Aluminium crystallises in a face-centred cubic lattice. The edge length of the unit cell of aluminium is 4.05 x 10-10m. What is the density of aluminium? (Atomic mass of Al=27)

Options

A. `2700 kg m^(-3)`
B. `3000 kg m^(-3)`
C. `2400 kg m^(-3)`
D. `2100 kg m^(-3)`

Answer

To calculate the density of aluminum crystallizing in a face-centered cubic (FCC) lattice, we can follow these steps:

### Step 1: Identify the formula for density
The formula for density (\( \rho \)) is given by:

\[ \rho = \frac{Z \cdot m}{a^3 \cdot N_A} \]

where:
- \( Z \) = number of atoms per unit cell
- \( m \) = molar mass of the element (in kg)
- \( a \) = edge length of the unit cell (in meters)
- \( N_A \) = Avogadro's number (\( 6.02 \times 10^{23} \) mol\(^{-1}\))

### Step 2: Gather the known values
From the problem statement, we have:
- Atomic mass of Al (\( m \)) = 27 g/mol = \( 27 \times 10^{-3} \) kg/mol (since 1 g = \( 10^{-3} \) kg)
- For FCC, \( Z = 4 \)
- Edge length of the unit cell (\( a \)) = \( 4.05 \times 10^{-10} \) m
- Avogadro's number (\( N_A \)) = \( 6.02 \times 10^{23} \) mol\(^{-1}\)

### Step 3: Substitute the values into the density formula
Now we can substitute these values into the density formula:

\[ \rho = \frac{4 \cdot (27 \times 10^{-3})}{(4.05 \times 10^{-10})^3 \cdot (6.02 \times 10^{23})} \]

### Step 4: Calculate \( a^3 \)
First, we need to calculate \( a^3 \):

\[ a^3 = (4.05 \times 10^{-10})^3 = 6.64 \times 10^{-29} \text{ m}^3 \]

### Step 5: Substitute \( a^3 \) back into the density formula
Now substituting \( a^3 \) back into the density formula:

\[ \rho = \frac{4 \cdot (27 \times 10^{-3})}{(6.64 \times 10^{-29}) \cdot (6.02 \times 10^{23})} \]

### Step 6: Calculate the denominator
Calculating the denominator:

\[ (6.64 \times 10^{-29}) \cdot (6.02 \times 10^{23}) = 3.99 \times 10^{-5} \]

### Step 7: Calculate the density
Now substituting back into the density formula:

\[ \rho = \frac{4 \cdot (27 \times 10^{-3})}{3.99 \times 10^{-5}} = \frac{0.108}{3.99 \times 10^{-5}} \approx 2700 \text{ kg/m}^3 \]

### Final Answer
Thus, the density of aluminum is approximately:

\[ \rho \approx 2700 \text{ kg/m}^3 \]