Question

Calculate the density of an element having molar mass \(225 \text{g mol}^{-1}\) forming bcc structure \([a^3\times N_A = 75 \text{cm}^3\text{mol}^{-1}]\)

• A. \(6.0\) g cm\(^{-3}\)
• B. \(2.81\) g cm\(^{-3}\)
• C. \(9.24\) g cm\(^{-3}\)
• D. \(11.36\) g cm\(^{-3}\)

Hint:

Use rho = Z M / (a^3 N_A) with Z = 2 for bcc.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Density of a crystal equals the mass of atoms in one unit cell divided by the volume of the cell.

Step 2: Key Formula:
\[ \rho = \frac{Z \times M}{a^3 \times N_A} \]
For bcc, \(Z = 2\) atoms per unit cell (8 corners at \(1/8\) each plus 1 body centre).

Step 3: Detailed Explanation:
Given \(M = 225\) g mol\(^{-1}\) and \(a^3 N_A = 75\) cm\(^3\) mol\(^{-1}\).
\[ \rho = \frac{2 \times 225}{75} = \frac{450}{75} = 6.0 \text{ g cm}^{-3} \]

Step 4: Why the other options are wrong.
Using \(Z = 1\) (simple cubic) would give 3.0, and \(Z = 4\) (fcc) would give 12. The other options 2.81, 9.24 and 11.36 do not follow from any valid value of \(Z\) with these data.

Final Answer:
The density is \(6.0\) g cm\(^{-3}\), option (A). \[ \boxed{6.0 \text{ g cm}^{-3}} \]