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: Mass in one cell:
A bcc cell has 2 atoms. One mole of atoms weighs 225 g, so the mass of 2 atoms is $2 \times 225 / N_A$ g.

Step 2: Volume of one cell:
$a^3$. The problem supplies $a^3 N_A = 75$ cm$^3$, meaning volume per mole of cells is 75.

Step 3: Divide:
\[ \rho = \frac{2 \times 225}{N_A a^3} = \frac{450}{75} = 6 \text{ g cm}^{-3} \]

Step 4: Check:
Doubling to fcc would give 12, so 6 is in line for bcc.

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