Question

Elements A and B have FCC and BCC structures respectively with unit cell edge length \( a = 3 \, \text{Å} \).
The number of atoms in 210 g of A = number of atoms in 594 g of B.
If density of A is \( 7 \, \text{g/cm}^3 \), what is the density of B?

• A. \( 9.9 \, \text{g/cm}^3 \)
• B. \( 4.5 \, \text{g/cm}^3 \)
• C. \( 6.8 \, \text{g/cm}^3 \)
• D. \( 11.2 \, \text{g/cm}^3 \)

Hint:

Use density formula for unit cells: \( \rho = \frac{Z M}{N_A a^3} \), relate masses via atom counts.

Answer 1

The Correct Option is A

Let molar mass of A = \( M_A \), B = \( M_B \) From data:
- A is FCC: 4 atoms/unit cell
- B is BCC: 2 atoms/unit cell
- \( a = 3 \, \text{Å} = 3 \times 10^{-8} \, \text{cm} \)
- Densities: \( \rho = \frac{Z \cdot M}{N_A \cdot a^3} \)
Given:
- Number of atoms in 210 g of A = in 594 g of B
So: \[ \frac{210}{M_A} \cdot N_A = \frac{594}{M_B} \cdot N_A \Rightarrow \frac{210}{M_A} = \frac{594}{M_B} \Rightarrow \frac{M_B}{M_A} = \frac{594}{210} = 2.83 \] Now, use density formula: \[ \rho_A = \frac{4 M_A}{N_A \cdot a^3} = 7 \Rightarrow M_A = \frac{7 \cdot N_A \cdot a^3}{4} \] Use it in: \[ \rho_B = \frac{2 M_B}{N_A \cdot a^3} = \frac{2 \cdot 2.83 \cdot M_A}{N_A \cdot a^3} = 2.83 \cdot \frac{2 M_A}{N_A \cdot a^3} = 2.83 \cdot \frac{2}{4} \cdot 7 = 2.83 \cdot 3.5 = 9.9 \, \text{g/cm}^3 \]