1 molal aqueous solution of an electrolyte $$A_2B_3$$ is 60% ionised. The boiling point of the solution at 1 atm is ______ K. (Rounded-off to the nearest integer) [Given $$K_b$$ for $$H_2O = 0.52$$ K kg mol$$^{-1}$$]

We are given a 1 molal aqueous solution of an electrolyte $$A_2B_3$$ that is 60% ionised, and we need to find the boiling point at 1 atm.

The electrolyte $$A_2B_3$$ dissociates as: $$A_2B_3 \to 2A^{3+} + 3B^{2-}$$. This produces a total of 5 ions per formula unit.

The van't Hoff factor is given by: $$i = 1 + \alpha(n - 1)$$, where $$\alpha$$ is the degree of ionisation and $$n$$ is the number of ions produced. Substituting $$\alpha = 0.60$$ and $$n = 5$$: $$i = 1 + 0.60 \times (5 - 1) = 1 + 0.60 \times 4 = 1 + 2.4 = 3.4$$

The elevation in boiling point is: $$\Delta T_b = i \times K_b \times m = 3.4 \times 0.52 \times 1 = 1.768 \text{ K}$$

The normal boiling point of water is 373 K (at 1 atm). Therefore, the boiling point of the solution is: $$T_b = 373 + 1.768 = 374.768 \text{ K} \approx 375 \text{ K}$$

The boiling point of the solution is $$\mathbf{375}$$ K.