Equal volumes of `0.2M KI` solution and `0.1M HgI_(2)` solution are mixed. Find the van't Hoff factor of the resulting solution.

Explanation

To find the van't Hoff factor of the resulting solution when equal volumes of 0.2M KI and 0.1M HgI₂ are mixed, we can follow these steps:

### Step 1: Understand the Reaction
When KI and HgI₂ are mixed, they react to form a complex. The reaction can be represented as:
\[ 2 \text{KI} + \text{HgI}_2 \rightarrow \text{K}_2\text{HgI}_4 \]

### Step 2: Determine the Dissociation of the Complex
The complex \( \text{K}_2\text{HgI}_4 \) can dissociate into ions:
\[ \text{K}_2\text{HgI}_4 \rightarrow 2 \text{K}^+ + \text{HgI}_4^{2-} \]

### Step 3: Count the Number of Particles Produced
From the dissociation, we see that one formula unit of \( \text{K}_2\text{HgI}_4 \) produces:
- 2 potassium ions (\( 2 \text{K}^+ \))
- 1 tetraiodomercurate ion (\( \text{HgI}_4^{2-} \))

Thus, the total number of ions produced from one unit of \( \text{K}_2\text{HgI}_4 \) is:
\[ n = 2 + 1 = 3 \]

### Step 4: Calculate the Degree of Dissociation
In this case, we assume that the degree of dissociation (\( \alpha \)) is 100%, which means \( \alpha = 1 \).

### Step 5: Use the van't Hoff Factor Formula
The van't Hoff factor (\( i \)) can be calculated using the formula:
\[ i = 1 + \alpha(n - 1) \]
Substituting the values we have:
- \( \alpha = 1 \)
- \( n = 3 \)

So,
\[ i = 1 + 1(3 - 1) \]
\[ i = 1 + 1 \times 2 \]
\[ i = 1 + 2 = 3 \]

### Conclusion
The van't Hoff factor of the resulting solution is:
\[ \boxed{3} \]