Gold crystallises in ccp structure. The number of voids present in 197 g of gold will be [Au = 197]

Options

A. `197xx N_(A)`
B. `2N_(A)`
C. `3N_(A)`
D. `4xx197 N_(A)`

Answer

To determine the number of voids present in 197 g of gold (Au) which crystallizes in a close-packed structure (CCP or FCC), we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the number of moles of gold:**
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{197 \, \text{g}}{197 \, \text{g/mol}} = 1 \, \text{mol} \]

2. **Determine the number of atoms in 1 mole of gold:**
\[ \text{Number of atoms} = N_A = 6.022 \times 10^{23} \, \text{atoms} \]
Here, \(N_A\) is Avogadro's number.

3. **Identify the structure type:**
Gold crystallizes in a close-packed structure, which is a face-centered cubic (FCC) structure.

4. **Calculate the number of atoms per unit cell in FCC:**
In FCC, there are 4 atoms per unit cell. Therefore, \(n = 4\).

5. **Calculate the number of voids in the FCC structure:**
- The number of octahedral voids = \(n\)
- The number of tetrahedral voids = \(2n\)
- Total voids = \(n + 2n = 3n\)

6. **Substitute \(n\) with the number of atoms:**
Since we have 1 mole of gold, the total number of atoms is \(N_A\):
\[ \text{Total voids} = 3n = 3 \times N_A = 3 \times 6.022 \times 10^{23} \, \text{voids} \]

7. **Final result:**
The total number of voids in 197 g of gold is:
\[ 3N_A = 3 \times 6.022 \times 10^{23} \approx 1.8066 \times 10^{24} \, \text{voids} \]

### Conclusion:
The number of voids present in 197 g of gold is \(3N_A\).