Cobalt has a density of 8.90 g/cm³ and a molecular mass of 58.9 g/mol.Its magnetic moment is 1.69 Bohr magnetons (1 Bohr magneton = 9.27 $\times$ 10^-24 A · m²).What is its saturation magnetization? A)1.43 $\times$ 10⁹ A/m B)9.10 $\times$ 10⁵ A/m C)1.43 $\times$ 10⁶ A/m D)9.10 $\times$ 10²⁸ A/m E)1.69 $\times$ 10⁵ A/m

Question

Quizplus · Accepted Answer

The saturation magnetization can be calculated using the formula: $M_s = \frac{\mu_0 \cdot N \cdot \mu}{V}$, where $\mu_0$ is the magnetic moment per atom (1.69 Bohr magnetons), $N$ is the number of atoms per unit volume, and $V$ is the volume. Given the density ($8.90 \, \text{g/cm}^3$) and the molecular mass ($58.9 \, \text{g/mol}$), we can find $N$ and then calculate $M_s$. The correct calculation leads to the answer in option C, which corresponds to the correct magnitude for saturation magnetization in the given units.

Cobalt Has a Density of 8 ×\times× 10-24 a · M2) ×\times×

Cobalt Has a Density of 8 $\times$ 10^-24 a · M²) $\times$