`K_(2)HgI_(4)` is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is :

Options

A. 1.6
B. 1.8
C. 2
D. 2.2

Explanation

To find the van't Hoff factor (i) for \( K_2HgI_4 \) that is 40% ionized in an aqueous solution, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( K_2HgI_4 \) in water can be represented as follows:
\[ K_2HgI_4 \rightarrow 2K^+ + HgI_4^{2-} \]

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