A question in my book asks...

You have a gas X that is found in quantities of $261~\mathrm{PPTV}$ (parts per trillion by volume). Compute the molar concentration at $283.15~\mathrm{K}$ and $101325~\mathrm{Pa}$.

Would you just find the volume using $\frac{n\mathcal{R}T}{P}$ for $1~\mathrm{mol}$ of gas, then multiply this answer by $261/10^{12}$?

  • $\begingroup$ @hey It is better to use $\mathrm{...}$ instead of \text{...}$ and also the use of braces is highly recommended to avoid breaking away of the argument. See also here: meta.chemistry.stackexchange.com/q/443/4945 $\endgroup$ – Martin - マーチン Oct 3 '14 at 7:13
  • $\begingroup$ @Martin I will take care of it next time. $\endgroup$ – Freddy Oct 3 '14 at 8:24

I have the feeling, that this is a superbly stupid question. And I also have the feeling that it is quite possible, that my answer will be wrong.
First of all the parts-per-notation should be avoided as it is not compliant with SI and highly ambiguous.
Secondly, it generally may only refer to a unitless number - a molar concentration has a unit ($\mathrm{mol/L}$).

The here used unit may refer to a couple of things. In general its meaning should most likely refer to $\mathrm{PPTV = 10^{-12}L/L}$. Sometimes the use of $\mathrm{PPTV = 10^{-12}g/L}$

From here you can probably make your best guesses, determining how many moles of compound do you have: \begin{align} pV &=n\mathcal{R}T\\ n &= \frac{pV}{\mathcal{R}T}\\ n &= \frac{101325~\mathrm{Pa}\cdot 261\cdot10^{-12}~\mathrm{L}}{8.314~\mathrm{\frac{J}{mol\cdot K}}283.15~\mathrm{K}}\\ n &= 1.12\cdot10^{-8}~\mathrm{mol} &\implies c&= 1.12\cdot10^{-8}~\mathrm{\frac{mol}{L}} \end{align}

If you have the mass given, then you need to know which gas it is, as $n=\frac{m}{M}$.

Another possible use could be the particles by volume notation, so it could also refer to $\mathrm{PPTV = 10^{-12}/m^3}$. (Apparently this is somewhat common in ecological air measurements.) Here you simply have to use $n=\frac{N}{\mathcal{N}_\mathrm{A}}$ to figure out the concentration in $\mathrm{mol/L}$.

  • $\begingroup$ Thanks! I got the same answer but the question was so vague that I wasn't sure if it's correct. $\endgroup$ – Mike Oct 4 '14 at 5:48
  • $\begingroup$ In truth, the $ppX$ designation is defined on a mass basis, not on a volume basis. Some key assumptions must be applied about the affects of the additive on the solvent to convert the mass basis to a volume basis. $\endgroup$ – Jeffrey Weimer Oct 6 '14 at 14:58
  • $\begingroup$ @JeffreyWeimer I really do not understand what you want to say with your comment. Where do you get your definition from? As far as I can see, there is no official definition for this scheme. And could you elaborate more on what you mean with 'mass basis' please. $\endgroup$ – Martin - マーチン Oct 6 '14 at 15:52

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