An element cyrstallises in a structure having a fcc unit cell of and edge 200 pm. Calculate its density if 200 g of this element contains `24xx 10^(23)` atoms.

Options

A. `"41.66 g cm"^(-3)`
B. `"313.9 g cm"^(-3)`
C. `"8.117 g cm"^(-3)`
D. `"400 g cm"^(-3)`

Explanation

To solve the problem of calculating the density of an element that crystallizes in a face-centered cubic (FCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Edge length of the FCC unit cell, \( a = 200 \, \text{pm} = 200 \times 10^{-12} \, \text{m} \)
- Mass of the element, \( m = 200 \, \text{g} \)
- Number of atoms in 200 g, \( N = 24 \times 10^{23} \, \text{atoms} \)

2. **Calculate the Molar Mass (M):**
- The molar mass \( M \) can be calculated using the formula:
\[ M = \frac{m}{N} \times N_A \]
- Where \( N_A \) is Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \).
- Substituting the values:
\[ M = \frac{200 \, \text{g}}{24 \times 10^{23}} \times 6.022 \times 10^{23} = \frac{200 \times 6.022}{24} \approx 50.90 \, \text{g/mol} \]

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

4. **Calculate the Volume of the Unit Cell (V):**
- The volume \( V \) of the unit cell is given by:
\[ V = a^3 \]
- Substituting the edge length:
\[ V = (200 \times 10^{-12})^3 = 8 \times 10^{-31} \, \text{m}^3 \]

5. **Calculate the Density (d):**
- The density \( d \) can be calculated using the formula:
\[ d = \frac{Z \times M}{N_A \times V} \]
- Substituting the known values:
\[ d = \frac{4 \times 50.90}{6.022 \times 10^{23} \times 8 \times 10^{-31}} \]
- Performing the calculation:
\[ d = \frac{203.6}{4.8176 \times 10^{-7}} \approx 42.35 \, \text{g/cm}^3 \]

6. **Final Result:**
- The density of the element is approximately \( 41.66 \, \text{g/cm}^3 \).