An element (atomic mass `= 100 g//mol`) having bcc structure has unit cell edge 400 pm .Them density of the element is

Options

A. `"10.37 g cm"^(-3)`
B. `"5.19 g cm"^(-3)`
C. `"7.29 g cm"^(-3)`
D. `"2.14 g cm"^(-3)`

Explanation

To find the density of the element with a body-centered cubic (BCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Atomic mass (m) = 100 g/mol
- Edge length (A) = 400 pm = 400 × 10^(-12) m

2. **Convert Edge Length to cm:**
- Since we need the density in g/cm³, we convert picometers to centimeters:
\[ A = 400 \, \text{pm} = 400 \times 10^{-10} \, \text{cm} = 4.00 \times 10^{-8} \, \text{cm} \]

3. **Calculate the Volume of the Unit Cell:**
- The volume (V) of the cubic unit cell is given by:
\[ V = A^3 = (4.00 \times 10^{-8} \, \text{cm})^3 = 6.4 \times 10^{-24} \, \text{cm}^3 \]

4. **Determine the Number of Atoms in the Unit Cell (Z):**
- In a BCC structure, the number of atoms (Z) is:
\[ Z = 2 \quad (\text{1 atom at the center and 8 corner atoms contributing } \frac{1}{8} \text{ each}) \]

5. **Use Avogadro's Number (Nₐ):**
- Avogadro's number (Nₐ) = \(6.022 \times 10^{23} \, \text{mol}^{-1}\)

6. **Calculate the Density (ρ):**
- The formula for density is:
\[ \rho = \frac{Z \cdot m}{N_a \cdot V} \]
- Plugging in the values:
\[ \rho = \frac{2 \cdot 100 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 6.4 \times 10^{-24} \, \text{cm}^3} \]

7. **Calculate the Density:**
\[ \rho = \frac{200}{6.022 \times 10^{23} \cdot 6.4 \times 10^{-24}} \approx 5.19 \, \text{g/cm}^3 \]

### Final Answer:
The density of the element is approximately **5.19 g/cm³**.