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Aluminum (atomic radius \( = 1.43{A^o}\) ) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum.

Short Answer

Expert verified

The density of aluminium metal is \(2.72\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3}\) with edge length \(4.04{A^o}\).

Step by step solution

01

Define the density of the solid

The edge length of face-centered cubic unit cell is directly related to atomic radius. Edge length can be further used to calculate the density of solid.

Edge length = Atomic radius \( \times 2\sqrt 2 \)

Density of solid \( = \dfrac{{Z \times M}}{{{a^3}{N_a}}}\)

02

Determine the edge length and density of aluminum

Atomic radius \( = 1.4{A^o}\)

For face centered cubic unit cell;

Edge length=Atomic radius \( \times 2\sqrt 2 = 1.43{A^o} \times 2\sqrt 2 = 4.04{A^o}\)

Density of solid \( = \dfrac{{Z \times M}}{{{a^3}{N_a}}}\)

Here:

\(Z = 4\); for face centred cubic structure

\(M = \)Molar mass \( = 26.98\;{\rm{g}}/{\rm{mol}}\)

\({N_a} = \)Avogadro constant \( = 6.023 \times {10^{23}}\)

\(a = \) edge length \( = 4.04{A^o} = 4.04 \times {10^{ - 8}}\;{\rm{cm}}\)

Plug all the given values in formula to calculate the density

Density of aluminium\( = \dfrac{{4 \times 26.98\;{\rm{g}}/{\rm{mo}}{{\rm{l}}^2}}}{{{{\left( {4.04 \times {{10}^{ - 8}}\;{\rm{cm}}} \right)}^3} \times 6.023 \times {{10}^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}\)

Density \( = 2.72\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3}\)

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