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).




[Raoult's law](https://en.wikipedia.org/wiki/Raoult%27s_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.


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