The Van't Hoff factor of `Hg_(2)Cl_(2)` in its aqueous solution will be (`Hg_(2)Cl_(2)` is `80%` ionized in the solution)
a.`1.6` , b.`2.6` ,c.`3.6` ,d.`4.6`

Explanation

To find the Van't Hoff factor (i) for `Hg_(2)Cl_(2)` in an aqueous solution where it is 80% ionized, we can follow these steps:

### Step 1: Understand the dissociation of `Hg_(2)Cl_(2)`
`Hg_(2)Cl_(2)` dissociates in solution as follows:
\[ Hg_2Cl_2 \rightarrow Hg_2^{2+} + 2Cl^{-} \]
This means that one formula unit of `Hg_(2)Cl_(2)` produces 3 ions in total (1 `Hg_2^{2+}` ion and 2 `Cl^{-}` ions).
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