An element having bcc geometry has atomic mass `60 g mol^(-1)`. Calculate the density of unit cell if edge length is 300 pm.

Explanation

To calculate the density of a body-centered cubic (BCC) unit cell with the given parameters, we can follow these steps:

### Step 1: Identify the parameters
- Atomic mass (m) = 60 g/mol
- Edge length (a) = 300 pm = 300 × 10^(-10) cm
- Number of atoms per unit cell (Z) in BCC = 2 (1 atom at the center and 8 corners with 1/8 contribution each)

### Step 2: Convert edge length to centimeters
Since the edge length is given in picometers, we need to convert it to centimeters for the density calculation:
\[ a = 300 \text{ pm} = 300 \times 10^{-10} \text{ cm} = 3 \times 10^{-8} \text{ cm} \]

### Step 3: Calculate the volume of the unit cell
The volume (V) of the unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of a:
\[ V = (3 \times 10^{-8} \text{ cm})^3 = 27 \times 10^{-24} \text{ cm}^3 = 2.7 \times 10^{-23} \text{ cm}^3 \]

### Step 4: Use the density formula
The density (d) of the unit cell can be calculated using the formula:
\[ d = \frac{Z \times m}{V \times N_0} \]
Where:
- Z = number of atoms per unit cell = 2
- m = molar mass = 60 g/mol
- V = volume of the unit cell = \(2.7 \times 10^{-23} \text{ cm}^3\)
- \(N_0\) = Avogadro's number = \(6.022 \times 10^{23} \text{ mol}^{-1}\)

### Step 5: Substitute the values into the density formula
Now substituting the values into the density formula:
\[ d = \frac{2 \times 60 \text{ g/mol}}{2.7 \times 10^{-23} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \]

### Step 6: Calculate the density
Calculating the numerator:
\[ 2 \times 60 = 120 \text{ g} \]
Calculating the denominator:
\[ 2.7 \times 10^{-23} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1} = 1.626 \text{ cm}^3 \]
Thus, the density is:
\[ d = \frac{120 \text{ g}}{1.626 \text{ cm}^3} \approx 73.8 \text{ g/cm}^3 \]

### Final Result
The density of the unit cell is approximately **73.8 g/cm³**.

---