If `NaCl` is doped with `10^(-4)mol%`of `SrCl_(2)` the concentration of cation vacancies will be `(N_(A)=6.02xx10^(23)mol^(-1))`

Options

A. `6.023xx10^(15)"mol"^(-1)`
B. `6.023xx10^(16)"mol"^(-1)`
C. `6.023xx10^(17)"mol"^(-1)`
D. `6.023xx10^(14)"mol"^(-1)`

Explanation

To solve the problem of determining the concentration of cation vacancies when NaCl is doped with SrCl₂, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Doping Concentration**:
- We are given that NaCl is doped with \(10^{-4}\) mol% of SrCl₂.
- Convert this percentage into moles:
\[ \text{Moles of SrCl₂} = \frac{10^{-4}}{100} = 10^{-6} \text{ moles} \]

2. **Identify the Contribution to Cation Vacancies**:
- Each mole of SrCl₂ contributes one mole of Sr²⁺ ions.
- Each Sr²⁺ ion creates one cation vacancy in the NaCl lattice.

3. **Calculate the Number of Cation Vacancies**:
- Since \(10^{-6}\) moles of Sr²⁺ are produced, the number of cation vacancies created will also be \(10^{-6}\) moles.

4. **Convert Moles to Number of Vacancies**:
- To find the total number of cation vacancies, we use Avogadro's number (\(N_A = 6.02 \times 10^{23} \text{ mol}^{-1}\)):
\[ \text{Number of cation vacancies} = 10^{-6} \text{ moles} \times 6.02 \times 10^{23} \text{ mol}^{-1} \]
\[ = 6.02 \times 10^{17} \text{ vacancies} \]

5. **Final Answer**:
- The concentration of cation vacancies in the doped NaCl is \(6.02 \times 10^{17}\).