Correction to molecular explanation
Dry air is composed of approx. 78% of diatomic nitrogen (N2 molecule) with a molecular weight of 14 (Atomic weight of N=7), and approx. 21% of diatomic oxygen (O2 molecule) with a molecular weight of 16 (Atomic weight of O=8).
If the air is humid, i.e., contains water vapor, the vapor molecules have a molecular weight of 10 (H2O molecular weight of 10 = 2 atom of Hydrogen, atomic weight 1, and one atom of oxygen, atomic weight 8).
From this you can see that a "particle" of water vapor is less dense than either a "particle" of nitrogen or oxygen.
OK. Let's see if we could come up with a "molecular weight" for dry air, i.e., treat the mixture that is air as if it were a compound. For simplicity, let's neglect the other gases that are in air, and also proportionately round up the percentages of N & O so that the total equals 100%:
Then: 78.75% N + 21.25% O = 100% Air.
(.7875 x 14) + (.2125 x 16) = 14.43. That is, if dry air were a chemical compound instead of a mixture, said compound would have a molecular weight of 14.43.
Therefore, neglecting specific measurement systems, we could say that a given parcel of air weighed 14.43 "units."
Now let's see what injecting some water vapor into the dry air would do.
John Dalton's law of partial pressures states that the Total Pressure of a gas mixture was the sum of the Partial Pressure of each gas. That is, for a given atmospheric pressure, that pressure is the sum total of the pressures of each gas in the mixture of gases that is air. In the case of humid air, one such gas in the mixture is water vapor.
OK, let's hypothesize that a parcel of humid air contains 3% water vapor.
(.97 x 14.43) + (.03 x 10) = 14.29. That is, a parcel of such air would weigh onl 14.29 units, a decrease of 0.14 units, or 1%.
Since the lift developed is proportional to the density of the air, this humid parcel of air would develop 1% less lift than dry air.