Question

What is the density of an element (Atomic mass 100 g mol\(^{-1}\)) having BCC structure with edge length 400 pm?

• A. 3.2 g cm\(^{-3}\)
• B. 8.2 g cm\(^{-3}\)
• C. 5.18 g cm\(^{-3}\)
• D. 4.8 g cm\(^{-3}\)

Hint:

In BCC, \(Z = 2\). Always convert pm to cm (1 pm = \(10^{-10}\) cm) for density in g/cm\(^3\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Given atomic mass = 100 g/mol, BCC structure (number of atoms per unit cell = 2), edge length a = 400 pm. We need density.

Step 2: Key Formula or Approach:
Density \(\rho = \frac{Z \times M}{N_A \times a^3}\), where \(Z\) = atoms per unit cell, \(M\) = molar mass, \(N_A\) = Avogadro’s number, \(a\) = edge length in cm.

Step 3: Detailed Explanation:
Convert a = 400 pm = \(400 \times 10^{-12}\) m = \(400 \times 10^{-10}\) cm = \(4.00 \times 10^{-8}\) cm. \(a^3 = (4.00 \times 10^{-8})^3 = 64.0 \times 10^{-24} = 6.40 \times 10^{-23}\) cm\(^3\). Mass of unit cell = \(\frac{2 \times 100}{6.022 \times 10^{23}} = \frac{200}{6.022 \times 10^{23}} = 3.322 \times 10^{-22}\) g. \(\rho = \frac{3.322 \times 10^{-22}}{6.40 \times 10^{-23}} = 5.19\) g cm\(^{-3}\) (approx).

Step 4: Final Answer:
Density ≈ 5.18 g cm\(^{-3}\), option (C).