The density of `CaF_(2)` (flourtie structure ) is `3.18 g// cm^(3)`. The length of the side of the unit cell is :

Options

A. 253 pm
B. 344 pm
C. 546 pm
D. 273 pm

Answer

To find the length of the side of the unit cell for calcium fluoride (CaF₂) with a fluoride structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
- Calcium fluoride (CaF₂) has a face-centered cubic (FCC) structure. In an FCC unit cell, there are 4 formula units of CaF₂.

2. **Determine the Molecular Weight**:
- The molecular weight (m) of CaF₂ can be calculated as follows:
- Atomic weight of Ca = 40.08 g/mol
- Atomic weight of F = 19.00 g/mol
- Therefore, molecular weight of CaF₂ = 40.08 + (2 × 19.00) = 40.08 + 38.00 = 78.08 g/mol

3. **Use the Density Formula**:
- The formula for density (ρ) is given by:
\[ \rho = \frac{z \cdot m}{N_a \cdot a^3} \]
where:
- \( z \) = number of formula units in the unit cell (for FCC, \( z = 4 \))
- \( m \) = molecular weight of CaF₂ (78.08 g/mol)
- \( N_a \) = Avogadro's number (approximately \( 6.022 \times 10^{23} \) mol⁻¹)
- \( a \) = edge length of the unit cell (in cm)

4. **Rearranging the Formula**:
- Rearranging the density formula to solve for \( a^3 \):
\[ a^3 = \frac{z \cdot m}{N_a \cdot \rho} \]

5. **Substituting the Values**:
- Substitute the known values into the equation:
\[ a^3 = \frac{4 \cdot 78.08 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 3.18 \, \text{g/cm}^3} \]

6. **Calculating \( a^3 \)**:
- Calculate the numerator:
\[ 4 \cdot 78.08 = 312.32 \, \text{g/mol} \]
- Calculate the denominator:
\[ 6.022 \times 10^{23} \cdot 3.18 = 1.915 \times 10^{24} \, \text{g/cm}^3 \]
- Now, calculate \( a^3 \):
\[ a^3 = \frac{312.32}{1.915 \times 10^{24}} \approx 162.9 \times 10^{-24} \, \text{cm}^3 \]

7. **Finding \( a \)**:
- Take the cube root of \( a^3 \):
\[ a = \sqrt[3]{162.9 \times 10^{-24}} \approx 5.46 \times 10^{-8} \, \text{cm} \]

8. **Convert to Picometers**:
- Convert centimeters to meters:
\[ a \approx 5.46 \times 10^{-10} \, \text{m} \]
- Convert meters to picometers (1 pm = \( 10^{-12} \) m):
\[ a \approx 546 \, \text{pm} \]

### Final Answer:
The length of the side of the unit cell is approximately **546 picometers**.