The freezing point depression of `0.001 m K_(x)` `[Fe(CN)_(6)]` is `7.10xx10^(-3) K`. Determine the value of x. Given, `K_(f)=1.86 K kg "mol"^(-1)` for water.

Explanation

To solve the problem, we need to determine the value of \( x \) in the complex \( K_x[Fe(CN)_6] \) given the freezing point depression and the molality of the solution.

### Step-by-Step Solution:

1. **Understand the given data**:
- Freezing point depression (\( \Delta T_f \)) = \( 7.10 \times 10^{-3} \, K \)
- Molality (m) = \( 0.001 \, mol/kg \)
- Freezing point depression constant (\( K_f \)) for water = \( 1.86 \, K \cdot kg/mol \)

2. **Use the formula for freezing point depression**:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where \( i \) is the Van't Hoff factor.

3. **Rearranging the formula to find \( i \)**:
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

4. **Substituting the known values**:
\[ i = \frac{7.10 \times 10^{-3}}{1.86 \cdot 0.001} \]

5. **Calculating \( i \)**:
\[ i = \frac{7.10 \times 10^{-3}}{1.86 \times 10^{-3}} \approx 3.817 \]

6. **Understanding the dissociation of the complex**:
The complex \( K_x[Fe(CN)_6] \) dissociates in solution as follows:
\[ K_x[Fe(CN)_6] \rightarrow x \, K^+ + [Fe(CN)_6]^{x-} \]
This means that for each formula unit of \( K_x[Fe(CN)_6] \), it produces \( x + 1 \) particles in solution (where \( x \) is the number of potassium ions and 1 is for the complex ion).

7. **Setting up the equation for \( i \)**:
\[ i = 1 + x \]
Given that \( i \approx 3.817 \), we can set up the equation:
\[ 1 + x = 3.817 \]

8. **Solving for \( x \)**:
\[ x = 3.817 - 1 = 2.817 \]
Rounding to the nearest whole number, we find:
\[ x \approx 3 \]

9. **Conclusion**:
The value of \( x \) in the complex \( K_x[Fe(CN)_6] \) is approximately 3.

### Final Answer:
The value of \( x \) is 3.