An ionic compound MCl is simultaneously doped with `10^(-6)` mole % `NCl_2` and `10^(-7)` mole % `RCl_3`.Calculate the concentration of cationic vacancies x. Express your answer as `x/9xx10^14`.
(take N & R =divalent & trivalent cations)

Explanation

To solve the problem, we need to calculate the concentration of cationic vacancies (x) created when the ionic compound MCl is doped with NCl₂ and RCl₃.

### Step-by-Step Solution:

1. **Identify the Contributions of Doping:**
- NCl₂ is a divalent cation (N²⁺), which creates 1 cationic vacancy per formula unit.
- RCl₃ is a trivalent cation (R³⁺), which creates 2 cationic vacancies per formula unit.

2. **Calculate the Moles of Doping:**
- The mole percent of NCl₂ is \(10^{-6}\) mole %, which means:
\[ \text{Moles of NCl₂} = \frac{10^{-6}}{100} = 10^{-8} \text{ moles} \]
- The mole percent of RCl₃ is \(10^{-7}\) mole %, which means:
\[ \text{Moles of RCl₃} = \frac{10^{-7}}{100} = 10^{-9} \text{ moles} \]

3. **Calculate the Number of Vacancies Created:**
- For NCl₂:
\[ \text{Cationic vacancies from NCl₂} = 10^{-8} \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \times 1 = 6.022 \times 10^{15} \text{ vacancies} \]
- For RCl₃:
\[ \text{Cationic vacancies from RCl₃} = 10^{-9} \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \times 2 = 1.2044 \times 10^{15} \text{ vacancies} \]

4. **Total Cationic Vacancies:**
- Adding the vacancies from both dopants:
\[ \text{Total vacancies} = 6.022 \times 10^{15} + 1.2044 \times 10^{15} = 7.2264 \times 10^{15} \text{ vacancies} \]

5. **Expressing the Result:**
- We need to express the concentration of cationic vacancies in the form \( \frac{x}{9 \times 10^{14}} \).
- First, convert \( 7.2264 \times 10^{15} \) to the required format:
\[ 7.2264 \times 10^{15} = 72.264 \times 10^{14} \]
- To express this as \( \frac{x}{9 \times 10^{14}} \):
\[ x = 72.264 \times 9 = 650.376 \]

6. **Final Answer:**
- Rounding \( x \) to the nearest whole number gives:
\[ x \approx 650 \]
- Therefore, the final answer is:
\[ \frac{650}{9 \times 10^{14}} \]