Question

KBr crystallises in NaCl type of crystal lattice and its density is 2.75 g/cm$^3$. Number of unit cells of KBr present in a 1.00 mm$^3$ grain of KBr are

• A. $3.6 \times 10^{18}$
• B. $3.6 \times 10^{16}$
• C. $2.34 \times 10^{23}$
• D. $6.02 \times 10^{24}$

Hint:

NaCl structure always has $Z = 4$.

Answer 1

The Correct Option is A

Concept: Density relation: \[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \] Number of unit cells $= \frac{\text{Total volume}}{\text{Volume of one unit cell}}$

Step 1: Convert volume:
\[ 1 \text{ mm}^3 = 10^{-3} \text{ cm}^3 \]

Step 2: Find volume of one unit cell using density relation:
For NaCl structure, $Z = 4$ \[ a^3 = \frac{Z \cdot M}{\rho \cdot N_A} \] Molar mass of KBr: \[ M = 39 + 80 = 119 \] \[ a^3 = \frac{4 \times 119}{2.75 \times 6.022 \times 10^{23}} \approx 2.87 \times 10^{-22} \text{ cm}^3 \]

Step 3: Calculate number of unit cells:
\[ \text{Number of unit cells} = \frac{10^{-3}}{2.87 \times 10^{-22}} \approx 3.48 \times 10^{18} \approx 3.6 \times 10^{18} \] Conclusion:
Number of unit cells $\approx 3.6 \times 10^{18}$