When AgCI is dopped with 0.01 mole `CdCl_(2)`, What is number of cationic vacancies ?

Options

A. `0.02 N_(A)`
B. `0.01N_(A)`
C. 0.01
D. 0.02

Explanation

To solve the problem of determining the number of cationic vacancies when AgCl is doped with 0.01 mole of CdCl2, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Doping**:
- Doping refers to the introduction of impurities into a semiconductor or ionic crystal to alter its properties. In this case, we are doping AgCl (silver chloride) with CdCl2 (cadmium chloride).

2. **Identifying the Cation**:
- In CdCl2, the cadmium ion (Cd²⁺) is the cation that will replace some of the silver ions (Ag⁺) in the AgCl lattice.

3. **Cationic Vacancy Creation**:
- When a Cd²⁺ ion replaces an Ag⁺ ion in the lattice, it creates a cationic vacancy. This is because the Cd²⁺ ion has a +2 charge, while the Ag⁺ ion has a +1 charge. Therefore, for every Cd²⁺ that replaces an Ag⁺, one cationic vacancy is created.

4. **Calculating the Number of Vacancies**:
- Given that we have 0.01 moles of CdCl2, and each mole of CdCl2 introduces one cationic vacancy per mole of Cd²⁺, we can conclude that:
- 0.01 moles of Cd²⁺ will create 0.01 moles of cationic vacancies.

5. **Converting Moles to Number of Vacancies**:
- To find the actual number of cationic vacancies, we multiply the number of moles of vacancies by Avogadro's number (approximately \(6.022 \times 10^{23}\) particles/mole):
\[ \text{Number of cationic vacancies} = 0.01 \, \text{moles} \times 6.022 \times 10^{23} \, \text{vacancies/mole} \]
\[ = 6.022 \times 10^{21} \, \text{cationic vacancies} \]

### Final Answer:
The number of cationic vacancies when AgCl is doped with 0.01 mole of CdCl2 is approximately \(6.022 \times 10^{21}\) vacancies.

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