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 solve the problem of calculating Avogadro's number based on the given data, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Edge length (a) of the face-centered cubic (FCC) cell: \( a = 352.39 \, \text{pm} = 352.39 \times 10^{-10} \, \text{m} = 3.5239 \times 10^{-8} \, \text{cm} \)
- Density (d) of nickel: \( d = 8.9 \, \text{g/cm}^3 \)
- Molar mass (M) of nickel: \( M = 58.7 \, \text{g/mol} \)

2. **Determine the Number of Atoms per Unit Cell (Z):**
- For a face-centered cubic (FCC) structure, the number of atoms per unit cell (Z) is 4. This is calculated as follows:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom.
- Thus, \( Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \).

3. **Calculate the Volume of the Unit Cell (V):**
- The volume of the unit cell can be calculated using the edge length:
\[ V = a^3 = (3.5239 \times 10^{-8} \, \text{cm})^3 = 4.36 \times 10^{-24} \, \text{cm}^3 \]

4. **Use the Density Formula to Relate Mass and Volume:**
- The density formula relates mass (m), volume (V), and density (d):
\[ d = \frac{m}{V} \]
- Rearranging gives us:
\[ m = d \times V = 8.9 \, \text{g/cm}^3 \times 4.36 \times 10^{-24} \, \text{cm}^3 = 3.87 \times 10^{-23} \, \text{g} \]

5. **Relate Mass to Molar Mass and Avogadro's Number:**
- The mass of the unit cell can also be expressed in terms of the molar mass and Avogadro's number (N₀):
\[ m = \frac{Z \times M}{N_0} \]
- Substituting the known values:
\[ 3.87 \times 10^{-23} \, \text{g} = \frac{4 \times 58.7 \, \text{g/mol}}{N_0} \]

6. **Solve for Avogadro's Number (N₀):**
- Rearranging gives:
\[ N_0 = \frac{4 \times 58.7 \, \text{g/mol}}{3.87 \times 10^{-23} \, \text{g}} \]
- Calculating the right-hand side:
\[ N_0 = \frac{234.8 \, \text{g/mol}}{3.87 \times 10^{-23} \, \text{g}} \approx 6.06 \times 10^{23} \, \text{mol}^{-1} \]

7. **Final Result:**
- Thus, Avogadro's number \( N_0 \approx 6.06 \times 10^{23} \, \text{mol}^{-1} \).