My teacher said that benzene and hexane do not form an ideal solution (with reference to Raoult's law) but did not explain any reason for it.

My approach: Cyclic hydrocarbons (benzene here) have higher van der waal's forces of attraction than straight chain hydrocarbons (hexane here) because of shorter intermolecular distances and greater surface area. So when hexane goes in between benzene, there will be weaker intermolecular bonds, causing positive deviation from the Raoult's law.

My doubts:

  1. Are the benzene-hexane interactions stronger than hexane-hexane interactions even though they are weaker than the benzene-benzene interactions? If so, is there still a net deviation from the Raoult's law?

  2. Do the same results apply to other solutions of cyclic and straight chain hydrocarbons?

  • $\begingroup$ Better question could be, if deviation from ideal solution is negligible. Strength of vdW bonds is hexane-hexane < hexane-benzene < benzene-benzene, unless there is strong deviation from the Raoult law. $\endgroup$
    – Poutnik
    Nov 5, 2021 at 7:30
  • $\begingroup$ Also, under My Approach, better jumping directly to Aromatics instead of Cyclics, as it's aromaticity here the more important feature of one component. Aromatics in solid states have quite a peculiar way of stacking (not as a "pile" BTW) and I guess some kind of residual interaction of the same nature can be present liquids, too. Also look at ncbi.nlm.nih.gov/pmc/articles/PMC7099588 about benzene interactions. $\endgroup$
    – Alchimista
    Nov 5, 2021 at 8:32
  • $\begingroup$ @Alchimista The source you mentioned says benzene-cyclohexane bonds are stronger. So that looks like a case of negative derivation. Am I right? But the question in my exam just said "hexane", so I think they were talking about the straight chain hexane only. I'm not sure how I could correlate the case of benzene-cyclohexane to this (I'm a class 12 student) $\endgroup$
    – Balu
    Nov 5, 2021 at 15:03
  • 1
    $\begingroup$ No, it does not point to saying that. Saying C>A and C<B does not mean C=AVG(A,B) $\endgroup$
    – Poutnik
    Nov 5, 2021 at 15:24
  • 1
    $\begingroup$ @Balu yes hexane is n-hexane. My point was to consider benzene for what it is, not a cyclic hydrocarbon only as it is in your question. Its aromaticity also dictates its intermolecular interactions, ergo the link. For the cyclohexane mentioned therein, it has a BP almost coincident to that of benzene but it does not mean that not effecti takes place by mixing with benzene. You can look for the data to get a clear picture. $\endgroup$
    – Alchimista
    Nov 6, 2021 at 7:31

1 Answer 1


A picture is worth a kiloword (1000 words). Then you can write 2 kilowords explaining the picture. But data are paramount, so here is a picture (Ref 1) :

enter image description here

According to the picture, which has three complete lines and one partial line, the straight line going from 0,0 to 1,1 designates the molar composition of the liquid (the independent variable). Then the two curved lines above it are the connected dots of measurements of the vapor-phase composition at approximately 1 atmosphere pressure and the boiling point of the mixture (which is varying, but called constant because it is uniformly the boiling point). The data were obtained by boiling the mixtures and taking vapor samples in an equilibrium still (Ref 2).

All the colored dots lie above the straight line, indicating that the vapor is enriched in hexane, or the liquid is impoverished in hexane, so hexane-benzene attraction not only does not keep the hexane in the liquid, but forcing hexane into benzene seems to be somewhat unattractive.

The partial line (a few red dots between 0.9, 0.9 and 1,1, at ~5 atmospheres pressure is about 60ºC hotter than at 1 atm, and does not add anything more to the conclusion. It was shown that enrichment of the vapor (in hexane) ceased at about 3.5 mole % benzene (Ref 2). (Note than on the graph, 0,0 corresponds to all benzene, which has the higher boiling point (80.1º vs 69º for n-hexane)). It had been thought that hexane and benzene formed a constant-boiling mixture at ~2.5% benzene in hexane, but this was disproven. It's just very difficult to get all the benzene out of n-hexane.

In a way, this suggests that there is an interaction between benzene and n-hexane, but only at very low concentrations of benzene, perhaps not well described by Van der Waals interactions. Maybe a little bit of benzene causes the n-hexane molecules to rearrange and use up their packing space more effectively.

Ref 1: http://www.ddbst.com/en/EED/VLE/Images/VLE%20Benzene;Hexane_002.png

Ref 2. https://pubs.acs.org/doi/abs/10.1021/ie50283a006


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