"Based on Figure 13.18, you might think that the reason volatile solvent molecules in a solution are less likely to escape to the gas phase, compared to the pure solvent, is that the solute molecules are physically blocking the solvent molecules from leaving at the surface. This is a common misconception. Design an experiment to test the hypothesis that solute blocking of solvent vaporization is not the reason that solutions have lower vapor pressures than pure solvents."

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The explanation (in the main chapter, not as an answer to this question) given in Chemistry: The Central Science (13th Edition) by Brown, LeMay, and Bursten (abbreviated as CTCS) is: "A solution consisting of a volatile liquid solvent and a nonvolatile solute forms spontaneously because of the increase in entropy that accompanies their mixing. In effect, the solvent molecules are stabilized in their liquid state by this process and thus have a lower tendency to escape into the vapor state. Therefore, when a nonvolatile solute is present, the vapor pressure of the solvent is lower than the vapor pressure of the pure solvent, as illustrated in ▲ Figure 13.18."

However, this is, I believe, the only textbook which gives this explanation for lowering in vapour pressure on adding a non-volatile solute. The other (reputed) textbooks I've read, like Zumdahl and NCERT Grade XII Chemistry, all give the explanation that CTCS calls a "common misconception". So, which explanation is correct? If possible, please give an experiment also.

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    $\begingroup$ Thanks for clearing my misconception!.. refer to the link: youtube.com/watch?v=CLZL4HqzMzA $\endgroup$ – Abdullah Jan 5 '15 at 6:23
  • $\begingroup$ Wow! Thanks for asking this question. I'm also following the Grade XII NCERT Chemistry textbook and I was also skeptical about the explanation given in that $\endgroup$ – Serotonin Jan 29 '18 at 5:41

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

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  • $\begingroup$ I edited the answer adding information from textbooks and a paper/article. $\endgroup$ – entropid Jan 5 '15 at 20:40
  • $\begingroup$ Thanks a lot for the excellent answer! It really cleared my concepts, especially the link you gave. $\endgroup$ – Leponzo Jan 6 '15 at 11:19
  • $\begingroup$ Its so sickening to see the dearth of correct explanation on the internet for this wonderful phenomena. But it is Very Nicely explained here. Thanks $\endgroup$ – Serotonin Jan 29 '18 at 8:07
  • $\begingroup$ very nicely explained! $\endgroup$ – Schwarz Kugelblitz Apr 7 at 16:00

This is similar to adsorption on a surface: van der Waals and other forces (e.g. hydrogen bond) bind the solute to the solvent, just as a gas may be bound to a solid. The lowest energy state depends on the strength of the solute-solvent bonds (adhesion) and the solute-solute bonds (cohesion) as opposed to random molecular motion causing evaporation. See http://en.wikipedia.org/wiki/Van_der_Waals_force and http://www.chem.purdue.edu/gchelp/solutions/eboil.html

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  • $\begingroup$ But if we consider an ideal solution (just assuming) then this logic cannot be applied. $\endgroup$ – jyoti proy Apr 30 '17 at 8:25

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