Thank you for visiting 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. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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⻹² m³
Volume of unit cell = 8 × 10⻲ⴠm³
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⻲ⴠm³
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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