Sodium crystallises in a cubic lattice and the edge length of the unit cell is 430 pm. Calculate the number of atoms in the unit cell. (Atomic mass Na = 23 amu, Density of Na = 0.9623 g `cm^(-3)`)

Explanation

To solve the problem of calculating the number of atoms in the unit cell of sodium, we will use the formula for density and rearrange it to find the number of atoms per unit cell (Z).

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Atomic mass of Sodium (m) = 23 amu
- Density of Sodium (D) = 0.9623 g/cm³
- Edge length of the unit cell (a) = 430 pm = 430 x 10⁻¹² m = 4.3 x 10⁻⁸ cm
- Avogadro's number (Nₐ) = 6.022 x 10²³ mol⁻¹

2. **Convert the Edge Length to cm:**
- Given a = 430 pm, we convert it to centimeters:
\[ a = 430 \, \text{pm} = 430 \times 10^{-12} \, \text{m} = 4.3 \times 10^{-8} \, \text{cm} \]

3. **Calculate the Volume of the Unit Cell (V):**
- The volume of the cubic unit cell is given by:
\[ V = a^3 = (4.3 \times 10^{-8} \, \text{cm})^3 = 7.97 \times 10^{-24} \, \text{cm}^3 \]

4. **Use the Density Formula to Find Z:**
- The formula for density is:
\[ D = \frac{Z \cdot m}{V} \]
- Rearranging this formula to solve for Z gives:
\[ Z = \frac{D \cdot V}{m} \]

5. **Substituting the Values:**
- Substitute the known values into the equation:
\[ Z = \frac{0.9623 \, \text{g/cm}^3 \cdot 7.97 \times 10^{-24} \, \text{cm}^3}{23 \, \text{g/mol}} \]
- Calculate Z:
\[ Z = \frac{0.9623 \cdot 7.97 \times 10^{-24}}{23} \]
\[ Z = \frac{7.66 \times 10^{-24}}{23} \approx 3.34 \times 10^{-25} \, \text{mol} \]

6. **Convert Z to Number of Atoms:**
- To find the number of atoms in the unit cell, multiply Z by Avogadro's number:
\[ \text{Number of atoms} = Z \cdot N_a = 3.34 \times 10^{-25} \, \text{mol} \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \approx 2 \]

### Final Answer:
The number of atoms in the unit cell of sodium is **2**.