An element crytallizes in the cubic lattice and the edge of the unit cell is 430 pm. Calculation the number of atoms in a unit cell. [ Atomic mass of Na = 23.0 amu. Density of sodium = 0.9623 g ` cm^(-3) , N_(A) = 6.023 xx 10^(23) mol^(-1)`

Answer

To solve the problem of calculating the number of atoms in a unit cell for sodium (Na) crystallizing in a cubic lattice, we can follow these steps:

### Step 1: Understand the given data
- Edge length of the unit cell (a) = 430 pm = \( 430 \times 10^{-10} \) cm
- Atomic mass of sodium (Na) = 23.0 amu
- Density of sodium = 0.9623 g/cm³
- Avogadro's number (N_A) = \( 6.023 \times 10^{23} \) mol⁻¹

### Step 2: Convert the atomic mass to grams per mole
Since the atomic mass is given in amu, we can directly convert it to grams per mole:
- Molecular weight of sodium = 23.0 g/mol

### Step 3: Use the density formula
The formula for density (d) in terms of the number of atoms in the unit cell (z), molecular weight (M), and volume of the unit cell (V) is given by:
\[ d = \frac{z \times M}{N_A \times V} \]
Where:
- \( V = a^3 \) (volume of the cubic unit cell)

### Step 4: Calculate the volume of the unit cell
Calculate the volume using the edge length:
\[ V = (430 \times 10^{-10} \, \text{cm})^3 = 7.97 \times 10^{-23} \, \text{cm}^3 \]

### Step 5: Substitute values into the density formula
Now we can rearrange the density formula to solve for z:
\[ z = \frac{d \times N_A \times V}{M} \]
Substituting the known values:
\[ z = \frac{0.9623 \, \text{g/cm}^3 \times (6.023 \times 10^{23} \, \text{mol}^{-1}) \times (7.97 \times 10^{-23} \, \text{cm}^3)}{23.0 \, \text{g/mol}} \]

### Step 6: Calculate z
Calculating the above expression:
\[ z = \frac{0.9623 \times 6.023 \times 7.97}{23.0} \]
\[ z = \frac{0.9623 \times 6.023 \times 7.97}{23.0} \approx 2 \]

### Conclusion
The number of atoms in a unit cell (z) for sodium crystallizing in a cubic lattice is **2**.