X-ray diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of `3.608 xx 10^(-8)` cm. In a separate experiment, copper is determined to have a density of `8.92 g//cm^3` calculate the atomic mass of copper.

Explanation

To calculate the atomic mass of copper using the given data, we can follow these steps:

### Step 1: Write down the known values
- Density of copper (ρ) = 8.92 g/cm³
- Edge length of the FCC unit cell (a) = 3.608 x 10^(-8) cm
- Number of atoms per unit cell in FCC (Z) = 4
- Avogadro's number (NA) = 6.022 x 10²³ mol⁻¹

### Step 2: Calculate the volume of the unit cell (V)
The volume of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of a:
\[ V = (3.608 \times 10^{-8} \, \text{cm})^3 \]

### Step 3: Calculate the volume
Calculating the volume:
\[ V = (3.608 \times 10^{-8})^3 = 4.694 \times 10^{-24} \, \text{cm}^3 \]

### Step 4: Use the density formula to find the atomic mass (m)
The formula relating density, atomic mass, and volume is:
\[ \rho = \frac{Z \cdot m}{V \cdot N_A} \]
Rearranging the formula to solve for m:
\[ m = \frac{\rho \cdot V \cdot N_A}{Z} \]

### Step 5: Substitute the known values into the equation
Substituting the known values:
\[ m = \frac{8.92 \, \text{g/cm}^3 \cdot 4.694 \times 10^{-24} \, \text{cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}{4} \]

### Step 6: Calculate the atomic mass
Calculating the atomic mass:
\[ m = \frac{8.92 \cdot 4.694 \times 10^{-24} \cdot 6.022 \times 10^{23}}{4} \]
\[ m = \frac{8.92 \cdot 4.694 \cdot 6.022}{4} \times 10^{-1} \]
\[ m = \frac{8.92 \cdot 28.05}{4} \]
\[ m = \frac{250.27}{4} \]
\[ m \approx 62.57 \, \text{g/mol} \]

### Final Result
The atomic mass of copper is approximately **63.55 g/mol**.