Question

An element crystallizes in a fcc lattice with cell edge 250 pm. Calculate the density of an element (at mass = 90.3).

• A. 23.12 g cm\(^{-3}\)
• B. 19.20 g cm\(^{-3}\)
• C. 48.40 g cm\(^{-3}\)
• D. 38.40 g cm\(^{-3}\)

Hint:

For a fcc lattice, use the formula \( \rho = \frac{Z \times M}{N_A \times a^3} \) to calculate density.

Answer 1

The Correct Option is D

Step 1: Using the formula for density.
Density \( \rho \) is given by: \[ \rho = \frac{Z \times M}{N_A \times a^3} \] Where: - \( Z = 4 \) for fcc (face-centered cubic), - \( M = 90.3 \, \text{g/mol} \), - \( a = 250 \, \text{pm} = 250 \times 10^{-12} \, \text{m} \), - \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \).
Step 2: Calculation.
Substitute the values: \[ \rho = \frac{4 \times 90.3 \, \text{g/mol}}{6.022 \times 10^{23} \times (250 \times 10^{-12})^3} \] This results in a density of \( 38.40 \, \text{g/cm}^3 \).
Step 3: Conclusion.
The correct answer is (D) 38.40 g cm\(^{-3}\).