$$K_2HgI_4$$ is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:

We have the salt $$K_2HgI_4$$ dissolved in water. The question tells us that it is 40 % ionised, that is, the degree of ionisation is $$\alpha = 0.40$$.

First, we write the dissociation equation for one mole of the salt:

$$K_2HgI_4 \;\longrightarrow\; 2K^+ \;+\; HgI_4^{2-}$$

From this balanced equation we see that, if dissociation were 100 % (i.e., $$\alpha = 1$$), one mole of the salt would give a total of

$$v = 2 \;+\; 1 \;=\; 3$$

ions in solution (two potassium ions and one tetraiodomercurate ion).

Now, let us introduce the formula for the van’t Hoff factor $$i$$ when the electrolyte is only partially ionised. For an electrolyte that would produce $$v$$ ions on complete dissociation, the factor is given by

$$i \;=\; 1 \;+\; \alpha\,(v-1).$$

This expression comes from the definition $$i = \dfrac{\text{total moles of solute particles present after dissociation}}{\text{initial moles of the solute added}}$$. Starting with one mole of the salt, a fraction $$\alpha$$ dissociates while a fraction $$(1-\alpha)$$ remains undissociated. Hence the total number of moles of particles becomes

$$\bigl(1-\alpha\bigr)\times 1 \;+\; \alpha \times v \;=\; 1 \;-\; \alpha \;+\; \alpha v \;=\; 1 \;+\; \alpha(v-1),$$

and dividing by the single mole originally taken gives the same expression for $$i$$ stated above.

We now substitute $$\alpha = 0.40$$ and $$v = 3$$ into the formula:

$$\begin{aligned} i &= 1 + \alpha\,(v-1) \\ &= 1 + 0.40\,(3 - 1) \\ &= 1 + 0.40 \times 2 \\ &= 1 + 0.80 \\ &= 1.80. \end{aligned}$$

Thus the van’t Hoff factor for this 40 % ionised solution of $$K_2HgI_4$$ is $$i = 1.8$$.

Hence, the correct answer is Option B.