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A metal exists as an FCC crystal. If the atomic radius is [tex]100.2 \, \text{pm}[/tex] and the density of the metal is [tex]12,500 \, \text{kg/m}^3[/tex], what is the atomic mass of the metal?

Answer :

The atomic mass of the metal is 52.65 g/mol is obtained by solving the equation relating density, atomic mass, and unit cell volume.

To find the atomic mass of the metal, we can utilize the formula:

Density = [tex]\frac{\text{Atomic mass} \times \text{Avogadro's number}}{\text{Volume of unit cell} \times \text{Volume occupied by one atom}}[/tex]

First, we need to calculate the volume of the unit cell:

Volume of unit cell = (2 × Atomic radius³

Volume of unit cell = (2 × 100 × 10⁻¹² m³

Volume of unit cell = 8 × 10⁻²â´ m³

Given that the metal is FCC (face-centered cubic), each unit cell contains 4 atoms. Hence, the volume occupied by one atom [tex](\( V_{\text{atom}} \))[/tex] is:

[tex]\[ V_{\text{atom}} = \frac{\text{Volume of unit cell}}{4}[/tex]

[tex]\[ V_{\text{atom}} = \frac{8 \times 10^{-24} \, \text{m}^3}{4} \][/tex]

[tex]\[ V_{\text{atom}} =[/tex] 2 × 10⁻²â´ m³

Now, substitute the given density (ρ) into the formula and solve for atomic mass:

[tex]\[ 12,500 \, \text{kg/m}^3 = \frac{\text{Atomic mass} \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{8 \times 10^{-24} \, \text{m}^3 \times 2 \times 10^{-24} \, \text{m}^3} \][/tex]

Atomic mass = [tex]\frac{12,500 \times 8 \times 10^{-24} \times 2 \times 10^{-24}}{6.022 \times 10^{23}}[/tex]

Atomic mass = 52.65 g/mol.

This calculation results in an atomic mass of approximately 52.65 g/mol, which is the final answer.

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Rewritten by : Jeany