An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.

Options

A. `95 x 10^23` atoms
B. `93.59 x 10^23` atoms
C. `92.59 x 10^23` atoms
D. `91.38 x 10^23` atoms

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Write down the given data
- Density (ρ) = 8.0 g/cm³
- Edge length of the unit cell (a) = 300 pm = 300 x 10^(-10) cm
- For FCC (Face-Centered Cubic) structure, the number of atoms per unit cell (Z) = 4
- Avogadro's number (N_A) = 6.02 x 10^23 atoms/mol
- Mass of the element (m) = 0.5 kg = 500 g

### Step 2: Convert the edge length to centimeters
The edge length is given in picometers (pm), we need to convert it to centimeters (cm):
\[ a = 300 \text{ pm} = 300 \times 10^{-10} \text{ cm} \]

### Step 3: Calculate the volume of the unit cell
The volume (V) of the unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of a:
\[ V = (300 \times 10^{-10} \text{ cm})^3 = 2.7 \times 10^{-29} \text{ cm}^3 \]

### Step 4: Calculate the molar mass (m) using the density formula
The density formula for a crystalline structure is given by:
\[ \rho = \frac{Z \cdot m}{V \cdot N_A} \]
Rearranging this to find the molar mass (m):
\[ m = \frac{\rho \cdot V \cdot N_A}{Z} \]
Substituting the known values:
\[ m = \frac{8.0 \, \text{g/cm}^3 \cdot 2.7 \times 10^{-29} \, \text{cm}^3 \cdot 6.02 \times 10^{23} \, \text{atoms/mol}}{4} \]

### Step 5: Calculate the molar mass
Calculating the above expression:
\[ m = \frac{8.0 \cdot 2.7 \times 10^{-29} \cdot 6.02 \times 10^{23}}{4} \]
\[ m \approx 32.5 \, \text{g/mol} \]

### Step 6: Calculate the number of moles in 500 g
To find the number of moles (n) in 500 g of the element:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{500 \, \text{g}}{32.5 \, \text{g/mol}} \]
Calculating this gives:
\[ n \approx 15.38 \, \text{mol} \]

### Step 7: Calculate the number of atoms in 500 g
Now, using Avogadro's number to find the total number of atoms:
\[ \text{Number of atoms} = n \cdot N_A = 15.38 \, \text{mol} \cdot 6.02 \times 10^{23} \, \text{atoms/mol} \]
Calculating this gives:
\[ \text{Number of atoms} \approx 9.25 \times 10^{24} \, \text{atoms} \]

### Final Answer
The number of atoms present in 0.5 kg of the element is approximately:
\[ \text{Number of atoms} \approx 9.25 \times 10^{24} \, \text{atoms} \]

---