To solve the problem of finding the number of unit cells in 108 g of a metallic element that crystallizes in a simple cubic lattice, we will follow these steps:
### Step 1: Understand the relationship between density, molar mass, and unit cells
The formula for density (d) in terms of the number of atoms per unit cell (z), molar mass (M), and volume of the unit cell (V) is given by:
\[
d = \frac{z \cdot M}{N_A \cdot V}
\]
where:
- \(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)),
- \(V\) is the volume of the unit cell.
### Step 2: Calculate the volume of the unit cell
In a simple cubic lattice, the volume of the unit cell (V) can be calculated using the edge length (a):
\[
V = a^3
\]
Given that the edge length \(a = 3 \, \text{Å} = 3 \times 10^{-8} \, \text{cm}\):
\[
V = (3 \times 10^{-8} \, \text{cm})^3 = 27 \times 10^{-24} \, \text{cm}^3 = 2.7 \times 10^{-23} \, \text{cm}^3
\]
### Step 3: Substitute values into the density formula
We know:
- Density \(d = 8 \, \text{g/cm}^3\),
- Molar mass \(M = 108 \, \text{g/mol}\),
- For a simple cubic lattice, \(z = 1\).
Substituting these values into the density formula:
\[
8 = \frac{1 \cdot 108}{6.022 \times 10^{23} \cdot (2.7 \times 10^{-23})}
\]
### Step 4: Rearranging the equation to find the number of unit cells
Rearranging the equation to solve for the number of unit cells \(N\):
\[
N = \frac{z \cdot M}{d \cdot V} = \frac{1 \cdot 108}{8 \cdot (2.7 \times 10^{-23})}
\]
### Step 5: Calculate the number of unit cells
Now, substituting the values:
\[
N = \frac{108}{8 \cdot 2.7 \times 10^{-23}} = \frac{108}{21.6 \times 10^{-23}} = 5 \times 10^{23}
\]
### Step 6: Final calculation for 108 g of metal
Since we are looking for the number of unit cells in 108 g of the metal, we can directly use the calculated value:
\[
N = 5 \times 10^{23}
\]
Thus, the number of unit cells in 108 g of the metallic element is:
\[
\boxed{5 \times 10^{23}}
\]
