If NaCl is doped with `10^(-2)` mol % `SrCl_2`, what is the concentration of cation vacancies?

Explanation

To find the concentration of cation vacancies when NaCl is doped with \(10^{-2}\) mol % \(SrCl_2\), we can follow these steps:

### Step 1: Understand the Doping Process
When \(SrCl_2\) is added to NaCl, the \(Sr^{2+}\) ions replace some of the \(Na^+\) ions in the NaCl lattice. Each \(Sr^{2+}\) ion that replaces a \(Na^+\) ion creates one cation vacancy because \(Sr^{2+}\) has a +2 charge, while \(Na^+\) has a +1 charge.

### Step 2: Calculate the Amount of Doping
The doping concentration is given as \(10^{-2}\) mol %. To convert this into moles, we can use the following formula:

\[ \text{Moles of } SrCl_2 = \frac{10^{-2}}{100} = 10^{-4} \text{ moles} \]

### Step 3: Calculate the Number of \(Sr^{2+}\) Ions
To find the number of \(Sr^{2+}\) ions, we can use Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles per mole. Thus, the number of \(Sr^{2+}\) ions can be calculated as follows:

\[ \text{Number of } Sr^{2+} \text{ ions} = 10^{-4} \text{ moles} \times 6.022 \times 10^{23} \text{ ions/mole} \]

Calculating this gives:

\[ \text{Number of } Sr^{2+} \text{ ions} = 10^{-4} \times 6.022 \times 10^{23} \approx 6.022 \times 10^{19} \text{ ions} \]

### Step 4: Determine the Number of Cation Vacancies
Since each \(Sr^{2+}\) ion creates one cation vacancy, the number of cation vacancies in the NaCl lattice is equal to the number of \(Sr^{2+}\) ions:

\[ \text{Number of cation vacancies} = 6.022 \times 10^{19} \]

### Final Answer
Thus, the concentration of cation vacancies in NaCl doped with \(10^{-2}\) mol % \(SrCl_2\) is:

\[ 6.022 \times 10^{19} \text{ cation vacancies} \]

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