# Using vapor mole fraction and pressure to determine liquid mole fraction

I am studying Chemistry by Zumdahl and Zumdahl (9th edition) and I have come across a question in Chapter 11 that I can't figure out the explanation for:

Liquid A has vapor pressure x , and liquid B has vapor pressure y . What is the mole fraction of the liquid mixture if the vapor above the solution is 30% A by moles?

I know that Raoult's Law tells us that the total vapor pressure is χx + (1-χ)y in this case. I also assume that the total vapor must be 0.3x + 0.7y given the other constraint of the question. This would mean the answer should be:

0.3x + 0.7y = χx + (1-χ)y

However this leads to the answer χ = 0.3, whereas the text gives the answer as 0.3y/(0.7x+0.3y)

What have I got wrong? (Also sorry for not formatting it properly but I'm a noob here and I can't figure out the equation formatting).

• Welcome to Chemistry.SE. A short introduction to the maths and chemistry notation using MathJax is here, with more information here. – Ben Norris Jul 16 '15 at 12:48

Raoult's law tells us the relationship between the composition of the liquid and the vapor pressure above it:

$$P_A=\chi_A P^\circ_A$$

Where:

• $P_A$ is the partial pressure of $A$ in the vapor above the solution
• $P^\circ_A$ is the vapor pressure of pure $A$
• $\chi_A$ is the mole fraction of $A$ in the solution

For a two component solution $A$ and $B$ (I don't want to use $x$ to avoid confusion with $\chi$) we can then use Dalton's Law to get:

$$P_T=P_A + P_B = \chi_A P^\circ_A +\chi_B P^\circ_B= \chi_A P^\circ_A +(1-\chi_A)P^\circ_B$$

However, from the definition of mole fraction we can derive a relation between the mole fraction in the vapor (from here on out $\phi$ to avoid confusion between vapor and solution, which is what your mistake is).

$$\phi_A=\frac{n_A}{n_T}$$

Where $n$ is the number of moles, and since at constant temperature and volume, $n\propto P$... $$\phi_A = \frac{n_A}{n_T}=\frac{P_A}{P_T}$$

So, we can rewrite $P_A = \phi_A P_T$ and $P_B=(1-\phi_A)P_T$

We are given $\phi_A=0.3$. Thus $\phi_B=0.7$

Here we go. Substitute $P_A=\chi_A P^\circ_A$ and $P_T=\chi_A P^\circ_A +(1-\chi_A)P^\circ_B$ into $P_A=0.3 P_T$:

$$\chi_A P^\circ_A=0.3\left(\chi_A P^\circ_A +(1-\chi_A)P^\circ_B\right)$$ $$0.7\chi_A P^\circ_A=0.3(1-\chi_A)P^\circ_B$$ $$0.7\chi_A P^\circ_A + 0.3\chi_A P^\circ_B=0.3P^\circ_B$$

I think you can get it from here.