If `NaCl` is doped with `10^(-3)` mol% of `SrCl_(2)`, what is the concentration of cation vacancies?

Explanation

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

### Step 1: Understand the Doping Process
When \(SrCl_2\) is introduced into the NaCl crystal, \(Sr^{2+}\) ions replace some of the \(Na^+\) ions. Since \(Sr^{2+}\) has a higher charge than \(Na^+\), to maintain electrical neutrality, a \(Na^+\) ion must be removed from the crystal lattice, creating a cation vacancy.

### Step 2: Calculate the Amount of \(SrCl_2\) Doped
Given that the doping concentration is \(10^{-3}\) mol%, we can express this in terms of moles.

- \(10^{-3}\) mol% means that in 100 moles of NaCl, there are \(10^{-3}\) moles of \(SrCl_2\).
- Therefore, in 1 mole of NaCl, the amount of \(SrCl_2\) is:
\[ \text{Moles of } SrCl_2 = \frac{10^{-3}}{100} = 10^{-5} \text{ moles} \]

### Step 3: Determine the Number of Cation Vacancies
For every mole of \(SrCl_2\) introduced, one \(Na^+\) ion is replaced by one \(Sr^{2+}\) ion, resulting in one cation vacancy. Thus, the number of cation vacancies created is equal to the moles of \(SrCl_2\) added.

- Therefore, the number of cation vacancies is:
\[ \text{Cation Vacancies} = 10^{-5} \text{ moles of } SrCl_2 \]

### Step 4: Convert Moles to Number of Vacancies
To find the concentration of cation vacancies in terms of number of vacancies, we use Avogadro's number (\(6.022 \times 10^{23}\) particles/mole):

- Number of cation vacancies:
\[ \text{Number of Cation Vacancies} = 10^{-5} \text{ moles} \times 6.022 \times 10^{23} \text{ particles/mole} \]
\[ = 6.022 \times 10^{18} \text{ cation vacancies} \]

### Conclusion
The concentration of cation vacancies in the NaCl crystal doped with \(10^{-3}\) mol% of \(SrCl_2\) is \(6.022 \times 10^{18}\) vacancies.

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