If `NaCl` is doped with `10^(-8)mol%`of `SrCl_(2)` the concentration of cation vacancies will be `(N_(A)=6.02xx10^(23)mol^(-1))`

Options

A. `6.022xx10^(16)" mol"^(-1)`
B. `6.022 xx 10^(17)"mol"^(-1)`
C. `6.022xx10^(13)"mol"^(-1)`
D. `6.022xx10^(15)"mol"^(-1)`

Explanation

To solve the problem of determining the concentration of cation vacancies in NaCl when doped with \( 10^{-8} \, \text{mol\%} \) of \( \text{SrCl}_2 \), we can follow these steps:

### Step 1: Understand the Doping Process
When \( \text{SrCl}_2 \) is added to \( \text{NaCl} \), the \( \text{Sr}^{2+} \) ions will replace the \( \text{Na}^{+} \) ions in the crystal lattice. Since \( \text{Sr}^{2+} \) has a +2 charge, it will replace two \( \text{Na}^{+} \) ions to maintain electrical neutrality. This replacement creates cation vacancies.

### Step 2: Determine the Ratio of Doping
Given that \( \text{NaCl} \) is doped with \( 10^{-8} \, \text{mol\%} \) of \( \text{SrCl}_2 \), we can express this in terms of moles.

If we consider 100 moles of \( \text{NaCl} \):
\[ \text{Moles of } \text{SrCl}_2 = \frac{10^{-8}}{100} = 10^{-10} \, \text{moles} \]

### Step 3: Calculate the Number of Cation Vacancies
Since one \( \text{Sr}^{2+} \) ion replaces two \( \text{Na}^{+} \) ions, it creates one cation vacancy. Therefore, the number of cation vacancies created by \( 10^{-10} \) moles of \( \text{SrCl}_2 \) is also \( 10^{-10} \).

### Step 4: Convert Moles to Number of Vacancies
To find the total number of cation vacancies, we multiply the number of moles of \( \text{SrCl}_2 \) by Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)):
\[ \text{Number of vacancies} = 10^{-10} \, \text{moles} \times 6.022 \times 10^{23} \, \text{mol}^{-1} \]
\[ = 6.022 \times 10^{13} \, \text{vacancies} \]

### Final Answer
The concentration of cation vacancies in \( \text{NaCl} \) doped with \( 10^{-8} \, \text{mol\%} \) of \( \text{SrCl}_2 \) is \( 6.022 \times 10^{13} \) vacancies.

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