How many unit cells are present in 5.0 gm of crystal AB (formula mass of AB =40) having rock salt type structure ? (`N_(A)` = Avogadro 's no.)

Options

A. `N_(A)`
B. `(N_(A))/(10)`
C. `4 N_(A)`
D. None of these

Answer

To solve the problem of how many unit cells are present in 5.0 g of crystal AB with a formula mass of 40 g/mol and a rock salt type structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula Mass and Given Mass**:
- The formula mass of AB is given as 40 g/mol.
- The mass of the crystal is given as 5.0 g.

2. **Calculate the Number of Moles of AB**:
- Use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
- Substitute the values:
\[ \text{Number of moles of AB} = \frac{5.0 \, \text{g}}{40 \, \text{g/mol}} = 0.125 \, \text{mol} \]

3. **Determine the Number of Formula Units**:
- Since 1 mole of AB contains Avogadro's number of formula units (N_A):
\[ \text{Number of formula units} = 0.125 \, \text{mol} \times N_A = 0.125 N_A \]

4. **Identify the Number of Unit Cells in Rock Salt Structure**:
- In a rock salt structure (NaCl type), the number of formula units per unit cell (z) is 4. This means:
\[ 1 \, \text{unit cell} = \frac{1}{4} \, \text{formula units of AB} \]

5. **Calculate the Number of Unit Cells**:
- To find the number of unit cells, we can use the relationship:
\[ \text{Number of unit cells} = \text{Number of formula units} \times \frac{1 \, \text{unit cell}}{1 \, \text{formula unit}} = 0.125 N_A \times 4 \]
- This simplifies to:
\[ \text{Number of unit cells} = \frac{0.125 N_A}{\frac{1}{4}} = 0.125 N_A \times 4 = 0.5 N_A \]

6. **Final Calculation**:
- Therefore, the total number of unit cells in 5.0 g of crystal AB is:
\[ \text{Number of unit cells} = \frac{N_A}{32} \]

### Conclusion:
The final answer is:
\[ \text{Number of unit cells} = \frac{N_A}{32} \]