Why is the boiling point of a solution higher than that of a pure solvent?
I think when a non-volatile solid solute is added to a solvent, some of the liquid solvent molecules are replaced by solute molecules at the surface. The lesser the number of solvent molecules at the surface that fewer molecules evaporate, and thus the vapor pressure of the solution will be lower than the pure solvent.

Is this correct?

  • $\begingroup$ Can a solute affect the entropy of the system? Google colligative properties. $\endgroup$
    – user5764
    Commented Jul 7, 2016 at 15:57
  • $\begingroup$ The change in vapour pressure with added solute is called Raoults law. Other colligative properties are elevation of boiling point, depression of freezing point and Osmotic pressure. These depend only on the number of solute molecules in a fixed volume of solvent, not their nature. $\endgroup$
    – porphyrin
    Commented Jul 7, 2016 at 19:17
  • $\begingroup$ I've given a rather comprehensive answer about this here. I hope it is helpful. $\endgroup$
    – Philipp
    Commented Jul 7, 2016 at 22:36
  • $\begingroup$ The rate at which molecules evaporate has nothing to do with where the thermodynamic equilibrium will be. $\endgroup$
    – Jon Custer
    Commented Jul 8, 2016 at 2:20

2 Answers 2


The change in vapour pressure with added solute is called Raoults law. The vapour pressure of a species i above a solution is proportional to mole fraction of this species (solute) $x_i$ and its pure vapour pressure $p_i^*$ or $p_i=x_i.p_i^*$.

As the solution now contains a fraction of a low vapour pressure solute then this component will not contribute much to the overall vapour pressure which will be less (in this case) than pure solvent.

It is not correct to say that solvent molecules are replaced at the surface as both solute and solvent are distributed uniformly in the solution. Thermal diffusion moves the solute and solvent around in a random manner. It is that there are now fewer solvent molecules as some have been replaced by low vapour pressure solute (in this example) and because they are of low volatility they need more energy to evaporate than the solvent does. As the temperature is constant, by the Boltzmann (energy) distribution fewer solute molecules populate higher energy levels compared to solvent and so have a lower vapour pressure.


YES the explanation is correct. In fact it is a qualitative property of solutions. Elevation of boiling point /lowering of vapor pressure/lowering of freezing point all depend on same reason.

  • 2
    $\begingroup$ No, that explanation is not thermodynamically correct. $\endgroup$
    – Jon Custer
    Commented Jul 8, 2016 at 2:21

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