Question

Platinum (atomic radius \( = 1.38{A^o}\) ) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of platinum.

Short answer

The density of platinum metal is \(21.8\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3}\) with edge length \(3.90{A^o}\)

Step-by-step solution

Define the edge length

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.

To calculate the edge length of unit cell and density of platinum metal.

Atomic radius \( = 1.38{A^o}\)

For face centred cubic unit cell;

Edge length = Atomic radius \( \times 2\sqrt 2 = 1.38{A^o} \times 2\sqrt 2 = 3.90{A^o}\)

Density of solid \( = \dfrac{{{\rm{Z}} \times {\rm{M}}}}{{{{\rm{a}}^3}\;{{\rm{N}}_{\rm{a}}}}}\)

Here:

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

\(M = \)Molar mass \( = 195.08\;{\rm{g}}/{\rm{mol}}\)

\(\;{{\rm{N}}_{\rm{a}}} = \)Avogadro constant \( = 6.023 \times {10^{23}}\)

\(a = \) edge length \( = 3.90{A^o} = 3.90 \times {10^{ - 8}}{\rm{cm}}\)

Plugging all the given values in the formula to calculate the density

Density of platinum \( = \dfrac{{4 \times 195.08\;{\rm{g}}/{\rm{mol}}}}{{{{\left( {3.90 \times {{10}^{ - 8}}\;{\rm{cm}}} \right)}^3} \times 6.022 \times {{10}^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}\)

Density \( = 21.8\;{\rm{g}}/{\rm{c}}{{\rm{m}}^3}\)