The Van't Hoff factor (i) for a dilute solution of `K_(3)[Fe(CN)_(6)]` is

Explanation

To find the Van't Hoff factor (i) for a dilute solution of \( K_3[Fe(CN)_6] \), we will follow these steps:

### Step 1: Identify the compound
The compound given is \( K_3[Fe(CN)_6] \). This is a coordination complex where potassium ions (K\(^+\)) and the complex ion \( [Fe(CN)_6]^{3-} \) are present.

### Step 2: Write the dissociation reaction
When \( K_3[Fe(CN)_6] \) is dissolved in water, it dissociates into its constituent ions. The dissociation can be represented as:
\[ K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-} \]

### Step 3: Count the number of particles produced
From the dissociation reaction, we can see that:
- 3 potassium ions (K\(^+\)) are produced.
- 1 complex ion \( [Fe(CN)_6]^{3-} \) is produced.

Thus, the total number of particles produced upon dissociation is:
\[ n = 3 + 1 = 4 \]

### Step 4: Determine the Van't Hoff factor (i)
The Van't Hoff factor (i) is defined as the number of particles into which a solute dissociates in solution. Therefore, for our case:
\[ i = n = 4 \]

### Final Answer
The Van't Hoff factor (i) for a dilute solution of \( K_3[Fe(CN)_6] \) is **4**.

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