By X-ray diffraction it is found that nickel `("at mass"=59g "mol"^(-1))`, crystallizes in ccp. The edge length of the unit cell is `3.5 Å`. If density of Ni crystal is `9.0g//cm^(3)`, then value of Avogadro's number from the data is:

Options

A. `6.05 xx 10^(23)`
B. `6.11 xx 10^(23)`
C. `6.02 xx 10^(23)`
D. `6.023 xx 10^(23)`

Answer

To find the value of Avogadro's number (NA) from the given data about nickel, we can follow these steps:

### Step 1: Identify the parameters given in the problem
- Atomic mass of nickel (m) = 59 g/mol
- Edge length of the unit cell (a) = 3.5 Å = 3.5 × 10^(-8) cm (since 1 Å = 10^(-8) cm)
- Density of nickel crystal (ρ) = 9.0 g/cm³
- Nickel crystallizes in cubic close packing (CCP), which is equivalent to face-centered cubic (FCC).

### Step 2: Determine the number of atoms per unit cell (Z)
In a face-centered cubic (FCC) structure:
- There are 8 corner atoms, each contributing 1/8 to the unit cell.
- There are 6 face-centered atoms, each contributing 1/2 to the unit cell.

Calculating the total number of atoms (Z):
\[ Z = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]

### Step 3: Use the formula for density
The formula for density (ρ) in terms of Z, m, NA, and a is given by:
\[ \rho = \frac{Z \cdot m}{N_A \cdot a^3} \]

Rearranging this to find NA:
\[ N_A = \frac{Z \cdot m}{\rho \cdot a^3} \]

### Step 4: Substitute the known values into the equation
Substituting the values we have:
- Z = 4
- m = 59 g/mol
- ρ = 9 g/cm³
- a = 3.5 × 10^(-8) cm

Calculating \(a^3\):
\[ a^3 = (3.5 \times 10^{-8} \text{ cm})^3 = 4.2875 \times 10^{-24} \text{ cm}^3 \]

Now substituting into the equation for NA:
\[ N_A = \frac{4 \cdot 59}{9 \cdot 4.2875 \times 10^{-24}} \]

### Step 5: Calculate NA
Calculating the numerator:
\[ 4 \cdot 59 = 236 \]

Calculating the denominator:
\[ 9 \cdot 4.2875 \times 10^{-24} = 38.5875 \times 10^{-24} \]

Now substituting these values:
\[ N_A = \frac{236}{38.5875 \times 10^{-24}} \approx 6.11 \times 10^{23} \text{ mol}^{-1} \]

### Conclusion
Thus, the value of Avogadro's number from the data is approximately:
\[ N_A \approx 6.11 \times 10^{23} \text{ mol}^{-1} \]