Question:
Ag crystallizes in fcc lattice. What is the total number of tetrahedral voids present in \(540 \, \text{g}\) of Ag metal? \((N_A=\text{Avogadro number}; \text{Ag atomic weight}=108 \, \text{g mol}^{-1})\)
Ag crystallizes in fcc lattice. What is the total number of tetrahedral voids present in \(540 \, \text{g}\) of Ag metal? \((N_A=\text{Avogadro number}; \text{Ag atomic weight}=108 \, \text{g mol}^{-1})\)
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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.
Updated On: Jun 26, 2026
- \(10N_A\)
- \(20N_A\)
- \(40N_A\)
- \(60N_A\)
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The Correct Option is C
Solution and Explanation
Step 1: Calculate the number of moles of Ag.
Given mass of Ag is \[ 540 \, \text{g} \] Atomic weight of Ag is \[ 108 \, \text{g mol}^{-1} \] Therefore, number of moles of Ag is \[ \frac{540}{108}=5 \] So, \(540 \, \text{g}\) of Ag contains \[ 5N_A \] Ag atoms.
Step 2: Relate atoms and tetrahedral voids in fcc lattice.
In a close-packed structure such as fcc lattice, the number of tetrahedral voids is twice the number of atoms present.
Therefore, \[ \text{Number of tetrahedral voids}=2 \times \text{number of atoms} \]
Step 3: Calculate total tetrahedral voids.
Here, number of Ag atoms is \[ 5N_A \] Hence, total number of tetrahedral voids is \[ 2 \times 5N_A \] \[ =10N_A \] But in an fcc unit cell, there are \(4\) atoms and \(8\) tetrahedral voids.
For \(5N_A\) atoms, the number of fcc unit cells is \[ \frac{5N_A}{4} \] Each fcc unit cell contains \[ 8 \] tetrahedral voids.
Thus, total tetrahedral voids are \[ \frac{5N_A}{4}\times 8 \] \[ =10N_A \]
Step 4: Final conclusion.
Therefore, the total number of tetrahedral voids present in \(540 \, \text{g}\) of Ag metal is \[ \boxed{10N_A} \]
Given mass of Ag is \[ 540 \, \text{g} \] Atomic weight of Ag is \[ 108 \, \text{g mol}^{-1} \] Therefore, number of moles of Ag is \[ \frac{540}{108}=5 \] So, \(540 \, \text{g}\) of Ag contains \[ 5N_A \] Ag atoms.
Step 2: Relate atoms and tetrahedral voids in fcc lattice.
In a close-packed structure such as fcc lattice, the number of tetrahedral voids is twice the number of atoms present.
Therefore, \[ \text{Number of tetrahedral voids}=2 \times \text{number of atoms} \]
Step 3: Calculate total tetrahedral voids.
Here, number of Ag atoms is \[ 5N_A \] Hence, total number of tetrahedral voids is \[ 2 \times 5N_A \] \[ =10N_A \] But in an fcc unit cell, there are \(4\) atoms and \(8\) tetrahedral voids.
For \(5N_A\) atoms, the number of fcc unit cells is \[ \frac{5N_A}{4} \] Each fcc unit cell contains \[ 8 \] tetrahedral voids.
Thus, total tetrahedral voids are \[ \frac{5N_A}{4}\times 8 \] \[ =10N_A \]
Step 4: Final conclusion.
Therefore, the total number of tetrahedral voids present in \(540 \, \text{g}\) of Ag metal is \[ \boxed{10N_A} \]
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