A crystalline solid AB adopts sodium chloride type structure with edge length of the unit cell as 745 pm and formula mass of 74.5 g The density of the crystalline compound is

Options

A. `2.16 g cm^(-3)`
B. `0.99 g cm^(-3)`
C. `1.88 g cm^(-3)`
D. `1.197 g cm^(-3)`

Answer

To find the density of the crystalline solid AB that adopts a sodium chloride type structure, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the edge length from picometers to centimeters**:
- Given edge length \( a = 745 \, \text{pm} \).
- Convert to meters: \( 745 \, \text{pm} = 745 \times 10^{-12} \, \text{m} \).
- Convert to centimeters:
\[ a = 745 \times 10^{-12} \, \text{m} \times 100 \, \text{cm/m} = 745 \times 10^{-10} \, \text{cm} \]

2. **Determine the formula mass**:
- Given formula mass \( M = 74.5 \, \text{g/mol} \).

3. **Identify the number of formula units per unit cell (Z)**:
- For sodium chloride (NaCl), the structure is face-centered cubic (FCC), which has \( Z = 4 \).

4. **Calculate the volume of the unit cell**:
- The volume \( V \) of the cubic unit cell is given by:
\[ V = a^3 = (745 \times 10^{-10} \, \text{cm})^3 = 4.136 \times 10^{-23} \, \text{cm}^3 \]

5. **Use Avogadro's number**:
- Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \).

6. **Calculate the density**:
- The density \( \rho \) can be calculated using the formula:
\[ \rho = \frac{Z \times M}{V \times N_A} \]
- Substituting the values:
\[ \rho = \frac{4 \times 74.5 \, \text{g/mol}}{4.136 \times 10^{-23} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}} \]
- Simplifying the calculation:
\[ \rho = \frac{298 \, \text{g}}{2.49 \times 10^{1} \, \text{cm}^3} \approx 1.197 \, \text{g/cm}^3 \]

7. **Final Result**:
- The density of the crystalline compound is approximately \( 1.197 \, \text{g/cm}^3 \).