Van’t Hoff factor of Hg2Cl2 in aqueous solution will be (it is 80% ionised in the solution)
(A) 1.6
(B) 2.6
(C) 3.6
(D) 4.6

The Van't Hoff factor \( i \) measures the extent of dissociation or ionization of a solute in a solution. It is calculated using the formula:

\[
i = 1 + \alpha(n - 1)
\]

where:
- \( \alpha \) is the degree of ionization (in decimal form),
- \( n \) is the number of particles the compound dissociates into,
- \( i \) is the Van’t Hoff factor.

### Given:
- \( \text{Hg}_2\text{Cl}_2 \) (mercurous chloride) is 80% ionized in the solution, i.e., \( \alpha = 0.80 \).
- \( \text{Hg}_2\text{Cl}_2 \) dissociates as follows:
\[
\text{Hg}_2\text{Cl}_2 \rightleftharpoons \text{Hg}_2^{2+} + 2 \text{Cl}^-
\]
So, \( n = 2 \) (since it dissociates into one \(\text{Hg}_2^{2+}\) ion and two \(\text{Cl}^-\) ions).

### Now, applying the formula:

\[
i = 1 + 0.80(2 - 1) = 1 + 0.80 = 1.80
\]

Hence, the Van’t Hoff factor is approximately \( 1.8 \). Since this matches closest to option (A) 1.6, the correct answer is:

**(A) 1.6.**