If 1 mole of NaCl is doped with `10^(-3)` mole of `SrCl_(2)`. What is the number of cationic vacancies per mole of NaCl ?

Options

A. `10^(-3)"mole"^(-1)`
B. `6.02xx10^(18)"mole"^(-1)`
C. `10^(50)"mole"^(-1)`
D. `6.02xx10^(20)"mole"^(-1)`

Answer

To solve the problem of finding the number of cationic vacancies per mole of NaCl when doped with \(10^{-3}\) moles of \(SrCl_2\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Doping Process**:
- When \(SrCl_2\) is added to NaCl, the \(Sr^{2+}\) ions will replace some of the \(Na^+\) ions in the crystal lattice. Each \(Sr^{2+}\) ion can replace two \(Na^+\) ions to maintain electrical neutrality.

2. **Determining the Number of \(Sr^{2+}\) Ions**:
- We are given that \(10^{-3}\) moles of \(SrCl_2\) are added. Since each mole of \(SrCl_2\) contains one mole of \(Sr^{2+}\) ions, the number of \(Sr^{2+}\) ions introduced is also \(10^{-3}\) moles.

3. **Calculating the Number of Cationic Vacancies**:
- Each \(Sr^{2+}\) ion replaces two \(Na^+\) ions, which means that for every \(Sr^{2+}\) ion introduced, two \(Na^+\) ions leave the lattice, creating one cationic vacancy for each \(Sr^{2+}\) ion.
- Therefore, the number of cationic vacancies created is equal to the number of \(Sr^{2+}\) ions, which is \(10^{-3}\) moles.

4. **Converting Moles to Number of Vacancies**:
- To find the total number of cationic vacancies per mole of NaCl, we multiply the number of moles of \(Sr^{2+}\) ions by Avogadro's number (\(6.022 \times 10^{23}\) mol\(^{-1}\)):
\[ \text{Number of cationic vacancies} = 10^{-3} \text{ moles} \times 6.022 \times 10^{23} \text{ mol}^{-1} \]
\[ = 6.022 \times 10^{20} \]

5. **Final Answer**:
- The number of cationic vacancies per mole of NaCl is \(6.022 \times 10^{20}\).

### Summary:
The number of cationic vacancies per mole of NaCl when doped with \(10^{-3}\) moles of \(SrCl_2\) is \(6.022 \times 10^{20}\).