If NaCl is doped with `10^(-3)` mol percent of `SrCI_(2)`, what is the concentration of cation vacancy?

Explanation

To find the concentration of cation vacancies when NaCl is doped with \(10^{-3}\) mol percent of \(SrCl_2\), 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 crystal lattice. Since \(Sr^{2+}\) has a +2 charge, it will create a charge imbalance in the lattice.

### Step 2: Determine the Effect of Doping
For every \(Sr^{2+}\) ion that replaces a \(Na^+\) ion, one \(Na^+\) ion will be removed, creating a cation vacancy to balance the charge. Therefore, the addition of one \(Sr^{2+}\) ion results in the creation of one cation vacancy.

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