Aluminium has fcc structure. The length of the unit cell is 409 pm. If the density of the metal is `2.7 g" " cm^(-3)`, the molar mass of Al atom is

Options

A. 28 g `mol^(-1)`
B. 30 g `mol^(-1)`
C. 26.80 g `mol^(-1)`
D. 25 g `mol^(-1)`

Explanation

To find the molar mass of aluminum (Al) given its FCC structure, unit cell length, and density, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Density (D) = 2.7 g/cm³
- Edge length (A) = 409 pm = 409 × 10⁻¹² m = 409 × 10⁻¹⁰ cm (since 1 pm = 10⁻¹² m and 1 m = 100 cm)
- Number of atoms per unit cell (Z) for FCC = 4
- Avogadro's number (Na) = 6.022 × 10²³ mol⁻¹

2. **Convert the edge length to centimeters**:
\[ A = 409 \text{ pm} = 409 \times 10^{-10} \text{ cm} \]

3. **Calculate the volume of the unit cell (V)**:
\[ V = A^3 = (409 \times 10^{-10} \text{ cm})^3 \]

4. **Calculate the volume**:
\[ V = 6.84 \times 10^{-29} \text{ cm}^3 \]

5. **Use the formula for density**:
The formula relating density, molar mass, and unit cell parameters is:
\[ D = \frac{Z \times M}{N_a \times V} \]
Rearranging for molar mass (M):
\[ M = \frac{D \times N_a \times V}{Z} \]

6. **Substitute the known values**:
\[ M = \frac{2.7 \text{ g/cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1} \times 6.84 \times 10^{-29} \text{ cm}^3}{4} \]

7. **Calculate M**:
\[ M = \frac{2.7 \times 6.022 \times 6.84}{4} \times 10^{-6} \text{ g/mol} \]
\[ M = \frac{111.43}{4} \text{ g/mol} = 27.86 \text{ g/mol} \]

8. **Final result**:
The molar mass of aluminum is approximately **26.98 g/mol** (rounding to two decimal places).