If NaCI is doped with `10^(-3)"mol"%SrCI_(2)` then the concentration of cation vacancies will be:

Options

A. `1xx10^(-3)mol%`
B. `2xx10^(-3)"mol"%`
C. `3xx10^(-3)"mol"%`
D. `4xx10^(-3)"mol"%`

Answer

To solve the problem of determining the concentration of cation vacancies in NaCl when doped with \(10^{-3} \text{ mol\%} \text{SrCl}_2\), we can follow these steps:

### Step 1: Understand the Doping Process
When SrCl₂ is added to NaCl, the strontium ions (Sr²⁺) will replace some of the sodium ions (Na⁺) in the NaCl lattice. Since Sr²⁺ has a +2 charge, it will create a charge imbalance in the crystal structure.

### Step 2: Calculate the Amount of SrCl₂ Added
Given that the doping level is \(10^{-3} \text{ mol\%}\), we can express this in terms of moles. If we assume we have 1 mole of NaCl, then:

\[ \text{Moles of SrCl}_2 = 10^{-3} \text{ mol\%} \times 1 \text{ mole} = 10^{-5} \text{ moles} \]

### Step 3: Determine the Number of Cation Vacancies Created
When Sr²⁺ replaces Na⁺, it creates a charge imbalance. For every Sr²⁺ that replaces Na⁺, one Na⁺ vacancy is created because Sr²⁺ needs to balance the charge by creating a vacancy for the missing Na⁺ ion.

Since each Sr²⁺ can replace 2 Na⁺ ions, we can say:

- For every 1 mole of Sr²⁺, 1 cation vacancy is created.

### Step 4: Calculate the Total Cation Vacancies
From the previous step, we know that:

\[ \text{Cation vacancies created} = \text{Moles of Sr}^{2+} = 10^{-5} \text{ moles} \]

### Step 5: Convert Moles to Concentration
To find the concentration of cation vacancies, we can express it in terms of molarity. Assuming the volume of the solution is 1 liter (for simplicity):

\[ \text{Concentration of cation vacancies} = 10^{-5} \text{ moles/L} \]

### Final Answer
Thus, the concentration of cation vacancies in NaCl when doped with \(10^{-3} \text{ mol\%} \text{SrCl}_2\) is:

\[ \text{Concentration of cation vacancies} = 10^{-5} \text{ mol/L} \]

---