Question

Elements A and B have fcc and bcc structures respectively with a unit cell edge length of \(3\mathring{A}\) for both elements. The number of atoms in \(210\,gm\) of A is equal to \(594\,gm\) of B. If density of A is \(7\,g\,cm^{-3}\), what is the density of B?

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

Hint:

For crystal structures, \[ \rho=\frac{ZM}{N_Aa^3} \] For fcc, \[ Z=4 \] and for bcc, \[ Z=2 \] If edge length is same, compare densities using \[ \rho \propto ZM \]

Answer 1

The Correct Option is A

Step 1: Establish the key relationship between moles.
Given: 210 g of A has the same number of atoms as 594 g of B. Same number of atoms means the same number of moles (since both are monatomic metals). Therefore: \[ \frac{210}{M_A} = \frac{594}{M_B} \Rightarrow \frac{M_B}{M_A} = \frac{594}{210} = \frac{594}{210} \]
Step 2: Write the density formula for both metals.
\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \] For A (FCC): $Z_A = 4$. For B (BCC): $Z_B = 2$. Both have $a = 3$ A = $3 \times 10^{-8}$ cm.
Step 3: Find the ratio of densities.
\[ \frac{\rho_B}{\rho_A} = \frac{Z_B \cdot M_B}{Z_A \cdot M_A} \cdot \frac{a_A^3}{a_B^3} \] Since $a_A = a_B = 3$ A, the $a^3$ terms cancel: \[ \frac{\rho_B}{\rho_A} = \frac{Z_B}{Z_A} \cdot \frac{M_B}{M_A} = \frac{2}{4} \times \frac{594}{210} \]
Step 4: Simplify the ratio.
\[ \frac{\rho_B}{\rho_A} = 0.5 \times \frac{594}{210} = \frac{594}{420} = \frac{99}{70} = 1.4143 \]
Step 5: Calculate $\rho_B$.
\[ \rho_B = \rho_A \times 1.4143 = 7 \times 1.4143 = 9.9 \text{ g/cm}^3 \]
Step 6: Verify and state the answer.
Note that $594/210 = 2.828 \approx 2\sqrt{2}$, so $\rho_B/\rho_A = \frac{2\sqrt{2}}{4} \times 2 = \frac{\sqrt{2}}{1} \approx 1.414$. Thus $\rho_B = 7\sqrt{2} \approx 9.9$ g/cm$^3$.
\[ \boxed{\rho_B = 9.9 \text{ g/cm}^3 \text{ (option 1)}} \]