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A quantity of benzene, $\ce{C6H6}$, and toluene, $\ce{C6H5CH3}$, is placed in a $1~\mathrm{L}$ evacuated vessel at $25~\mathrm{^\circ C}$. At equilibrium, a small volume of liquid is visible at the bottom of the container. A sample of the vapour phase is analysed and found to contain $53~\mathrm{mol-\%}$ benzene. What is the mole fraction of benzene in the liquid phase?

Vapour pressures at 25 °C: benzene = 0.125 atm, toluene = 0.037 atm

My attempt at the problem:
First of all, I am not even certain if this a Raoult's law problem. I just assumed so, since Henry's constant was not given in the problem.

First I'm getting the total liquid pressure using Raoult's law. \begin{align} P(\text{benzene}) &= X(\text{benzene}) \cdot P_\mathrm{vap}(\text{benzene}) & &= 0.53 \times 0.125 &&= 0.06625~\mathrm{atm}\\ P(\text{toluene}) &= X(\text{toluene}) \cdot P_\mathrm{vap}(\text{toluene}) & &= 0.47 \times 0.037 &&= 0.01739~\mathrm{atm}\\ P_\mathrm{total} &&&= 0.06625 + 0.01739 &&= 0.08364~\mathrm{atm}\\ \end{align}

Then using Dalton's law, I am getting the mole fraction of Benzene in the liquid phase.
\begin{align} P(\text{benzene}) &= Y(\text{benzene}) \cdot P_\mathrm{total}\\ Y(\text{benzene}) &= \frac{P(\text{benzene})}{P_\mathrm{total}}\\ &= \frac{0.06625}{0.08364} = 0.792. \end{align}

However, my answer is incorrect according to the answer key that has $0.25$ as the answer. I have a feeling my whole approach may be wrong since my answer is off by a lot. What am I doing wrong?

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The question gives you the mol percentage of benzene ($53 \%$) in the vapour phase, and not in the liquid phase.

It's always good to write down what you have first. (note: I use $y$ to denote mole fractions in the vapour phase and $x$ to denote mole fractions in the liquid state)

First of all, we have $y_\text{benzene} = 0.53$. Additionally, we have been given the vapour pressure of pure benzene and toluene (denoted with a superscript $*$):

$$p^*_\text{benzene} = 0.125\ \mathrm{atm} \qquad p^*_\text{toluene} = 0.037\ \mathrm{atm} $$

and we require $x_\text{benzene}$

We also, know $p_\text{benzene} = x_\text{benzene}\ p^*_\text{benzene}$, and the total vapor pressure is given by $$P = p_\text{benzene} + p_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene} + x_\text{toluene}\ p^*_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}$$

Finally,

$$y_\text{benzene} = \frac{p_\text{benzene}}{P} = \frac{x_\text{benzene}\ p^*_\text{benzene}}{x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}}$$

It is now, trivial to solve this equation. Since I am lazy, and was not in the mood to write down two lines of algebra and reach for a calculator, I chucked it into Mathematica, and the result is:

enter image description here

Thus, $x_\text{benzene} = 0.25$, which is indeed the result given in your answer key.

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You solved the problem assuming that 0.53 is the mole fraction of benzene in the liquid phase. However, the problem statement says that "the vapor phase is analyzed and found to contain 53 mole percent benzene." Call P the total pressure. What is the partial pressure of benzene in the vapor? What is the partial pressure of toluene in the vapor? In terms of P, what is the mole fraction of benzene in the liquid? In terms of P, what is the mole fraction toluene in the liquid. In terms of P, what is the sum of the mole fractions of benzene and toluene in the liquid? From this equation, what is the total pressure P?

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  • $\begingroup$ Okay, so I retried again and was able to arrive at the correct answer with your steps, but I don't quite understand the chemistry behind it. Partial pressure of benzene in vapour is 0.53P, 0.47P for toluene. Mole fraction of benzene in liquid state is 0.53P/0.125 and for toluene it is 0.47P/0.037. This part, I am confused by why the partial pressure in vapour is used in Raoult's law when I am calculating mole fraction in liquid. $\endgroup$ – M.M Dec 18 '15 at 5:53
  • $\begingroup$ Raoult's law says that the partial pressure in the vapor is equal to the equilibrium vapor pressure of the pure substance at the given temperature times the mole fraction in the liquid. So, the mole fraction in the liquid is equal to the partial pressure in the vapor divided by the equilibrium vapor pressure of the pure substance. $\endgroup$ – Chet Miller Dec 18 '15 at 12:18
  • $\begingroup$ Thanks a lot! To clarify on the states, is it always the case that Raoult's law says partial pressure in vapor equals equilibrium vapor pressure times mole fraction of liquid? The partial pressure part is always in vapor right? However the mole fraction part can be liquid/liquid solution or liquid/solid solution? $\endgroup$ – M.M Dec 18 '15 at 14:36
  • $\begingroup$ Yes, for vapor-liquid equilibrium. For liquid/liquid or liquid/solid solution, a different law applies. $\endgroup$ – Chet Miller Dec 18 '15 at 22:10
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According to Raoults law, it should be 0,25. Lets say, you have the molar fraction of each component in vapor, and lets say you have the saturated pressure at some temperature, than according to Raoults law you can say: 0,53(molar fraction of benzene in vapor)Pi(which is equal to Xbpb(saturated, benzene)+Xtpt(saturated, toluene) = Xbpb than: Xb*0,125=0,53*(Xb*0,125+(1-Xb)*0,037) Xb=0,25

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