Iron (II) oxide has a cubic strcuture and each unit cell has side 5 Å. If the density of the oxide is 4 g `cm^(-3)`, the number of oxide ions present in each unit cell is ( Molar mass of FeO = ` 72 "g mol"^(-1), N_(A) = 6.02 xx 10^(23) "mol"^(-1)`

Explanation

To find the number of oxide ions present in each unit cell of Iron (II) oxide (FeO), we can use the formula relating density, molar mass, and the number of formula units in a unit cell.

### Step-by-Step Solution:

1. **Identify the given values**:
- Density (d) = 4 g/cm³
- Molar mass of FeO (M) = 72 g/mol
- Side length of the unit cell (a) = 5 Å = 5 x 10⁻⁸ cm
- Avogadro's number (Nₐ) = 6.02 x 10²³ mol⁻¹

2. **Calculate the volume of the unit cell (V)**:
\[ V = a^3 = (5 \times 10^{-8} \text{ cm})^3 = 1.25 \times 10^{-22} \text{ cm}^3 \]

3. **Use the density formula**:
The formula relating density, mass, and volume is:
\[ d = \frac{z \times M}{N_a \times V} \]
where:
- \( z \) = number of formula units (FeO) in the unit cell
- \( M \) = molar mass of the compound
- \( N_a \) = Avogadro's number
- \( V \) = volume of the unit cell

4. **Rearranging the formula to solve for \( z \)**:
\[ z = \frac{d \times N_a \times V}{M} \]

5. **Substituting the values**:
\[ z = \frac{4 \text{ g/cm}^3 \times 6.02 \times 10^{23} \text{ mol}^{-1} \times 1.25 \times 10^{-22} \text{ cm}^3}{72 \text{ g/mol}} \]

6. **Calculating \( z \)**:
\[ z = \frac{4 \times 6.02 \times 10^{23} \times 1.25 \times 10^{-22}}{72} \]
\[ z = \frac{30.1 \times 10^{1}}{72} \]
\[ z \approx 4.18 \]

Since \( z \) must be a whole number, we round it to 4.

7. **Conclusion**:
The number of oxide ions (O²⁻) present in each unit cell of FeO is 4.