2
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • 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. $\endgroup$
    – Mathew Mahindaratne
    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

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.