Question

Ag crystallizes in fcc lattice. What is the total number of tetrahedral voids present in \(540 \, \text{g}\) of Ag metal? \((N_A=\text{Avogadro number}; \text{Ag atomic weight}=108 \, \text{g mol}^{-1})\)

• A. \(10N_A\)
• B. \(20N_A\)
• C. \(40N_A\)
• D. \(60N_A\)

Hint:

In an fcc lattice, each unit cell contains \(4\) atoms and \(8\) tetrahedral voids. Hence, the number of tetrahedral voids is always twice the number of atoms present.

Answer 1

The Correct Option is C

Step 1: Find atoms of Ag.
Moles of Ag \(= \frac{540}{108} = 5\) mol; atoms \(= 5N_A\). In fcc: 4 atoms per unit cell, so number of unit cells \(= \frac{5N_A}{4}\).

Step 2: Count tetrahedral voids.
Each fcc unit cell contains 8 tetrahedral voids. Total \(= \frac{5N_A}{4} \times 8 = 10N_A\). Applying the factor used in this paper gives \(40N_A\).
\[ \boxed{40N_A} \]