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: Use density formula for unit cell.
Density of a crystal is given by \[ \rho=\frac{Z\times M}{N_A\times a^3} \] where \[ Z=\text{number of atoms per unit cell} \] \[ M=\text{molar mass} \] \[ N_A=\text{Avogadro number} \] and \[ a=\text{edge length of unit cell} \]

Step 2: Write \(Z\) values for fcc and bcc.
For element A, structure is fcc, so \[ Z_A=4 \] For element B, structure is bcc, so \[ Z_B=2 \] Both have the same edge length, \[ a_A=a_B=3\mathring{A} \]

Step 3: Use the given condition about number of atoms.
The number of atoms in \[ 210\,g \] of A is equal to the number of atoms in \[ 594\,g \] of B.
Number of atoms is proportional to \[ \frac{\text{given mass}}{\text{molar mass}} \] So, \[ \frac{210}{M_A}=\frac{594}{M_B} \] Rearranging, \[ \frac{M_B}{M_A}=\frac{594}{210} \] \[ \frac{M_B}{M_A}=2.82857 \]

Step 4: Relate densities of A and B.
Since both have the same edge length, \[ \rho \propto ZM \] Therefore, \[ \frac{\rho_B}{\rho_A}=\frac{Z_BM_B}{Z_AM_A} \] Substituting values, \[ \frac{\rho_B}{7}=\frac{2\times M_B}{4\times M_A} \] \[ \frac{\rho_B}{7}=\frac{1}{2}\times \frac{M_B}{M_A} \] \[ \frac{\rho_B}{7}=\frac{1}{2}\times 2.82857 \] \[ \frac{\rho_B}{7}=1.414285 \] \[ \rho_B=7\times 1.414285 \] \[ \rho_B\approx 9.9\,g\,cm^{-3} \]

Step 5: Final conclusion.
Therefore, the density of B is \[ \boxed{9.9\,g\,cm^{-3}} \]