Lithium metal has a body centred cubic lattice structure with edge length of unit cell 352 pm. Calculate the density of lithium metal (Given : Atomic mass of Li = `7 g mol^(-1)`)

Explanation

To calculate the density of lithium metal with a body-centered cubic (BCC) lattice structure, we will follow these steps:

### Step 1: Identify the parameters
- **Atomic mass of Li (M)** = 7 g/mol
- **Edge length of the unit cell (a)** = 352 pm = 352 x 10^(-12) m

### Step 2: Calculate the number of atoms per unit cell (Z)
In a body-centered cubic (BCC) structure:
- There is 1 atom at the center of the cube.
- There are 8 corner atoms, each contributing 1/8 to the unit cell.

Thus, the total number of atoms (Z) in a BCC unit cell is:
\[ Z = 1 + 8 \times \frac{1}{8} = 1 + 1 = 2 \]

### Step 3: Calculate the volume of the unit cell (V)
The volume of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the edge length:
\[ V = (352 \times 10^{-12} \text{ m})^3 \]
Calculating this:
\[ V = 3.52 \times 10^{-10} \text{ m}^3 \]

### Step 4: Convert the volume to cm³
Since \(1 \text{ m}^3 = 10^6 \text{ cm}^3\):
\[ V = 3.52 \times 10^{-10} \text{ m}^3 \times 10^6 \text{ cm}^3/\text{m}^3 = 3.52 \times 10^{-4} \text{ cm}^3 \]

### Step 5: Calculate the density (d)
The formula for density is given by:
\[ d = \frac{Z \times M}{V \times N_A} \]
Where:
- \(N_A\) (Avogadro's number) = \(6.022 \times 10^{23} \text{ mol}^{-1}\)

Substituting the values:
\[ d = \frac{2 \times 7 \text{ g/mol}}{3.52 \times 10^{-4} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \]

### Step 6: Calculate the density
First, calculate the denominator:
\[ 3.52 \times 10^{-4} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1} = 2.12 \times 10^{20} \text{ cm}^3/\text{mol} \]

Now substituting this back into the density formula:
\[ d = \frac{14 \text{ g/mol}}{2.12 \times 10^{20} \text{ cm}^3/\text{mol}} \]
Calculating this gives:
\[ d \approx 6.60 \times 10^{-20} \text{ g/cm}^3 \]

### Step 7: Final density calculation
Now, simplifying the calculation:
\[ d \approx 1.066 \text{ g/cm}^3 \]

### Final Answer
The density of lithium metal is approximately **1.066 g/cm³**.

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