Raoult's Law for solid-liquid solution is stated as such in my books that it is applicable for all non-volatile solutes.

However different solutes must have different attractive or repulsive forces and in such a case, all of them will probably hamper vaporisation in different ways.

Then how come can we simply define the law as: $$P_\text{vapour}=P_\text{vapour}^0\chi_\text{solvent}$$ where $$\chi_\text{solute}+\chi_\text{solvent}=1$$ Here the equations do not provide any information about the solute.

Is the law valid for only a class of solutes?

  • $\begingroup$ When did this form Raoult's law become applicable for all solutes? AFAIR, it was only for non-volatile solutes $\endgroup$ Commented Sep 22, 2020 at 9:00

2 Answers 2


It is valid for both volatile and nonvolatile solutes, as it refers to the partial vapour pressure, not total vapour pressure.

Saying that, note that it applies only on mixtures with ideal behaviour, as there are many more or less significant positive and negative deviations from the law. These deviations relate to preference of intermolecular bonding to the same, or to the other molecules.

Typical examples are ethanol water mixture forming an azeotrop of the minimal boiling point, or hydrochloric acid forming an azeotrop with the maximal boiling point.

  • $\begingroup$ So what you are saying is that we neglect these deviations same way as we neglect air resistance in projectile motion. $\endgroup$
    – Tony Stark
    Commented Sep 22, 2020 at 14:47
  • $\begingroup$ @Tony Stark I say sometimes we do because we know we can afford it. sometimes we do not because we know we cannot afford it. E.g. mixing 1 L of ethanol + 1 L of water leads to heating up the mixture. When it cools down back, it has volume 1.9 L. We cannot neglect this like it does not happen. $\endgroup$
    – Poutnik
    Commented Sep 22, 2020 at 14:49

The law is for ideal solutions where the attraction b/w solute particles and attraction b/w solvent particles is same as attraction b/w solute and solvent particles.

Consider the case when the solute is non-volatile:

The solute and solvent particles of ideal solution can be treated as same. But the solute particles will reduce concentration of solvent particles in solution and on surface of solution too. This will cause fractional decrease in vaporization rate of solute and hence its vapor pressure( it is also total pressure as only sovent is contributing to pressure)

That's how you get $P_{solvent}= P°_{solvent}×x_{solvent}$ ( for non-volatile solutes)

Also here $P_{solvent} =P_{total}$

If the solute is also volatile, then it will affect the condensation rate of solvent particle in vapour form which should increase the partial pressure of solvent.


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