Question

A compound forms bcc unit cell with edge length \(400\) pm. Determine the molar mass of the compound if the density of compound is \(3.5 \text{g cm}^{-3}\) [\(N_A = 6.022\times 10^{23}\)]

• A. \(65.2 \text{g mol}^{-1}\)
• B. \(67.4 \text{g mol}^{-1}\)
• C. \(69.8 \text{g mol}^{-1}\)
• D. \(71.6 \text{g mol}^{-1}\)

Hint:

Use density = Z x M / (a^3 x NA) with Z = 2 for bcc, and convert pm to cm.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The density of a crystal is the mass of atoms in the unit cell divided by the volume of the cell.

Step 2: Key Formula:
\[ \rho = \dfrac{Z \cdot M}{a^3 \cdot N_A} \]
For bcc, \(Z = 2\).

Step 3: Convert units:
\(a = 400\) pm \(= 400 \times 10^{-10}\) cm \(= 4 \times 10^{-8}\) cm, so \(a^3 = 64 \times 10^{-24}\) cm\(^3\) \(= 6.4 \times 10^{-23}\) cm\(^3\).

Step 4: Solve for M:
\[ M = \dfrac{\rho\,a^3 N_A}{Z} = \dfrac{3.5 \times 6.4 \times 10^{-23} \times 6.022 \times 10^{23}}{2} \]
\[ M = \dfrac{3.5 \times 38.54}{2} = \dfrac{134.9}{2} = 67.4 \text{ g/mol} \]

Step 5: Why the other options are wrong.
Using Z = 4 (fcc) would give 33.7 g/mol times 4 = 134.9 g/mol, and the wrong Z changes the answer by a whole-number factor. The other choices are close values that do not follow from the calculation.

Final Answer:
The molar mass of the compound is 67.4 g/mol. \[ \boxed{\text{(B) }67.4\ \text{g mol}^{-1}} \]