The density of aluminium (atomic mass = 27) is 2700 kg `m^(-3)`. If Al has feestructure, calculate its atomic radius.

Explanation

To calculate the atomic radius of aluminum (Al), given its density and that it has a face-centered cubic (FCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Atomic mass of aluminum (M) = 27 g/mol
- Density of aluminum (D) = 2700 kg/m³ = 2.7 g/cm³
- For FCC structure, the number of atoms per unit cell (Z) = 4.

2. **Use the Density Formula**:
The formula relating density (D), number of atoms per unit cell (Z), molar mass (M), and volume of the unit cell (a³) is:
\[ D = \frac{Z \cdot M}{N_A \cdot a^3} \]
where \(N_A\) is Avogadro's number (\(6.02 \times 10^{23}\) mol⁻¹).

3. **Rearranging the Formula**:
Rearranging the formula to solve for \(a^3\):
\[ a^3 = \frac{Z \cdot M}{D \cdot N_A} \]

4. **Substituting the Values**:
Substitute the known values into the equation:
\[ a^3 = \frac{4 \cdot 27 \text{ g/mol}}{2.7 \text{ g/cm}^3 \cdot 6.02 \times 10^{23} \text{ mol}^{-1}} \]

5. **Calculating \(a^3\)**:
Calculate the numerator and denominator:
- Numerator: \(4 \cdot 27 = 108\)
- Denominator: \(2.7 \cdot 6.02 \times 10^{23} \approx 1.622 \times 10^{24}\)

Now, substituting these values:
\[ a^3 = \frac{108}{1.622 \times 10^{24}} \approx 6.66 \times 10^{-24} \text{ cm}^3 \]

6. **Finding \(a\)**:
To find \(a\), take the cube root:
\[ a = (6.66 \times 10^{-24})^{1/3} \approx 1.88 \times 10^{-8} \text{ cm} \]

7. **Calculating Atomic Radius**:
In an FCC structure, the relationship between the atomic radius (r) and the edge length (a) is given by:
\[ r = \frac{a}{2\sqrt{2}} \]
Substituting the value of \(a\):
\[ r = \frac{1.88 \times 10^{-8}}{2\sqrt{2}} \approx 6.64 \times 10^{-9} \text{ cm} = 6.64 \text{ Å} \]

### Final Answer:
The atomic radius of aluminum is approximately \(6.64 \, \text{Å}\).