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) of \( Hg_2Cl_2 \) in its aqueous solution, given that it is 80% ionized, we can follow these steps:

### Step 1: Understand the ionization
The compound \( Hg_2Cl_2 \) dissociates in water as follows:
\[ Hg_2Cl_2 \rightarrow Hg_2^{2+} + 2Cl^- \]
From this dissociation, we can see that 1 mole of \( Hg_2Cl_2 \) produces a total of 3 moles of ions (1 mole of \( Hg_2^{2+} \) and 2 moles of \( Cl^- \)).
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