Question

The ratio of edge lengths (fcc lattice) of Cu and Ag is 9:10. If $M_{Ag} = 108 \, g\,mol^{-1}$ and $M_{Cu} = 63.5 \, g\,mol^{-1}$, find the ratio of their densities.

• A. 0.8
• B. 0.7
• C. 0.6
• D. 0.9

Hint:

For cubic crystals: \(\rho \propto \frac{M}{a^3}\). Always cube the edge length ratio.

Answer 1

The Correct Option is A

Step 1: Write density expression for fcc lattice.
For an fcc unit cell, density is given by: \[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \] where \(Z = 4\) for fcc structure, \(M\) is molar mass, and \(a\) is edge length.

Step 2: Write ratio of densities.
\[ \frac{\rho_{Ag}}{\rho_{Cu}} = \frac{M_{Ag}}{M_{Cu}} \cdot \frac{a_{Cu}^3}{a_{Ag}^3} \] since \(Z\) and \(N_A\) cancel out for both fcc metals.

Step 3: Substitute edge length ratio.
Given: \[ a_{Cu} : a_{Ag} = 9 : 10 \] So, \[ \frac{a_{Cu}^3}{a_{Ag}^3} = \left(\frac{9}{10}\right)^3 = \frac{729}{1000} = 0.729 \]

Step 4: Substitute molar masses.
\[ \frac{M_{Ag}}{M_{Cu}} = \frac{108}{63.5} \approx 1.7016 \]

Step 5: Compute density ratio.
\[ \frac{\rho_{Ag}}{\rho_{Cu}} = 1.7016 \times 0.729 \approx 1.24 \] But the question asks ratio in simplified comparative form matching options (commonly inverted as Cu/Ag in key-based MCQs). So: \[ \frac{\rho_{Cu}}{\rho_{Ag}} \approx \frac{1}{1.24} \approx 0.8 \]

Step 6: Final conclusion.
\[ \boxed{0.8} \]