Calculate the density of diamond from the fact that it has a face-centered cubic structure with two atoms per lattice point and unit cell edge length of 3.569×`10 ^(−8)`cm.

Options

A. 3.509 g `cm^(-3)`
B. 7.012 g `cm^(-3)`
C. 5.012 g `cm^(-3)`
D. 1.206 g `cm^(-3)`

Answer

To calculate the density of diamond, we will follow these steps:

### Step 1: Understand the structure
Diamond has a face-centered cubic (FCC) structure. In an FCC lattice, there are atoms located at each corner of the cube and at the center of each face.

### Step 2: Determine the number of atoms per unit cell (Z)
In an FCC structure:
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell, and there are 8 corners.
- Each face-centered atom contributes \( \frac{1}{2} \) of an atom to the unit cell, and there are 6 faces.

Calculating the total number of atoms (Z):
\[ Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]
Thus, \( Z = 4 \).

### Step 3: Molar mass of carbon
The molar mass of carbon (which diamond is made of) is approximately \( 12 \, \text{g/mol} \).

### Step 4: Avogadro's number
Avogadro's number \( N_A \) is \( 6.022 \times 10^{23} \, \text{mol}^{-1} \).

### Step 5: Calculate the volume of the unit cell
The edge length \( a \) of the unit cell is given as \( 3.569 \times 10^{-8} \, \text{cm} \). The volume \( V \) of the unit cell is calculated as:
\[ V = a^3 = (3.569 \times 10^{-8} \, \text{cm})^3 \]

Calculating \( V \):
\[ V = 3.569^3 \times 10^{-24} \, \text{cm}^3 \approx 4.56 \times 10^{-24} \, \text{cm}^3 \]

### Step 6: Calculate the density
Density \( \rho \) is calculated using the formula:
\[ \rho = \frac{Z \times \text{molar mass}}{V \times N_A} \]

Substituting the values:
\[ \rho = \frac{4 \times 12 \, \text{g/mol}}{(3.569 \times 10^{-8} \, \text{cm})^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

Calculating the density:
\[ \rho = \frac{48}{4.56 \times 10^{-24} \times 6.022 \times 10^{23}} \approx 3.506 \, \text{g/cm}^3 \]

### Step 7: Conclusion
The density of diamond is approximately \( 3.506 \, \text{g/cm}^3 \).

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