${p_s} = p_B^0.{X_A}$
B. ${p^0} = {p_B}.{X_B}$
C. ${p_s} = p_B^0.{X_B}$
D. $p_B^0 = {p_s}.{X_A}$
Answer
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Hint: In a binary solution of two volatile liquids, let’s denote the two components as 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually an equilibrium would be attained between the vapour phase and the liquid phase.
Complete step by step answer:
Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2. The French chemist, Francois Marta Raoult (1886) gave the quantitative relationship between them. The relationship is known as the Raoult’s law which states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. Thus, for component 1.
\[{p_1} \propto {x_1}\]
\[{p_1} = p_1^0{x_1}\]
where \[p_1^0\] is the vapour pressure of pure component 1 at the same temperature.
Similarly, for component 2.
\[{p_2} = p_2^0{x_2}\]
where $p_2^0$ represents the vapour pressure of the pure component 2.
According to Dalton’s law of partial pressures, the total pressure over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as:
\[{p_{total}}{\rm{ }} = {\rm{ }}{p_1}{\rm{ }} + {\rm{ }}{p_2}\] Substituting the values of ${p_1}$ and${p_2}$, we get,
\[{p_{total}}{\rm{ }} = {\rm{ }}{x_1}p_1^0{\rm{ }} + {\rm{ }}{x_2}p_2^0\]
For non-volatile solute:
$P_1^0 = 0$
Similarly, in our question we have A non-volatile solute dissolved in volatile solvent B.
So, by Raoult’s Law,
\[{p_s}{\rm{ }} = {\rm{ P}}_B^0{{\rm{X}}_B}{\rm{ }} + {\rm{ P}}_A^0{X_A}\]
Since A is non-volatile,
$P_A^0 = 0$
So, \[{p_s}{\rm{ }} = {\rm{ P}}_B^0{{\rm{X}}_B}\]
Therefore, from the above explanation the correct option is (C).
Note:
The total vapour pressure over the solution can be associated to the mole fraction of any one component and it is also related to the mole fraction of the other component.
Complete step by step answer:
Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2. The French chemist, Francois Marta Raoult (1886) gave the quantitative relationship between them. The relationship is known as the Raoult’s law which states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. Thus, for component 1.
\[{p_1} \propto {x_1}\]
\[{p_1} = p_1^0{x_1}\]
where \[p_1^0\] is the vapour pressure of pure component 1 at the same temperature.
Similarly, for component 2.
\[{p_2} = p_2^0{x_2}\]
where $p_2^0$ represents the vapour pressure of the pure component 2.
According to Dalton’s law of partial pressures, the total pressure over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as:
\[{p_{total}}{\rm{ }} = {\rm{ }}{p_1}{\rm{ }} + {\rm{ }}{p_2}\] Substituting the values of ${p_1}$ and${p_2}$, we get,
\[{p_{total}}{\rm{ }} = {\rm{ }}{x_1}p_1^0{\rm{ }} + {\rm{ }}{x_2}p_2^0\]
For non-volatile solute:
$P_1^0 = 0$
Similarly, in our question we have A non-volatile solute dissolved in volatile solvent B.
So, by Raoult’s Law,
\[{p_s}{\rm{ }} = {\rm{ P}}_B^0{{\rm{X}}_B}{\rm{ }} + {\rm{ P}}_A^0{X_A}\]
Since A is non-volatile,
$P_A^0 = 0$
So, \[{p_s}{\rm{ }} = {\rm{ P}}_B^0{{\rm{X}}_B}\]
Therefore, from the above explanation the correct option is (C).
Note:
The total vapour pressure over the solution can be associated to the mole fraction of any one component and it is also related to the mole fraction of the other component.
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