Finding the density of nitrogen at 0 degrees Celsius and 1 atmosphere 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.
Using the ideal gas law
\begin{align}
  pV &amp;= nRT, &amp;
  \frac nV &amp;= \frac{p}{RT},
\end{align}
where
\begin{align}
  T &amp;= \pu{273 K},\\
  p &amp;= \pu{1 atm},\\
  R &amp;= \pu{0.0821 L atm mol-1 K-1},\\
  M(\ce{N2}) &amp;= \pu{28.0 g mol-1},
\end{align}
we get
\begin{align}
  \frac{n}{V} &amp;= 
    \frac{\pu{1 atm}}{\pu{0.0821 L atm//mol K} \times\pu{273 K}},\\
  \frac{n}{V} &amp;= \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.