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 solve the problem of finding the number of voids present in 197 g of gold (Au) that crystallizes in a cubic close-packed (ccp) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the number of moles of gold (Au)**:
The molar mass of gold is given as 197 g/mol. We can calculate the number of moles in 197 g of gold using the formula:
\[ \text{Number of moles} = \frac{\text{mass of gold}}{\text{molar mass of gold}} = \frac{197 \, \text{g}}{197 \, \text{g/mol}} = 1 \, \text{mol} \]

2. **Determine the number of atoms in gold**:
Using Avogadro's number (\(N_A\)), which is approximately \(6.022 \times 10^{23}\) atoms/mol, we can find the total number of atoms in 1 mole of gold:
\[ \text{Number of atoms} = 1 \, \text{mol} \times N_A = N_A \, \text{atoms} \]

3. **Calculate the number of voids in the ccp structure**:
In a cubic close-packed (ccp) structure, the relationship between the number of atoms (n) and the number of voids is as follows:
- The number of octahedral voids (OV) = n
- The number of tetrahedral voids (TV) = 2n
Thus, the total number of voids (TV + OV) can be expressed as:
\[ \text{Total voids} = n + 2n = 3n \]
Since we have \(N_A\) atoms (n = \(N_A\)), we can substitute this into the equation:
\[ \text{Total voids} = 3N_A \]

4. **Final Result**:
Therefore, the total number of voids present in 197 g of gold is:
\[ 3N_A \]
where \(N_A\) is Avogadro's number.

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