The osmotic pressure of a dilute solution of an compound XY in water is four times that of a solution of 0.01 M `BaCl_(2)` in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY in solution is `x xx 10^(–2)` mol/L. The numerical value of x is____.

Explanation

To solve the problem, we need to determine the concentration of the compound XY in the solution based on the given information about osmotic pressure and the dissociation of ionic compounds.

### Step-by-Step Solution:

1. **Understand Osmotic Pressure**: The osmotic pressure (π) of a solution can be expressed using the formula:
\[ \pi = iCRT \]
where:
- \(i\) = Van't Hoff factor (number of particles the solute dissociates into)
- \(C\) = molar concentration of the solute
- \(R\) = ideal gas constant
- \(T\) = temperature (assumed constant)

2. **Identify the Given Information**:
- The osmotic pressure of the solution of compound XY is 4 times that of a 0.01 M BaCl₂ solution.
- For BaCl₂, it dissociates into 3 ions (1 Ba²⁺ and 2 Cl⁻), so its Van't Hoff factor \(i\) is 3.

3. **Calculate the Osmotic Pressure of BaCl₂**:
\[ \pi_{\text{BaCl}_2} = i \cdot C \cdot R \cdot T = 3 \cdot 0.01 \cdot R \cdot T \]

4. **Express the Osmotic Pressure of XY**:
- Let the concentration of XY be \(C_{XY}\) and since XY dissociates into 2 ions (X⁺ and Y⁻), its Van't Hoff factor \(i\) is 2.
\[ \pi_{XY} = 2 \cdot C_{XY} \cdot R \cdot T \]

5. **Set Up the Relationship**:
- According to the problem, the osmotic pressure of XY is 4 times that of BaCl₂:
\[ \pi_{XY} = 4 \cdot \pi_{\text{BaCl}_2} \]
Substituting the expressions for osmotic pressure:
\[ 2 \cdot C_{XY} \cdot R \cdot T = 4 \cdot (3 \cdot 0.01 \cdot R \cdot T) \]

6. **Simplify the Equation**:
- Since \(R\) and \(T\) are constants, they can be canceled from both sides:
\[ 2 \cdot C_{XY} = 4 \cdot 3 \cdot 0.01 \]
\[ 2 \cdot C_{XY} = 0.12 \]

7. **Solve for \(C_{XY}\)**:
\[ C_{XY} = \frac{0.12}{2} = 0.06 \text{ mol/L} \]

8. **Express in Required Format**:
- The problem asks for the concentration in the form \(x \times 10^{-2}\) mol/L:
\[ C_{XY} = 6 \times 10^{-2} \text{ mol/L} \]
Therefore, the numerical value of \(x\) is 6.

### Final Answer:
The numerical value of \(x\) is **6**.