Introduction:
When acetaldehyde (CH3CHO) is treated with a catalytic amount of H2SO4 (sulfuric acid), a stable association product X is formed. We are asked to determine the number of lone pair of electrons present in XI, the association product.

Explanation:
To understand the formation of the association product X, we need to examine the reaction between acetaldehyde and H2SO4.

1. Reaction:
The reaction between acetaldehyde and H2SO4 can be represented as follows:

CH3CHO + H2SO4 → XI

2. Formation of XI:
The catalytic amount of H2SO4 acts as a proton donor, and acetaldehyde acts as a proton acceptor. The acidic nature of H2SO4 allows it to donate a proton (H+) to the carbonyl group of acetaldehyde. This results in the formation of a new compound, XI.

3. Structure of XI:
In XI, the proton from H2SO4 is attached to the oxygen atom of the carbonyl group. The oxygen atom also retains its lone pair of electrons. Additionally, the carbon atom of the carbonyl group forms a double bond with the oxygen atom from another acetaldehyde molecule. This forms a stable association product with a cyclic structure.

4. Number of Lone Pairs:
To determine the number of lone pairs of electrons in XI, we need to analyze its structure. In XI, there are two oxygen atoms - one with a proton attached and the other involved in the double bond. Each oxygen atom has 2 lone pairs of electrons. Therefore, the total number of lone pairs in XI is 2 (from the oxygen with the proton) + 2 (from the oxygen involved in the double bond) = 4 + 2 = 6.

Conclusion:
When acetaldehyde is treated with a catalytic amount of H2SO4, a stable association product XI is formed. XI has a total of 6 lone pairs of electrons, which are distributed between the two oxygen atoms present in the structure.