Gold has cubic crystals whose unic cell has edge length of 407.9 pm. Density of gold is 19.3 g `cm^(-3)`. Calculate the number of atoms per unit cell. Also predict the type of crystal lattice of gold (Atomic mass of gold = 197 amu)

Explanation

To solve the problem, we need to calculate the number of atoms per unit cell (Z) for gold, given its edge length and density. We will also determine the type of crystal lattice it forms.

### Step-by-Step Solution:

1. **Given Data:**
- Edge length (a) of the unit cell = 407.9 pm = \(4.079 \times 10^{-8}\) cm
- Density (D) of gold = 19.3 g/cm³
- Atomic mass (M) of gold = 197 amu
- Avogadro's number (N_A) = \(6.022 \times 10^{23}\) mol⁻¹

2. **Convert Atomic Mass to grams:**
- Since 1 amu = \(1.66 \times 10^{-24}\) g, we convert the atomic mass of gold:
\[ M = 197 \, \text{amu} = 197 \times 1.66 \times 10^{-24} \, \text{g} = 3.27 \times 10^{-22} \, \text{g} \]

3. **Calculate the Volume of the Unit Cell:**
- The volume (V) of the cubic unit cell can be calculated using the edge length:
\[ V = a^3 = (4.079 \times 10^{-8} \, \text{cm})^3 = 6.78 \times 10^{-24} \, \text{cm}^3 \]

4. **Use the Density Formula to Find Z:**
- The density formula relates density, mass, and volume:
\[ D = \frac{Z \cdot M}{V} \]
Rearranging this gives:
\[ Z = \frac{D \cdot V}{M} \]
Substituting the values:
\[ Z = \frac{19.3 \, \text{g/cm}^3 \cdot 6.78 \times 10^{-24} \, \text{cm}^3}{3.27 \times 10^{-22} \, \text{g}} \]

5. **Calculate Z:**
\[ Z = \frac{19.3 \cdot 6.78 \times 10^{-24}}{3.27 \times 10^{-22}} \approx \frac{1.308 \times 10^{-22}}{3.27 \times 10^{-22}} \approx 4 \]

6. **Determine the Type of Crystal Lattice:**
- Since Z = 4, gold crystallizes in a face-centered cubic (FCC) lattice structure.

### Final Answers:
- The number of atoms per unit cell (Z) = 4
- The type of crystal lattice of gold = Face-Centered Cubic (FCC)