Chemistry - What causes the lowering of vapour pressure in volatile/nonvolatile solvent mixtures?

I think that the second explanation is the correct one. A simple experiment can prove it: if you place a porous cover over the pure solvent, its vapour pressure will not change. In alternative, you can place a few corks and observe the same result (no variation in the vapour pressure). In both cases surface sites are blocked, but the vapor pressure will obviously remain unaltered. Therefore the reason cannot be that the solute molecules are hindering the solvent molecule escape by occupying the solution surface.

The force that is driving the solvent molecules to the vapour phase is the difference between the entropy of the liquid and the vapour phases: when the molecule escape the liquid there is an entropy gain. Adding a non-volatile solute will increase the entropy of the liquid phase without having an effect on the entropy of the vapour phrase (given that the solute is non-volatile). This will lower the difference in entropy between the two phases, which is the driving force that make vaporisation happen, therefore there will be a lower number of solvent particles in the vapour phase and a consequent lower vapour pressure.

Textbooks like Chemistry: The Molecular Nature of Matter and Change (Silberberg, Amateis) or Physical Chemistry for the Biosciences (Chang) provides the same explanation. In addition, there is a short paper/article on this topic^[1] where the author highlights how only very few freshman chemistry texts give an adequate explanation of this phenomenon by making use of a entropy-based argument.

     _{¹ Vapor Pressure Lowering by Nonvolatile Solutes, G. D. Peckham, J. Chem. Educ., 1998, 75 (6), 787}