Answer
$μ=5.15 \times 10^{-24} \mathrm{~A} \cdot \mathrm{m}^2 .
$$
Work Step by Step
$
\begin{aligned}
n & =\frac{\rho N_A}{M}=\frac{\left(8.90 \mathrm{~g} / \mathrm{cm}^3\right)\left(6.02 \times 10^{23} \text { atoms } / \mathrm{mol}\right)}{58.71 \mathrm{~g} / \mathrm{mol}}=9.126 \times 10^{22} \text { atoms } / \mathrm{cm}^3 \\
& =9.126 \times 10^{23} \text { atoms } / \mathrm{m}^3
\end{aligned}
$
The dipole moment of a single atom of nickel is
$
\mu=\frac{M_{\text {stt }}}{n}=\frac{4.70 \times 10^5 \mathrm{~A} / \mathrm{m}}{9.126 \times 10^{28} \mathrm{~m}^3}=5.15 \times 10^{-24} \mathrm{~A} \cdot \mathrm{m}^2 .
$
