Question

How do you calculate X (acetone) and X (cyclohexane) in the vapor above the solution. Po acetone is $229.5torr$ and Po cyclohexane is $97.6torr$.

Answer 1

Hint: We must know about Roult’s law before solving this question. Assume that we have a closed container filled with a volatile liquid A. After some time, due to evaporation, vapor particles of A will start to form. Then as time passes, the vapor particles of A will be in dynamic equilibrium with the liquid particles.

Complete answer:
Vapor pressure is exhibited by all solids and liquids and depends only on the type of liquid and temperature. Now imagine we are adding another liquid B (solute) to this container. This will result in B particles occupying the space between A particles on the surface of the solution. For any given liquid there are a fraction of molecules on the surface which will have sufficient energy to escape to the vapor phase. Since now we have a lesser number of A particles on the surface, the number of vapor particles of A in the vapor phase will be lesser. This will result in lower vapor pressure of A. Now if we assume that B is volatile as well, we will have a lesser number of B particles in the vapor phase as compared to pure liquid B.
To start solving this question we must first calculate the mole fractions for both acetone and cyclohexane.
In case of acetone:
$ = \dfrac{{2.88}}{{\left( {2.88 + 1.45} \right)}}$
$ = \dfrac{{2.88}}{{4.33}} = 0.665mol$
In case of cyclohexane:
$ = \dfrac{{1.45}}{{\left( {2.88 + 1.45} \right)}}$
On simplification we get,
$ = \dfrac{{1.45}}{{4.33}} = 0.335mol$
Raoult's law states that:
${P_{\left( {Total} \right)}} = {P_{\left( {acetone} \right)}} \times {X_{\left( {acetone} \right)}} + {P_{\left( {cyclohexane} \right)}} \times {X_{\left( {cyclohexane} \right)}}$
Now we can substitute the known given values we get,
\[{P_{\left( {Total} \right)}} = (229.5 \times 0.665) + \left( {97.6 \times 0.335} \right)\]
${P_{\left( {Total} \right)}} = 152.61 + 32.696$
On addition we get,
P(total) \[ = 185.3{\text{ }}torr\]
x(mol fraction of acetone)\[ = \dfrac{{\left( {229.5 \times 0.665} \right)}}{{185.3torr}}\]
\[{X_{\left( {acetone} \right)}} = 0.823\]
x(mol fraction of cyclohexane) $ = \dfrac{{\left( {97.6 \times 0.335} \right)}}{{185.3torr}}$
\[{X_{\left( {cyclohexane} \right)}} = 0.177\]

Note:
Few limitations of Roult’s law can be that Raoult's law is apt for describing ideal solutions. However, ideal solutions are hard to find and they are rare. Different chemical components have to be chemically identical equally. Since many of the liquids that are in the mixture do not have the same uniformity in terms of attractive forces, these types of solutions tend to deviate away from the law.