NaCl is doped with `2xx10^(-3)` mol % `SrCl_2` , the concentration of cation vacancies is

Options

A. `3.01xx10^(18) "mol"^(-1)`
B. `12.04 xx10^(18) "mol"^(-1)`
C. `6.02xx10^(18) "mol" ^(-1)`
D. `12.04 xx10^(20) "mol"^(-1) `

Answer

To solve the problem of finding the concentration of cation vacancies in NaCl doped with \(2 \times 10^{-3}\) mol % of \(SrCl_2\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Doping**:
- When \(SrCl_2\) is added to NaCl, it introduces cation vacancies because \(Sr^{2+}\) ions replace \(Na^+\) ions. Each \(Sr^{2+}\) ion replaces two \(Na^+\) ions, creating one cation vacancy.

2. **Calculating the Number of Cation Vacancies**:
- The mol percentage of \(SrCl_2\) is given as \(2 \times 10^{-3}\) mol %. This means that for every 100 moles of NaCl, there are \(2 \times 10^{-3}\) moles of \(SrCl_2\).

3. **Finding the Concentration of Cation Vacancies**:
- The concentration of cation vacancies can be calculated using the formula:
\[ \text{Cation Vacancies} = \left(\frac{2 \times 10^{-3}}{100}\right) \times N_A \]
- Here, \(N_A\) (Avogadro's number) is approximately \(6.02 \times 10^{23}\) mol\(^{-1}\).

4. **Substituting Values**:
- Plugging in the values:
\[ \text{Cation Vacancies} = \left(\frac{2 \times 10^{-3}}{100}\right) \times (6.02 \times 10^{23}) \]

5. **Calculating the Result**:
- First, calculate \(\frac{2 \times 10^{-3}}{100}\):
\[ = 2 \times 10^{-5} \]
- Now multiply by \(N_A\):
\[ \text{Cation Vacancies} = 2 \times 10^{-5} \times (6.02 \times 10^{23}) = 12.04 \times 10^{20} \]

6. **Final Answer**:
- The concentration of cation vacancies is:
\[ 12.04 \times 10^{20} \text{ vacancies per mole of NaCl} \]

### Conclusion:
The concentration of cation vacancies in NaCl doped with \(2 \times 10^{-3}\) mol % of \(SrCl_2\) is \(12.04 \times 10^{20}\) vacancies per mole of NaCl.