= Solution
{c}
The leading coefficient fixes the atom count:
$$
N=\frac{25}{3k_B}=6.04\times10^{23},\qquad n_{\mathrm{mol}}=\frac{25}{3R}=1.002\,\mathrm{mol}.
$$
With molar mass $40\,\mathrm{g\,mol^{-1}}$, the sample mass is
$$
\boxed{m_{\mathrm{sample}}\simeq40.1\,\mathrm g.}
$$
One atom belongs to each <primitive unit cell> of monatomic <argon> on a <face-centered cubic lattice>, but the conventional cube contains four atoms. If its side is $a$, $N=4V/a^3$, and nearest neighbors lie across half a face diagonal. Therefore
$$
\boxed{d=\frac a{\sqrt2}=\frac1{\sqrt2}\left(\frac{4V}{N}\right)^{1/3}\simeq3.66\times10^{-10}\,\mathrm m=3.66\,\text{\AA},}
$$
using $V=21\times10^{-6}\,\mathrm m^3$. The corresponding conventional lattice constant is $a\simeq5.18\,\text{\AA}$. Treating the source's one-atom cell as the conventional cube would omit the factor four and give a wrong nearest-neighbor distance.
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