Lime reacts exothermally with water to give 'A' which has low solubility in water. Aqueous solution of 'A' is often used for the test of CO$$_2$$, a test in which insoluble B is formed. If B is further reacted with CO$$_2$$ then soluble compound is formed. 'A' is

Lime (CaO, quicklime) reacts exothermally with water:

$$CaO + H_2O \to Ca(OH)_2$$ (slaked lime)

$$Ca(OH)_2$$ has low solubility in water. Its aqueous solution is lime water.

Test for CO₂: $$Ca(OH)_2 + CO_2 \to CaCO_3 \downarrow + H_2O$$

CaCO₃ (B) is the white insoluble precipitate (milky appearance).

Further reaction with CO₂: $$CaCO_3 + CO_2 + H_2O \to Ca(HCO_3)_2$$ (soluble)

Therefore, 'A' is slaked lime [Ca(OH)₂].

The correct answer is Option 2.