The density of `KBr` is `2.75 g cm^(-3)`. The length of the unit cell is `654` pm. Atomic mass of `K = 39, Br = 80`. Then what is true about the predicted nature of the solid?

Options

A. Unit cell is fcc type
B. No. of formula units =4
C. Solid has bcc type
D. No. of formula units =2

Answer

To determine the predicted nature of the solid KBr based on the given data, we will follow these steps:

### Step 1: Gather the Given Data
- Density of KBr (D) = 2.75 g/cm³
- Length of the unit cell (a) = 654 pm = 654 × 10^(-10) cm
- Atomic mass of K (Potassium) = 39 g/mol
- Atomic mass of Br (Bromine) = 80 g/mol
- Avogadro's number (N_A) = 6.022 × 10^23 mol^(-1)

### Step 2: Calculate the Molar Mass of KBr
The molar mass of KBr can be calculated as follows:
\[ \text{Molar Mass of KBr} = \text{Atomic Mass of K} + \text{Atomic Mass of Br} = 39 + 80 = 119 \text{ g/mol} \]

### Step 3: Use the Density Formula
The formula for density in terms of the number of atoms per unit cell (Z), molar mass (M), volume of the unit cell (V), and Avogadro's number (N_A) is:
\[ D = \frac{Z \cdot M}{V \cdot N_A} \]
Where:
- Volume of the unit cell (V) = \(a^3\)
- \(a = 654 \times 10^{-10} \text{ cm}\)

### Step 4: Calculate the Volume of the Unit Cell
\[ V = a^3 = (654 \times 10^{-10} \text{ cm})^3 = 2.80 \times 10^{-29} \text{ cm}^3 \]

### Step 5: Rearrange the Density Formula to Solve for Z
Rearranging the formula gives:
\[ Z = \frac{D \cdot V \cdot N_A}{M} \]

### Step 6: Substitute the Values into the Equation
Substituting the known values:
\[ Z = \frac{2.75 \text{ g/cm}^3 \cdot 2.80 \times 10^{-29} \text{ cm}^3 \cdot 6.022 \times 10^{23} \text{ mol}^{-1}}{119 \text{ g/mol}} \]

### Step 7: Calculate Z
Calculating the above expression:
\[ Z = \frac{2.75 \cdot 2.80 \times 10^{-29} \cdot 6.022 \times 10^{23}}{119} \]
\[ Z \approx \frac{4.25 \times 10^{-5}}{119} \approx 3.57 \approx 4 \]

### Step 8: Determine the Nature of the Solid
Since Z is approximately 4, this indicates that there are 4 formula units of KBr per unit cell. In solid-state chemistry, a structure with 4 atoms per unit cell typically corresponds to a face-centered cubic (FCC) lattice.

### Conclusion
The predicted nature of the solid KBr is that it has a face-centered cubic (FCC) lattice structure with 4 formula units per unit cell.

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