Answer :
(a) The edge length of the unit cell is 286 pm.
(b) The the density of aluminum metal is 1.27 x 10⁻²³ g/cm³.
What is edge length of the aluminum atom?
The edge length of the aluminum atom is calculated as follows;
a = 2r
where;
- r is the radius of the atom
- a is edge length
a = 2 x 143 pm = 286 pm
Volume of the aluminum atom
V = a³
V = (286 x 10⁻¹²)³
V = 2.34 x 10⁻²⁹ m³
Density of the aluminum metal
ρ = ZM/VN
where;
- Z is 4 for face-centered cubic
- M is mass of aluminum atom (g/mol), 26.982 amu = (1.66 x 10⁻²⁴ x 26.982) = 4.479 x 10⁻²³ g/mol
- V is volume
- N is Avogadro's number
ρ = (4 x 4.479 x 10⁻²³) / ( 2.34 x 10⁻²⁹ x 6.023 x 10²³)
ρ = 1.27 x 10⁻¹⁷ g/m³
ρ = 1.27 x 10⁻¹⁷ g/m³ x (1 m³ / 10⁶ cm³)
ρ = 1.27 x 10⁻²³ g/cm³
Thus, the edge length of the unit cell is 286 pm and the the density of aluminum metal is 1.27 x 10⁻²³ g/cm³.
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