The density of nitrogen at $\pu{0 ^\circ C}$ and $\pu{1 atm}$ is most nearly equal to which of the following quantities?
(a) $\pu{0.001 g/L}$; (b) $\pu{0.01 g/L}$; (c) $\pu{0.1 g/L}$; (d) $\pu{1 g/L}$; (e) $\pu{10 g/L}$.

The answer is (d), $\pu{1 g L-1}$.

I think this is how you do it, but please correct me if I'm missing something. I didn't end up using the standard molar volume, so this might be incorrect. $$pV = nRT$$

We know the pressure is $\pu{1 atm}$, the temperature is about $\pu{273 K}$ and $R = 0.08$.

Solving for $\frac {n}{V}$, we find $\frac {n}{V} = \frac {1}{22}$. Now, this is in moles per litre, so by multiplying by the molar mass of nitrogen (approximately $\pu{28 g mol-1}$), we find that we have about $\pu{1 g}$ per litre as the density.

  • 2
    $\begingroup$ "… the following quantities" – These are not "quantities", they are "quantity values", "values of a quantity", or simply "values". $\endgroup$
    – user7951
    Jun 13, 2020 at 15:51
  • $\begingroup$ Your approach is correct. But use units whenever needed and follow correct significant figures. $\endgroup$ Jun 14, 2020 at 7:02
  • $\begingroup$ @FadedGiant According to my reading of IUPAC's defintion of "quantity" (goldbook.iupac.org/terms/view/Q04982), the HW question's use of the term is acceptable: "The term quantity may refer to a quantity in a general sense, for example length, mass, or to a particular quantity, for example length of a particular rod, mass of a specified object". Hence quantity could also refer to the density of a particular sample of gas, which is precisely what the HW question is listing. $\endgroup$
    – theorist
    Jun 15, 2020 at 3:38
  • $\begingroup$ Also, "quantity value" and "value" are problematic, since if you take quantity in the first sense IUPAC uses then for, say, "0.01 g/L" the quantity would be "density", the value would be "0.01", and the units would be "g/L". See, for instance, how NIST distinguishes quantity, value, and units here: webbook.nist.gov/cgi/cbook.cgi?ID=C24203369&Mask=40 . Alas, this is inconsistent with how NIST uses "value of quant." here, which seems along the lines you're suggesting: nist.gov/pml/special-publication-811/… $\endgroup$
    – theorist
    Jun 15, 2020 at 3:53

1 Answer 1


Using the ideal gas law \begin{align} pV &= nRT, & \frac nV &= \frac{p}{RT}, \end{align} where \begin{align} T &= \pu{273 K},\\ p &= \pu{1 atm},\\ R &= \pu{0.0821 L atm mol-1 K-1},\\ M(\ce{N2}) &= \pu{28.0 g mol-1}, \end{align}
we get \begin{align} \frac{n}{V} &= \frac{\pu{1 atm}}{\pu{0.0821 L atm//mol K} \times\pu{273 K}},\\ \frac{n}{V} &= \pu{0.0446 mol//L}, \end{align}
and finally
$$ \rho(\ce{N2}) = \pu{0.0446 mol//L} \times \pu{28 g//mol} = \pu{1.2 g//L}. $$
Therefore answer (d) is correct.


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