Question

Nickel crystallizes in a face-centered cubic lattice. If the density of the metal is 8.908 g/cm3, what is the unit cell edge length in pm?

Answer 1

1) Calculate the average mass of one atom of Ni:

58.6934 g mol¯¹ ÷ 6.022 x 10²³ atoms mol¯¹ = 9.746496 x 10¯²³ g/atom

2) Calculate the mass of the 4 nickel atoms in the face-centered cubic unit cell:

9.746496 x 10¯²³ g/atom times 4 atoms/unit cell = 3.898598 x 10¯²² g/unit cell

3) Use density to get the volume of the unit cell:

3.898598 x 10¯²² g ÷ 8.908 g/cm³ = 4.376514 x 10¯²³ cm³

4) Determine the edge length of the unit cell:

[cube root of] 4.376514 x 10¯²³ cm³ = 3.524 x 10¯⁸ cm

5) Convert cm to pm:

cm = 10¯² m; pm = 10¯¹² m.

Consequently, there are 10¹⁰ pm/cm

(3.524 x 10¯⁸ cm) (10¹⁰ pm/cm) = 352.4 pm