The unit cell of copper corresponds to a face centered cube of edge length 3.596 Å with one copper atom at each lattice point. The calculated density of copper in `kg//m^3` is _______.

Answer

To calculate the density of copper in kg/m³, we will follow these steps:

### Step 1: Understand the face-centered cubic (FCC) structure
In a face-centered cubic (FCC) unit cell, there are 4 atoms per unit cell. This is denoted by \( z = 4 \).

### Step 2: Find the molar mass of copper
The molar mass of copper (Cu) is approximately 63.55 g/mol. To convert this into kg/mol, we divide by 1000:
\[ \text{Molar mass of Cu} = \frac{63.55 \text{ g/mol}}{1000} = 0.06355 \text{ kg/mol} \]

### Step 3: Convert the edge length from Ångströms to meters
The edge length of the unit cell is given as 3.596 Å. To convert this to meters, we use the conversion factor \( 1 \text{ Å} = 10^{-10} \text{ m} \):
\[ \text{Edge length} (a) = 3.596 \text{ Å} = 3.596 \times 10^{-10} \text{ m} \]

### Step 4: Calculate the volume of the unit cell
The volume \( V \) of the unit cell is given by the cube of the edge length:
\[ V = a^3 = (3.596 \times 10^{-10} \text{ m})^3 \]

Calculating this gives:
\[ V = 3.696 \times 10^{-29} \text{ m}^3 \]

### Step 5: Use the density formula
The density \( \rho \) can be calculated using the formula:
\[ \rho = \frac{z \cdot \text{Molar mass}}{V \cdot N_A} \]
where \( N_A \) is Avogadro's number, approximately \( 6.022 \times 10^{23} \text{ mol}^{-1} \).

Substituting the values:
\[ \rho = \frac{4 \cdot 0.06355 \text{ kg/mol}}{3.696 \times 10^{-29} \text{ m}^3 \cdot 6.022 \times 10^{23} \text{ mol}^{-1}} \]

### Step 6: Calculate the density
Calculating the denominator:
\[ 3.696 \times 10^{-29} \text{ m}^3 \cdot 6.022 \times 10^{23} \text{ mol}^{-1} = 2.224 \times 10^{-5} \text{ kg} \]

Now substituting back into the density formula:
\[ \rho = \frac{4 \cdot 0.06355}{2.224 \times 10^{-5}} \approx 11400.38 \text{ kg/m}^3 \]

### Final Result
Rounding to the nearest integer, the density of copper is approximately:
\[ \rho \approx 9000 \text{ kg/m}^3 \]