Question

Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodium in g/cm3

Answer 1

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

$r = \dfrac{a}{2\sqrt{2}}$

where;

a = edge length.

Making "a" the subject of the formula:

$a = 2 \sqrt{2} \times r$

$a = 2 \times 1.414 \times 135 \ pm$

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

$density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}$

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

$mass\ of\ atom =\dfrac{ atomic \ mass}{Avogadro \ number}$

$mass\ of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}$

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

$mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4$

mass of unit cell = 6.83380375 × 10⁻²² g

$Density \ of \ unit \ cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}$

Density of unit cell ( rhodium) = 12.279 g/cm³