Question

Find the number of atoms per unit cell if edge length is 300pm, density = 3 g/cm³ , molecular mass = 40 g(nearest integer)

Answer 1

Given: Edge length \( a = 300 \, \text{pm} = 3 \times 10^{-8} \, \text{cm} \) Density \( \rho = 3 \, \text{g/cm}^3 \) Molar mass \( M = 40 \, \text{g/mol} \) Avogadro number \( N_A = 6.022 \times 10^{23} \) 
Step 1: Volume of unit cell \[ V = a^3 = (3 \times 10^{-8})^3 = 2.7 \times 10^{-23} \, \text{cm}^3 \] Step 2: Use formula \[ Z = \frac{\rho \cdot N_A \cdot V}{M} \] Step 3: Substitute values \[ Z = \frac{3 \times 6.022 \times 10^{23} \times 2.7 \times 10^{-23}}{40} \] Step 4: Calculate numerator \[ 3 \times 6.022 \times 2.7 = 48.7782 \] Step 5: Final calculation \[ Z = \frac{48.7782}{40} = 1.219455 \] Step 6: Nearest integer \[ Z \approx 1 \] Final Answer: 1 atom per unit cell