Question

A solution of two liquids A and B with $30$ mole $%$ of A is in equilibrium with its vapor which contains $60$ mole $%$ of A. Calculate the ratio of vapor pressure of pure A and pure B. Assuming ideal behavior in liquid and vapor state.

Answer 1

Hint: Raoult’s law states that the vapor pressure of the solution is equal to the product of vapor pressure of solvent and the mole fraction of the solvent. All solid and liquid have vapor pressure and this pressure is constant regardless of the fact that how much the substance is present.

Complete step-by-step answer:According to the question,
It is given that mole $%$ of A is $30$
Therefore, the mole of A will be $0.3$
And the mole $%$ of B is $70$
Therefore, the mole of B will be $0.7$
As we know that,
${{x}_{A}}=\dfrac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}}$
Where, ${{x}_{A}}$ is the mole fraction of A and ${{n}_{A}}$ is the moles of A, ${{n}_{B}}$ is the moles of B and ${{n}_{A}}+{{n}_{B}}$ is the total number of moles.
In liquid phase, mole fraction of A $({{x}_{A}})=\dfrac{0.3}{0.3+0.7}$
$\Rightarrow {{x}_{A}}=0.3$
And mole fraction of B $({{x}_{B}})=\dfrac{0.7}{0.3+0.7}$
$\Rightarrow {{x}_{B}}=0.7$
In vapor phase, mole fraction of A is $x{{'}_{A}}$ and mole fraction of $x{{'}_{B}}$
Similarly, in vapor phase the mole fraction of A is,
\[x{{'}_{A}}=\dfrac{0.6}{0.6+0.4}\]
\[\Rightarrow x{{'}_{A}}=0.6\]
And the mole fraction of B is,
\[x{{'}_{B}}=\dfrac{0.4}{0.6+0.4}\]
\[\Rightarrow x{{'}_{B}}=0.4\]
By using Raoult’s law,
\[{{P}_{A}}=P_{A}^{{}^\circ }{{x}_{A}}\]
\[{{P}_{B}}=P_{B}^{{}^\circ }{{x}_{B}}\]
Where, $P$ is the vapor pressure of solution, $x$ is the mole fraction of solvent, ${{P}^{{}^\circ }}$ is the vapor pressure of solvent.
In vapor phase, $x{{'}_{A}}=\dfrac{{{P}_{A}}}{{{P}_{T}}}$
Where, ${{P}_{T}}$ is the total pressure.
And $x{{'}_{B}}=\dfrac{{{P}_{B}}}{{{P}_{T}}}$
Now, if we put the value of ${{P}_{A}}$ and ${{P}_{B}}$ in the above formula, we get,
$x{{'}_{A}}=\dfrac{P_{A}^{{}^\circ }{{x}_{A}}}{{{P}_{T}}}$
$x{{'}_{B}}=\dfrac{P_{B}^{{}^\circ }{{x}_{B}}}{{{P}_{T}}}$
Now, if we divide $x{{'}_{A}}$ by $x{{'}_{B}}$ , we get,
$\dfrac{x{{'}_{A}}}{x{{'}_{B}}}=\dfrac{P_{A}^{{}^\circ }{{x}_{A}}{{P}_{T}}}{P_{B}^{{}^\circ }{{x}_{B}}{{P}_{T}}}$
$\Rightarrow \dfrac{x{{'}_{A}}}{x{{'}_{B}}}=\dfrac{P_{A}^{{}^\circ }{{x}_{A}}}{P_{B}^{{}^\circ }{{x}_{B}}}$
Hence, $\dfrac{P_{A}^{{}^\circ }}{P_{B}^{{}^\circ }}=\dfrac{x{{'}_{A}}{{x}_{B}}}{x{{'}_{B}}{{x}_{A}}}$
Now, if we substitute the values of \[{{x}_{A}},{{x}_{B}},x{{'}_{A}},x{{'}_{B}}\] , we get,
$\dfrac{P_{A}^{{}^\circ }}{P_{B}^{{}^\circ }}=\dfrac{0.6\times 0.3}{0.4\times 0.7}$
On further solving, we get,
$\dfrac{P_{A}^{{}^\circ }}{P_{B}^{{}^\circ }}=\dfrac{7}{2}=3.5$
The ratio of pure A and pure B is $3.5$

Note:There are some limitations of Raoult’s law:
Raoult’s law is suitable for describing ideal solutions but it is very difficult to find ideal solutions because they are very rare.
Mixtures contain different types of liquids that do not show the same uniformity in terms of attractive forces therefore, this type of solution deviates away from the law. This deviation can be either negative or positive. The negative deviation occurs when the vapor pressure is lower than the expectation. Positive deviation occurs when the cohesion between the similar molecule is greater or it exceeds adhesion between dissimilar molecules.