Calculate the number of formula units of NaCl per unit cell of NaCl. Given : (i)Internuclear distance of adjacent ions = 0.282 nm
(ii)Density of solid NaCl= `2.17xx10^3 "kg/m"^3`

Answer

To calculate the number of formula units of NaCl per unit cell (denoted as Z), we can follow these steps:

### Step 1: Convert the given internuclear distance to meters
The internuclear distance (d) is given as 0.282 nm. We need to convert this to meters:
\[ d = 0.282 \, \text{nm} = 0.282 \times 10^{-9} \, \text{m} = 2.82 \times 10^{-10} \, \text{m} \]

### Step 2: Calculate the edge length (a) of the unit cell
In NaCl, the edge length (a) is twice the internuclear distance because it is the distance between the centers of the Na+ and Cl- ions:
\[ a = 2d = 2 \times 0.282 \times 10^{-9} \, \text{m} = 0.564 \times 10^{-9} \, \text{m} = 5.64 \times 10^{-10} \, \text{m} \]

### Step 3: Use the density formula to find Z
The formula relating density (ρ), the number of formula units per unit cell (Z), molar mass (M), and the volume of the unit cell (a³) is given by:
\[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \]
Where:
- ρ = density = \(2.17 \times 10^3 \, \text{kg/m}^3\)
- M = molar mass of NaCl = 58.5 g/mol = \(58.5 \times 10^{-3} \, \text{kg/mol}\)
- \(N_A\) = Avogadro's number = \(6.022 \times 10^{23} \, \text{mol}^{-1}\)

### Step 4: Rearranging the formula to solve for Z
Rearranging the density formula to solve for Z gives:
\[ Z = \frac{\rho \cdot a^3 \cdot N_A}{M} \]

### Step 5: Calculate a³
First, we calculate \(a^3\):
\[ a^3 = (5.64 \times 10^{-10} \, \text{m})^3 = 1.795 \times 10^{-28} \, \text{m}^3 \]

### Step 6: Substitute the values into the equation
Now we can substitute the values into the rearranged formula:
\[ Z = \frac{(2.17 \times 10^3 \, \text{kg/m}^3) \cdot (1.795 \times 10^{-28} \, \text{m}^3) \cdot (6.022 \times 10^{23} \, \text{mol}^{-1})}{58.5 \times 10^{-3} \, \text{kg/mol}} \]

### Step 7: Calculate Z
Calculating the above expression:
\[ Z = \frac{(2.17 \times 10^3) \cdot (1.795 \times 10^{-28}) \cdot (6.022 \times 10^{23})}{58.5 \times 10^{-3}} \]
\[ Z \approx \frac{2.17 \times 1.795 \times 6.022}{58.5} \times 10^{-3} \approx 3.99 \approx 4 \]

### Conclusion
The number of formula units of NaCl per unit cell (Z) is approximately 4.

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