We learnt to apply Raoult's law for the ideal solutions. When the questions are asked, they tell that the intermolecular forces of toluene and benzene are equal. But in the same question, they provide two vapor (saturated) pressures for toluene and benzene.

Question: If the intermolecular forces of toluene and benzene are equal, how do they have different vapor pressure?


3 Answers 3


Most likely your "they" are exam-paper setters in some public exams. The fact is that the moment you deal with a different molecule, the intermolecular interactions cannot be the same. Benzene and toluene are different molecules just like two different persons on Earth, they have different molecular weights and not surprisingly different molecular interactions. Benzene boils at 81 °C and toluene boils at 110 °C. In general, a bigger/heavier organic molecule has a lower vapor pressure and a higher boiling point. Intermolecular forces is an umbrella term, so there is no single number, hence this can never be equal for two different molecules.

In the comments you asked about molar mass. In general heavier molecules have a higher boiling point but one cannot make a prediction. Compare benzene with a molar mass of 78 g/mol, and another metal such as platinum with a molar mass of 195 g/mol. The former boils below 100 °C and platinum boils at several thousand degrees. Molar mass alone cannot let you predict a number related to melting or boiling. Many software do predict the boiling and melting point of compounds, more often that is quite wrong.

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    $\begingroup$ thank you so much for your answer. I have one more question related to this and hope you can help me. How does the molar weight affects the boiling point. I guess that it increases the london forces. Am I correct? $\endgroup$ Dec 2, 2022 at 5:32
  • $\begingroup$ The benzene/platinum example doesn't seem very helpful. I would expect the boiling point of a "lighter" liquid to be less than the boiling point of the "heavier" liquid. With only two data points, it's not which of the two is surprising: are you saying benzene should have a higher boiling point than it does or that platinum should have a lower boiling point, given the two molar masses? (If you wanted to use lithium, an element with a molar mass of just ~7 g/mol, as the liquid with a high boiling point, that might make more sense for the point you are trying to make.) $\endgroup$
    – chepner
    Dec 2, 2022 at 20:01
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    $\begingroup$ @KavinduChamodhNethsara Yes, the mass itself does not affect the vapor pressure directly. A larger mass typically correlates with larger molecules that have stronger interactions per molecule. Krypton pairs have stronger London dispersion forces than neon pairs because Kr has more electrons. Octane pairs have stronger dispersion interactions than butane pairs, because they have twice as many carbon atoms. However, for other types of molecules, other interactions might matter. Larger polysaccharides or polypeptides can form more intermolecular hydrogen bonds than smaller ones. $\endgroup$ Dec 2, 2022 at 21:06
  • $\begingroup$ @chepner I think the point is that the masses are not dramatically different but the bp are. Then again the intermolecular interactions in the two cases are very different. In the OPs example of toluene and benzene, intermolecular interactions would be dominated in both cases by London dispersion. This is what confused the OP. $\endgroup$
    – Buck Thorn
    Dec 3, 2022 at 8:50

Perhaps, by saying benzene and toluene have "equal intermolecular forces", maybe they meant they both have London dispersion forces, NOT that they have the same degree of LDF. Then it would be up to you to explain why they have different vapor pressures by saying the toluene would have more sites for LDF than benzene.


Benzene and toluene do not have exactly the same intermolecular forces. Both have Van der Waals - London forces from temporary induced dipoles and toluene has also a slight dipole-dipole interaction from its very small dipole moment[toluene is small while benzene is vanishingly small effectively zero]. The difference in intermolecular attractions is not the force but the energy of attraction or the work required to separate the molecules. This energy is expressed in the heat of fusion and evaporation. If the forces are about equal, then the area in contact is determined by the molecular size and shape and these affect the energy of attraction. The differences in molecular sizes between toluene and benzene are apparent. Therefore very similar intermolecular forces result in different energies of attraction.

Finally, the boiling point or temperature of equilibrium depends on the heats and entropies of the transitions: at equilibrium delta H = T Delta S; T = Delta H/Delta S


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