An element of atomic mass 90 occurs in fce structure with cell edge of 500 pm. Calculate the Avogadro's number if the density is 4.2g `cm^(-3)`

Explanation

To calculate Avogadro's number (Na) for the given element in a face-centered cubic (FCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Atomic mass (M) = 90 g/mol
- Density (D) = 4.2 g/cm³
- Edge length of the FCC unit cell (a) = 500 pm = \(500 \times 10^{-10}\) cm

2. **Determine the Number of Atoms per Unit Cell (Z)**:
- For an FCC structure, the number of atoms per unit cell (Z) = 4.

3. **Convert Edge Length to Cubic Centimeters**:
- Convert the edge length from picometers to centimeters:
\[ a = 500 \text{ pm} = 500 \times 10^{-10} \text{ cm} \]

4. **Calculate the Volume of the Unit Cell (V)**:
- The volume of the unit cell (V) is given by:
\[ V = a^3 = (500 \times 10^{-10} \text{ cm})^3 \]

5. **Calculate the Volume**:
- Calculate \(a^3\):
\[ V = (500 \times 10^{-10})^3 = 1.25 \times 10^{-22} \text{ cm}^3 \]

6. **Use the Density Formula**:
- The formula for density is:
\[ D = \frac{Z \times M}{V \times N_a} \]
- Rearranging for Avogadro's number (Na):
\[ N_a = \frac{Z \times M}{D \times V} \]

7. **Substitute the Values**:
- Substitute the known values into the equation:
\[ N_a = \frac{4 \times 90 \text{ g/mol}}{4.2 \text{ g/cm}^3 \times 1.25 \times 10^{-22} \text{ cm}^3} \]

8. **Calculate Na**:
- Calculate the numerator:
\[ 4 \times 90 = 360 \text{ g/mol} \]
- Calculate the denominator:
\[ 4.2 \times 1.25 \times 10^{-22} = 5.25 \times 10^{-22} \text{ g} \]
- Now calculate \(N_a\):
\[ N_a = \frac{360}{5.25 \times 10^{-22}} \approx 6.857 \times 10^{23} \text{ mol}^{-1} \]

9. **Final Result**:
- Thus, Avogadro's number \(N_a \approx 6.85 \times 10^{23} \text{ mol}^{-1}\).