= Solution
{c}
The number of formula units per volume follows from the mass <density>:
$$
n_f=\frac{5.9\times10^3\,\mathrm{kg\,m^{-3}}}{(69.7+74.9)(1.66053906892\times10^{-27}\,\mathrm{kg})}
=2.4572\times10^{28}\,\mathrm{m^{-3}}.
$$
Each neutral formula unit contains $31+33=64$ <electrons>, so $\rho_e=64n_f=1.5726\times10^{30}\,\mathrm{m^{-3}}$. Let $R^2$ be the electron-weighted <mean-square orbital radius>, averaged over all these <electrons>. The supplied <Larmor-Langevin susceptibility> then gives the dimensionless SI <volume magnetic susceptibility>
$$
\chi_0=-\frac{\mu_0\rho_e e^2}{6m_e}R^2,
\qquad
\boxed{\chi_0=-2.5990\times10^{-5}\left(\frac{R}{a_0}\right)^2,}
$$
where $a_0=0.0529177\,\mathrm{nm}$ is the <Bohr radius>. An explicit rough atomic-scale assumption $R\simeq a_0$ gives $\chi_0\simeq-2.6\times10^{-5}$.
The quoted bulk data do not determine $R$: a nuclear charge does not specify an orbital radius in a screened many-<electron> <chemical atom>. Consequently the numerical estimate just made has atomic-model uncertainty; it is not a deduction with guaranteed 30 percent accuracy. For example, the alternative assumed $R=2a_0$ gives four times the magnitude while leaving every supplied bulk number unchanged. Atomic <wavefunctions>, an explicit screened-orbital model, or measured atomic <magnetic susceptibilities> of materials exhibiting <diamagnetism> would be needed to support that accuracy. Keeping $R$ explicit makes the later comparison independent of an unadvertised radius assumption. The bare <electron mass> belongs in this atomic formula; the <conduction band> <effective mass> is relevant instead to the large donor envelope in part (c).
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