Question

A metallic element crystallizes in simple cubic lattice. If edge length of the unit cell is 34Å, with density 8 g/cm\(^3\), what is the number of unit cells in 100 g of the metal? (Molar mass of metal = 108 g/mol)

• A. \( 1.33 \times 10^{20} \)
• B. \( 2 \times 10^{24} \)
• C. \( 2.7 \times 10^{22} \)
• D. \( 5 \times 10^{23} \)

Hint:

For calculating the number of unit cells, use the relationship between molar mass, density, and the volume of the unit cell.

Answer 1

The Correct Option is D

Step 1: Understanding the simple cubic lattice.
In a simple cubic lattice, the number of atoms per unit cell is 1. The formula to calculate the number of unit cells is: \[ \text{Number of unit cells} = \frac{\text{Mass of metal}}{\text{Mass of one unit cell}} \]
Step 2: Calculating the volume of the unit cell.
The volume of the unit cell is given by: \[ V_{\text{unit cell}} = a^3 = (34 \times 10^{-10})^3 \, \text{cm}^3 \] Where \(a = 34 \, \text{Å} = 34 \times 10^{-10} \, \text{cm}\).

Step 3: Finding the number of unit cells.
Using the density formula: \[ \text{Density} = \frac{M}{V_{\text{unit cell}} \times N_{\text{A}}} \] We calculate the number of unit cells in 100 g of metal. The final result is \( 5 \times 10^{23} \) unit cells.

Step 4: Conclusion.
The correct answer is (D) \( 5 \times 10^{23} \).