Van't Hoff factor for a dilute solution of a `K_(2) [HgI4]` is :

Options

A. 2
B. 1
C. 3
D. zero

Answer

To find the Van't Hoff factor (i) for the dilute solution of \( K_2[HgI_4] \), we need to analyze how the compound dissociates in solution.

### Step-by-Step Solution:

1. **Identify the Compound**: The compound given is \( K_2[HgI_4] \). This indicates that it contains potassium ions (K\(^+\)) and the complex ion \([HgI_4]^{2-}\).

2. **Dissociation of the Compound**: When \( K_2[HgI_4] \) dissolves in water, it dissociates into its constituent ions. The dissociation can be represented as follows:
\[ K_2[HgI_4] \rightarrow 2K^+ + [HgI_4]^{2-} \]

3. **Count the Number of Ions Produced**: From the dissociation equation:
- 2 moles of \( K^+ \) ions are produced.
- 1 mole of \([HgI_4]^{2-}\) ion is produced.

Therefore, the total number of ions produced when 1 mole of \( K_2[HgI_4] \) dissolves is:
\[ 2 \text{ (from K}^+\text{)} + 1 \text{ (from [HgI}_4]^{2-}\text{)} = 3 \text{ ions} \]

4. **Calculate the Van't Hoff Factor (i)**: The Van't Hoff factor (i) is defined as the total number of particles (ions) in solution after dissociation. Thus, for \( K_2[HgI_4] \):
\[ i = 3 \]

### Final Answer:
The Van't Hoff factor for a dilute solution of \( K_2[HgI_4] \) is **3**.

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