To determine the predicted nature of the solid KBr based on the given data, we will follow these steps:
### Step 1: Understand the formula for density
The density (d) of a crystalline solid can be expressed using the formula:
\[
d = \frac{Z \cdot M}{V \cdot N_0}
\]
where:
- \( Z \) = number of formula units per unit cell
- \( M \) = molar mass of the compound (g/mol)
- \( V \) = volume of the unit cell (cm³)
- \( N_0 \) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))
### Step 2: Calculate the molar mass of KBr
The molar mass \( M \) of KBr can be calculated as follows:
\[
M = \text{Atomic mass of K} + \text{Atomic mass of Br} = 39 \, \text{g/mol} + 80 \, \text{g/mol} = 119 \, \text{g/mol}
\]
### Step 3: Calculate the volume of the unit cell
The length of the unit cell is given as \( 654 \, \text{pm} \) (picometers). We need to convert this to centimeters:
\[
654 \, \text{pm} = 654 \times 10^{-10} \, \text{cm}
\]
The volume \( V \) of the cubic unit cell is given by:
\[
V = a^3 = (654 \times 10^{-10} \, \text{cm})^3
\]
Calculating this gives:
\[
V = 654^3 \times 10^{-30} \, \text{cm}^3 = 28.1 \times 10^{-30} \, \text{cm}^3
\]
### Step 4: Substitute known values into the density formula
Now we can rearrange the density formula to solve for \( Z \):
\[
Z = \frac{d \cdot V \cdot N_0}{M}
\]
Substituting the known values:
- \( d = 2.75 \, \text{g/cm}^3 \)
- \( V = 28.1 \times 10^{-30} \, \text{cm}^3 \)
- \( N_0 = 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- \( M = 119 \, \text{g/mol} \)
Calculating \( Z \):
\[
Z = \frac{2.75 \, \text{g/cm}^3 \times 28.1 \times 10^{-30} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{119 \, \text{g/mol}}
\]
Calculating this gives:
\[
Z \approx 4
\]
### Step 5: Determine the nature of the solid
The value of \( Z = 4 \) indicates that there are 4 formula units of KBr per unit cell. This is characteristic of a face-centered cubic (FCC) structure.
### Conclusion
The predicted nature of the solid KBr is that it has a face-centered cubic (FCC) unit cell.
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