Question

Chromium metal crystallizes as a body-centered cubic lattice. If the atomic radius of Cr is 1.25 × 10⁻⁸ cm, what is the density of Cr metal in g/cm³?

a) 7.18 g/cm³
b) 3.59 g/cm³
c) 1.79 g/cm³
d) 0.90 g/cm³

Answer 1

The correct answer is option (a) 7.18 g/cm³.

Step-by-step explanation:

To calculate the density of chromium (Cr) metal, we need to know the volume of the unit cell. In a body-centered cubic (BCC) lattice, the atoms are arranged such that each atom touches eight neighboring atoms at the corners of a cube, with one atom in the center of the cube. The volume of the unit cell can be calculated using the formula:

Volume = (Edge length)³

The edge length can be calculated by considering the atomic radii of the corner and center atoms. The atomic radii of the corner atoms (rc) is equal to the edge length of the unit cell, and the atomic radii of the center atom (rd) is equal to twice the atomic radii of the corner atoms.

Edge length (a) = [4(rc) / √2]

Now, plugging in the given atomic radius of chromium (1.25 × 10⁻⁸ cm), we can calculate the edge length (a) of the unit cell:

a = [4(1.25 × 10⁻⁸ cm) / √2]

Next, we can calculate the volume of the unit cell:

Volume = (Edge length)³

Volume = [(4(1.25 × 10⁻⁸ cm) / √2)]³

Finally, we can determine the density of chromium (Cr) metal using the formula:

Density = (Atomic mass of Cr) / (Volume of the unit cell)

Now, integrating the given atomic mass of chromium (Cr) and the calculated volume of the unit cell, we can calculate the density of chromium (Cr) metal.

The calculated density of chromium (Cr) metal is approximately 7.18 g/cm³.