Q.An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 $g\,cm^{-3}$. Calculate the number of atoms in 108 g of the element.

Concept: Relating crystal structure parameters (unit cell edge, lattice type, density) to molar mass, then using Avogadro's number to find atom count.

Step 1: For an fcc lattice, $Z = 4$ atoms per unit cell. The unit cell volume is $V = a^3 = (300 \times 10^{-10}\,\text{cm})^3 = 2.7 \times 10^{-23}\,\text{cm}^3$.

Step 2: Density relates mass and volume at the unit cell level:

where $M$ is the molar mass. Rearranging:

Step 3: Number of moles in 108 g: …

The density of a crystal connects macroscopic mass to the microscopic arrangement of atoms in the unit cell. For an fcc (face-centered cubic) lattice, we know the geometry: each unit cell contains a specific number of atoms, and the cell edge length tells us the volume. By combining density with unit-cell parameters, we can extract the molar mass of the element, which then lets us count atoms in any given mass.

The key insight is that density $\rho = \frac{\text{mass of unit cell}}{\text{volume of unit cell}}$, and the mass of a unit cell depends on how many atoms it holds and the mass of each atom.

Step 1: Identify the number of atoms per unit cell in fcc

In a face-centered cubic lattice, atoms sit at each corner (shared by 8 cells, contributing $\frac{1}{8}$ each) and at the center of each face (shared by 2 cells, contributing $\frac{1}{2}$ each). The total number of atoms per unit cell is:

Step 2: Calculate the volume of the unit cell

The cell edge $a = 300 \text{ pm} = 300 \times 10^{-10} \text{ cm} = 3 \times 10^{-8} \text{ cm}$.

The volume is:

Step 3: Relate density to molar mass

The density formula for a crystal is:

where $Z = 4$ (atoms per cell), $M$ is the molar mass (g/mol), $N_A = 6.022 \times 10^{23}$ (Avogadro's number), and $V$ is the unit-cell volume.

Rearranging for $M$:

Substitute the values: …