If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ______ molal. (Rounded-off to the nearest integer)
[$$K_b = 0.52$$ K kg mol$$^{-1}$$].

The compound AB dissociates in aqueous solution as: $$AB \rightarrow A^+ + B^-$$. Given that the degree of dissociation $$\alpha = 0.75$$ (75%), the van't Hoff factor is $$i = 1 + \alpha(n - 1)$$, where $$n = 2$$ (number of particles formed per formula unit). So $$i = 1 + 0.75(2 - 1) = 1 + 0.75 = 1.75$$.

The elevation in boiling point is given by $$\Delta T_b = i \times K_b \times m$$, where $$\Delta T_b = 2.5$$ K, $$K_b = 0.52$$ K kg mol$$^{-1}$$, and $$m$$ is the molality.

Substituting: $$2.5 = 1.75 \times 0.52 \times m = 0.91 \times m$$.

Solving for molality: $$m = \frac{2.5}{0.91} = 2.747$$ molal.

Rounding to the nearest integer, the molality is $$\boxed{3}$$ molal.