An element (At. Mass =50 g/mol) having fcc structure has unit cell edge length 400 pm. The density of element is

Options

A. 5.188 g/cc
B. 10.376 g/cc
C. 2.56 g/cc
D. 1.2 g/cc

Answer

To calculate the density of an element with a face-centered cubic (FCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Atomic mass (M) = 50 g/mol
- Structure = FCC
- Edge length (a) = 400 pm = 400 × 10^(-10) cm

2. **Determine the Number of Atoms in the Unit Cell (Z):**
- For FCC structure, the number of atoms per unit cell (Z) = 4.

3. **Convert Edge Length to Centimeters:**
- Edge length (a) = 400 pm = 400 × 10^(-10) cm.

4. **Calculate the Volume of the Unit Cell (V):**
- The volume of the unit cell (V) = a³
- V = (400 × 10^(-10) cm)³ = 6.4 × 10^(-23) cm³.

5. **Use Avogadro's Number (N_A):**
- Avogadro's number (N_A) = 6.022 × 10^(23) mol^(-1).

6. **Calculate the Density (ρ):**
- The formula for density is:
\[ \rho = \frac{Z \times M}{V \times N_A} \]
- Substitute the known values:
\[ \rho = \frac{4 \times 50 \text{ g/mol}}{6.4 \times 10^{-23} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \]

7. **Perform the Calculation:**
- Calculate the numerator: \( 4 \times 50 = 200 \text{ g} \).
- Calculate the denominator:
\[ 6.4 \times 10^{-23} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1} = 3.84 \text{ cm}^3 \]
- Now, calculate the density:
\[ \rho = \frac{200 \text{ g}}{3.84 \text{ cm}^3} \approx 52.08 \text{ g/cm}^3 \]

8. **Final Result:**
- The density of the element is approximately 5.2 g/cm³.