Question

Edge length of a unit cell of a crystal is 288 pm. If its density is 7.2 g/cm3, then determine the type of unit cell assuming mass = 52 g.

Answer 1

We can use the formula for the density of a crystal in terms of its unit cell parameters:

\[ \rho = \frac{ZM}{V N_A} \]

where:

• \(\rho\) is the density of the crystal,
• Z is the number of atoms per unit cell,
• M is the molar mass of the substance,
• V is the volume of the unit cell,
• \(N_A\) is Avogadro's constant.

Step 1: Calculate the volume of the unit cell:
To determine the type of unit cell, we first need to calculate the volume of the unit cell using the edge length. For a cubic unit cell, the volume is given by:

\[ V = a^3 \]

where \(a\) is the edge length of the unit cell.

Substituting the given values, we get:

\[ V = (288 \, \text{pm})^3 = (288 \times 10^{-12} \, \text{m})^3 = 2.359 \times 10^{-23} \, \text{m}^3 \]

Step 2: Use the formula for density to solve for Z:
Now that we know the volume, we can use the formula for density to solve for \(Z\), the number of atoms per unit cell:

\[ \rho = \frac{ZM}{V N_A} \]

Substituting the given values into the equation:

\[ 7.2 \, \text{g/cm}^3 = \frac{Z \times 52 \, \text{g/mol}}{2.359 \times 10^{-23} \, \text{m}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

Simplifying the equation to solve for \(Z\):

\[ Z = \frac{7.2 \, \text{g/cm}^3 \times 2.359 \times 10^{-23} \, \text{m}^3 \times 6.022 \times 10^{23}}{52 \, \text{g/mol}} \]

After evaluating the equation, we get:

\[ Z \approx 4 \]

Step 3: Determine the unit cell type:
The value of \(Z = 4\) suggests that the crystal has a face-centered cubic (FCC) unit cell. In an FCC structure, there are 4 atoms per unit cell, with atoms located at the corners and at the center of each face of the cube.

Conclusion:
Therefore, we can conclude that the crystal has a face-centered cubic (FCC) structure.