If an ionic solid XY (X & Y are monovalent ions) is doped with `10^-2` moles % of another ionic solid `AY_3` , then the concentration of the cation vacancies created is

Options

A. `6.023xx10^19 mol^-1`
B. `60.23xx10^18 mol^-1`
C. `12.05xx10^21 mol^-1`
D. `1.205xx10^21 mol^-1`

Answer

To solve the problem, we need to determine the concentration of cation vacancies created when an ionic solid XY is doped with another ionic solid AY₃.

### Step-by-Step Solution:

1. **Understanding the Doping Process**:
- The ionic solid XY consists of monovalent ions X⁺ and Y⁻.
- The ionic solid AY₃ consists of A³⁺ (a trivalent cation) and Y³⁻ (a monovalent anion).
- When AY₃ is doped into XY, the trivalent cation A³⁺ replaces some of the X⁺ ions, leading to the creation of cation vacancies.

2. **Calculating the Amount of AY₃ Doped**:
- We are given that the doping concentration is \(10^{-2}\) moles %.
- This means that for every 100 moles of XY, we have \(10^{-2}\) moles of AY₃.

3. **Total Moles of XY**:
- Assume we have 100 moles of XY. Therefore, the amount of AY₃ doped is:
\[ \text{Moles of AY₃} = 10^{-2} \text{ moles % of 100 moles} = 10^{-2} \text{ moles} \]

4. **Understanding the Cation Vacancy Creation**:
- Each A³⁺ ion replaces three X⁺ ions, which means that for every mole of AY₃, three cation vacancies are created.
- Therefore, the number of cation vacancies created by doping with AY₃ is:
\[ \text{Cation vacancies} = 3 \times \text{Moles of AY₃} \]
- Substituting the value of AY₃:
\[ \text{Cation vacancies} = 3 \times 10^{-2} \text{ moles} = 3 \times 10^{-2} \text{ moles} \]

5. **Calculating the Concentration of Cation Vacancies**:
- To find the concentration of cation vacancies, we need to convert moles to number of vacancies using Avogadro's number (\(6.022 \times 10^{23}\)):
\[ \text{Number of cation vacancies} = 3 \times 10^{-2} \text{ moles} \times 6.022 \times 10^{23} \text{ vacancies/mole} \]
- Performing the calculation:
\[ \text{Number of cation vacancies} = 3 \times 10^{-2} \times 6.022 \times 10^{23} \approx 1.81 \times 10^{22} \text{ vacancies} \]

### Final Answer:
The concentration of cation vacancies created is approximately \(1.81 \times 10^{22}\) vacancies.