NaCl is doped with `2xx10^(-3)` mol % `SrCl_2` , the concentration of cation vacancies is

Options

A. `3.01xx10^(18) "mol"^(-1)`
B. `12.04 xx10^(18) "mol"^(-1)`
C. `6.02xx10^(18) "mol" ^(-1)`
D. `12.04 xx10^(20) "mol"^(-1) `

Answer

To solve the problem of finding the concentration of cation vacancies in NaCl doped with \(2 \times 10^{-3}\) mol % \(SrCl_2\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Doping Concentration**:
The concentration of \(SrCl_2\) is given as \(2 \times 10^{-3}\) mol %. This means that for every 100 moles of NaCl, there are \(2 \times 10^{-3}\) moles of \(SrCl_2\).

2. **Convert Mol % to Moles**:
To convert mol % to moles, we can use the formula:
\[ \text{Moles of } SrCl_2 = \frac{2 \times 10^{-3}}{100} = 2 \times 10^{-5} \text{ moles} \]

3. **Determine the Number of \(Sr^{2+}\) Ions**:
Each mole of \(SrCl_2\) produces one mole of \(Sr^{2+}\) ions. Therefore, the number of \(Sr^{2+}\) ions from \(2 \times 10^{-5}\) moles of \(SrCl_2\) is:
\[ \text{Number of } Sr^{2+} = 2 \times 10^{-5} \text{ moles} \times 6.022 \times 10^{23} \text{ ions/mole} \]
\[ = 1.2044 \times 10^{19} \text{ ions} \]

4. **Vacancy Creation**:
When \(Sr^{2+}\) replaces \(Na^+\) in the NaCl lattice, it removes two \(Na^+\) ions (due to its +2 charge) but only occupies one lattice site. This creates one cation vacancy for every \(Sr^{2+}\) ion introduced. Therefore, the number of cation vacancies created is equal to the number of \(Sr^{2+}\) ions:
\[ \text{Number of cation vacancies} = 1.2044 \times 10^{19} \]

5. **Final Answer**:
The concentration of cation vacancies in the NaCl crystal doped with \(SrCl_2\) is:
\[ \text{Cation vacancies} = 1.2044 \times 10^{19} \]