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You have a mixture of $\ce{He, Ne, Ar}$ gases with a total pressure of $8.40$ atmospheres. $\ce{He}$ has a pressure of $1.50$ and $\ce{Ne}$ has $2.00$. What is the molar fraction of $\ce{Ar}$?

The molar fraction is

$$\frac{\text{moles of Ar}}{\text{total moles of mixture}}$$

However, to know the number of moles of any of these components, I need to know their mass in the mixture. How can I know if all I am given is their pressures?

The answer is $0.583$ and I am not sure how was this concluded. All I can tell is that $\ce{Ar}$ has a pressure of $4.9$ atmospheres by Dalton's law but that's it. By the same law I know that all gases there have the same temperature and volume, but since I don't know either value for any of them it isn't useful at all.

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The molar fraction of a gas can be calculated as

$$ x_a=\frac{n_a}{n}= \frac{V_a}{V} = \frac{p_a}{p} $$

In your case, firstly you need to calculate the partial pressure of Argon. Dalton's law states that in a mixture of non-reacting gases the total pressure exerted equals the sum of the partial pressures of the individual gases. Thus $p = p_{He} + p_{Ne} + p_{Ar} $, from which you get $ p_{Ar} $ = 4,9 atm.

So the molar fraction of Argon is $ x_a = \frac{p_a}{p} =\frac{4.90 atm}{8.40 atm} = 0,583 $

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  • $\begingroup$ Hi! Ok, I understand that the pressure of $\ce{Ar}$ is $4.9$. However, my understanding is that the molar fraction is $\text{moles of Ar} / \text{total moles}$, so I need the number of moles for each substance in the mixture, not the pressures. Or perhaps the molar fraction can also be calculated by $\text{pressure of Ar} / \text{total pressure}$? $\endgroup$ – Voldemort Jul 15 '15 at 16:48
  • $\begingroup$ As you said it yourself, the molar fraction can also be calculated from pressure data. If I remember correctly, it is so beacuse of the combined gas law. $\endgroup$ – user17508 Jul 15 '15 at 16:54
  • $\begingroup$ Can it be calculated with something else? Other than mole and pressure data? $\endgroup$ – Voldemort Jul 15 '15 at 16:55
  • $\begingroup$ Also from volume data. But in this exercice just from pressure data. $\endgroup$ – user17508 Jul 15 '15 at 16:59
  • $\begingroup$ @midi-chlorian this is a homework question that should not to be answered completely as you did because this site is no homework portal. Instead the questioner should be guided to get to the answer themself. $\endgroup$ – pH13 - Yet another Philipp Jul 15 '15 at 17:07
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Your mole fraction is $$x_\ce{Ar}=\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}$$ and you somehow want to insert your pressures. You can try to convert all moles that you don't know into pressures that you know by using the ideal gas law $$p~V=n~R~T$$ which will make you find that everything you have (1.5, 2.0, 4.9 atm) is all what you need to come up with $x_\ce{Ar}=0.583$.


As your question was answered by midi-chlorian without further explanation, here is why you don't need more than those pressures given to you.

I said that you can use the ideal gas law to find out the amount of substance of each gas. Therefor we need to rearrange it to give $$n=\frac{p~V}{R~T}$$ We know the gas constant is constant ... that's where the name comes from ... and we can assume the temperature to be constant at whatever temperature we want ... it's not important to know the temperature here. On top of that the missing volume of each gas is equal for all three ideal gases, as Avogadro's law says.

This means that not only R and T are constant but that the volume will also not change, which in turn simplifies everything a lot. $$n=p~\frac{V_m}{R~T}=p~k$$ Let's insert it into the mole fraction (as I told you to do) and see what happens. $$\begin{align} x_\ce{Ar}=&~\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}\\ =&~\frac{p_\ce{Ar}~k}{p_\ce{He}~k+p_\ce{Ne}~k+p_\ce{Ar}~k}\\=&~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\cdot\frac{k}{k}\\=&~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\end{align}$$

And this is the simple reason why you do not need more than your pressures.

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  • $\begingroup$ But to use the ideal gas law I need the temperature and volume to be able to calculate the number of moles for each gas? $\endgroup$ – Voldemort Jul 15 '15 at 16:44
  • $\begingroup$ You don't. Just write it down and you'll see. $\endgroup$ – pH13 - Yet another Philipp Jul 15 '15 at 16:50
  • $\begingroup$ Hmm... let's try for Ar: $$4.9 \cdot V = n \cdot 0.0821 \cdot T$$ then $$\frac{4.9 \cdot V}{0.0821 \cdot T} = n$$ To find the number of moles $n$ I would need $V$ and $T$? (Clearly there's something I don't quite grasp about the usage of this formula sorry!) $\endgroup$ – Voldemort Jul 15 '15 at 16:54
  • $\begingroup$ Ideal gases do all have the same volume. R is a constant by definition and as the temperature will not change it is also a constant. Now don't enter any numbers, just write down what you get after rearranging the ideal gas law and inserting it into the mole fraction. $\endgroup$ – pH13 - Yet another Philipp Jul 15 '15 at 17:01

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