= Solution
The <mass> per <lead> atom is
$$
m=\frac{0.2070\,\mathrm{kg\,mol^{-1}}}{N_A}=3.4373\times10^{-25}\,\mathrm{kg}.
$$
Let $u_n$ be its longitudinal displacement. <Newton's second law> gives the <harmonic chain> equation
$$
m\ddot u_n=\kappa(u_{n+1}+u_{n-1}-2u_n).
$$
Insert the <normal mode> $u_n=u\,e^{i(kna-\omega t)}$ to obtain
$$
\omega^2=\frac{4\kappa}{m}\sin^2\frac{ka}{2}.
$$
At long <wavelength>, $\omega\simeq a\sqrt{\kappa/m}\,|k|$. The <sound speed> is the corresponding low-<frequency> <group velocity>:
$$
\boxed{v_s=a\sqrt{\frac\kappa m}=2.1155\times10^3\,\mathrm{m\,s^{-1}}.}
$$
The relevant inertia in this chain equation is one atomic <mass>. The reduced <mass> $m/2$ belongs to the relative vibration of an isolated pair and should not be substituted for $m$ in the chain's <dispersion relation>.
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