Question

Chromium (atomic mass 52.00 g/mol) crystallizes in a body-centered cubic unit cell. If the length of an edge of the unit cell is 289 pm, what is the density (in g/cm³) of chromium?

a. 3.58 g/cm³
b. 7.15 g/cm³
c. 13.7 g/cm³
d. 14.3 g/cm³
e. 21.3 g/cm³

Answer 1

The correct option for the density of chromium, assuming a body-centered cubic unit cell and given dimensions, is approximately 7.15 g/cm³. The calculation involves converting the given edge length from picometers to centimeters, calculating the volume of the unit cell, and then determining the mass of the two atoms present in a bcc cell.

Step-by-step explanation:

To calculate the density of chromium which crystallizes in a body-centered cubic (bcc) unit cell, we can use the following steps:

1. Determine the number of atoms per unit cell in a bcc structure, which is 2.
2. Calculate the volume of the unit cell in cm³. Since the edge length is given in picometers (1 pm = 10-12 m), we must convert it to cm (1 cm = 10-2 m) before calculating the volume:
3. Volume = (edge length in cm)³ = (289 x 10-10 cm)³
4. Now, use the formula Density = (mass of atoms in the cell) / (volume of the cell) to calculate the density:
5. Mass of atoms in the cell = (2 atoms/unit cell) x (52 g/mol) / (Avogadro's number)
6. Density = (mass of 2 atoms in the cell) / (volume of the unit cell)

After calculating, the correct option for the density of chromium is c. 7.19 g/cm³. Therefore, option b. 7.15 g/cm³ is the closest approximation and the correct option in the final answer.