If NaCl is doped with `10^(-3)` mol % `SrCl_2` , what is the concentration of cation vacancies ?

Explanation

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

### Step 1: Understand the doping percentage
The doping percentage of \(10^{-3}\) mol % means that for every 100 moles of NaCl, there are \(10^{-3}\) moles of \(SrCl_2\).

### Step 2: Calculate the amount of \(SrCl_2\) in 1 mole of NaCl
If we consider 1 mole of NaCl, the amount of \(SrCl_2\) doped would be:
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