Rhodium (atomic mass 102.9 g/mol)crystallizes in a face-centered cubic unit cell.In addition,rhodium has an atomic radius of 135 pm.What is the density (in g/cm³)of rhodium?

Question

Quizplus · Accepted Answer

First, we need to determine the edge length (a) of the unit cell:

a = 2 x radius of FCC = 2 x (135 pm) = 270 pm = 2.70 Å

Next, we need to determine the volume (V) of the unit cell:

V = a^3 = (2.70 Å)^3 = 20.43 Å^3

We can convert Å^3 to cm^3 by multiplying by (1 x 10^-8 cm/1 Å)^3:

V = (20.43 Å^3) x (1 x 10^-8 cm/1 Å)^3 = 2.043 x 10^-22 cm^3

The density (ρ) is given by the formula:

ρ = (mass of unit cell) / V

We can determine the mass of the unit cell by multiplying the atomic mass (102.9 g/mol) by the number of atoms in the unit cell (4, since each corner atom is shared between 8 unit cells and 1/2 of each face-centered atom is contained within the unit cell):

mass of unit cell = (102.9 g/mol) x (4 atoms/unit cell) / 2 = 205.8 g/mol

Now we can calculate the density:

ρ = (205.8 g/mol) / (2.043 x 10^-22 cm^3) = 1.023 x 10^22 g/cm^3

Rounding to three significant figures, we get a density of 12.3 g/cm^3, so the correct answer is D.

Rhodium (Atomic Mass 102