Density of a unit cell is respresented as
`rho = (" Effective no. of atoms (s)" xx "Mass of a unit cell ")/("Volume of a unit cell ")=(Z.M)/(N_(A).a^(3))`
where , Z = effective no . of atoms(s) or ion (s).
M= At . mass// formula
`N_(A) `= Avogadro' s no . `rArr 6.0323 xx 10^(23)`
a= edge length of unit cell
The density of KBr is `2.75 g //cm^(-3)` . The length of the edge of the unit cell is 645 pm .To which type of cubic crystal , KBr belongs ?

Options

A. Simple cubic
B. bcc
C. fcc
D. None of these

Answer

To determine the type of cubic crystal that KBr belongs to, we will use the formula for the density of a unit cell:

\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \]

Where:
- \(\rho\) = density of the substance
- \(Z\) = effective number of atoms (or ions) per unit cell
- \(M\) = molar mass of the substance
- \(N_A\) = Avogadro's number
- \(a\) = edge length of the unit cell

### Step 1: Identify the given values
- Density of KBr, \(\rho = 2.75 \, \text{g/cm}^3\)
- Edge length of the unit cell, \(a = 645 \, \text{pm} = 645 \times 10^{-10} \, \text{cm}\)
- Molar mass of KBr, \(M = 119 \, \text{g/mol}\)
- Avogadro's number, \(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)

### Step 2: Convert the edge length to centimeters
\[ a = 645 \, \text{pm} = 645 \times 10^{-12} \, \text{m} = 645 \times 10^{-10} \, \text{cm} \]

### Step 3: Substitute the values into the density formula to solve for \(Z\)
Rearranging the density formula to solve for \(Z\):
\[ Z = \frac{\rho \cdot N_A \cdot a^3}{M} \]

Now substituting the values:
\[ Z = \frac{2.75 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot (645 \times 10^{-10} \, \text{cm})^3}{119 \, \text{g/mol}} \]

### Step 4: Calculate \(a^3\)
\[ a^3 = (645 \times 10^{-10} \, \text{cm})^3 = 2.695 \times 10^{-28} \, \text{cm}^3 \]

### Step 5: Substitute \(a^3\) back into the equation for \(Z\)
\[ Z = \frac{2.75 \cdot 6.022 \times 10^{23} \cdot 2.695 \times 10^{-28}}{119} \]

### Step 6: Perform the calculation
Calculating the numerator:
\[ 2.75 \cdot 6.022 \times 10^{23} \cdot 2.695 \times 10^{-28} \approx 4.42 \times 10^{-4} \]

Now dividing by \(119\):
\[ Z \approx \frac{4.42 \times 10^{-4}}{119} \approx 3.71 \approx 4 \]

### Step 7: Determine the type of cubic crystal
Since \(Z \approx 4\), KBr has an effective number of atoms per unit cell equal to 4. This corresponds to a face-centered cubic (FCC) structure, where there are 4 atoms per unit cell (1 from each of the 8 corners and 3 from the 6 faces).

### Conclusion
KBr belongs to the face-centered cubic (FCC) type of cubic crystal.

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