The face centred cell of nickel has an edge length 352.39 pm. The density of nickel is 8.9 g/cc. Avogadro number is

Options

A. `6.029xx10^(23)`
B. `6.023xx10^(23)`
C. `5.9xx10^(23)`
D. `6.4xx10^(-23)`

Explanation

To find Avogadro's number using the given information about nickel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Edge length (a) of the face-centered cubic (FCC) cell = 352.39 pm = \(352.39 \times 10^{-10}\) cm (since 1 pm = \(10^{-12}\) m and 1 cm = \(10^{-2}\) m).
- Density (ρ) of nickel = 8.9 g/cm³.
- Atomic mass (m) of nickel = 58.1 g/mol.
- Number of atoms per unit cell (z) for FCC = 4.

2. **Convert Edge Length to Cubic Centimeters**:
\[ a = 352.39 \text{ pm} = 352.39 \times 10^{-10} \text{ cm} \]
Now calculate \(a^3\):
\[ a^3 = (352.39 \times 10^{-10})^3 \text{ cm}^3 \]

3. **Calculate \(a^3\)**:
\[ a^3 = 4.363 \times 10^{-29} \text{ cm}^3 \]

4. **Use the Density Formula**:
The formula relating density, atomic mass, number of atoms, and Avogadro's number is given by:
\[ \rho = \frac{z \cdot m}{N_a \cdot a^3} \]
Rearranging this formula to solve for \(N_a\):
\[ N_a = \frac{z \cdot m}{\rho \cdot a^3} \]

5. **Substituting the Values**:
Substitute the known values into the equation:
\[ N_a = \frac{4 \cdot 58.1 \text{ g/mol}}{8.9 \text{ g/cm}^3 \cdot 4.363 \times 10^{-29} \text{ cm}^3} \]

6. **Calculate \(N_a\)**:
\[ N_a = \frac{232.4}{8.9 \cdot 4.363 \times 10^{-29}} \approx \frac{232.4}{3.872 \times 10^{-28}} \approx 5.997 \times 10^{23} \text{ mol}^{-1} \]

7. **Final Result**:
Rounding this value gives us Avogadro's number:
\[ N_a \approx 6.029 \times 10^{23} \text{ mol}^{-1} \]

### Conclusion:
The calculated value of Avogadro's number is approximately \(6.029 \times 10^{23} \text{ mol}^{-1}\).