Step 1: Understanding the Concept:
The density of a crystal unit cell is given by the formula:
\[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \]
where \( Z \) is the number of atoms per unit cell, \( M \) is molar mass, \( a \) is edge length, and \( N_A \) is Avogadro's number. We need to solve for \( M \).
Step 2: Key Formula or Approach:
For FCC lattice, \( Z = 4 \).
\( M = \frac{\rho \cdot a^3 \cdot N_A}{Z} \).
Ensure \( a \) is in cm. \( 1 \, \text{pm} = 10^{-10} \, \text{cm} \).
Step 3: Detailed Explanation:
Given:
Density \( \rho = 2 \, \text{g/cm}^3 \).
Edge length \( a = 600 \, \text{pm} = 600 \times 10^{-10} \, \text{cm} = 6 \times 10^{-8} \, \text{cm} \).
Avogadro's number \( N_A = 6 \times 10^{23} \).
\( Z = 4 \).
Calculate \( a^3 \):
\[ a^3 = (6 \times 10^{-8})^3 = 216 \times 10^{-24} \, \text{cm}^3 \]
Calculate \( M \):
\[ M = \frac{2 \times (216 \times 10^{-24}) \times (6 \times 10^{23})}{4} \]
\[ M = \frac{2 \times 216 \times 6 \times 10^{-1}}{4} \]
\[ M = \frac{12 \times 216 \times 0.1}{4} \]
\[ M = 3 \times 216 \times 0.1 \]
\[ M = 3 \times 21.6 \]
\[ M = 64.8 \, \text{g/mol} \]
Step 4: Final Answer:
The molar mass is 64.8 g mol\(^{-1}\).