Raoult's Law states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

As we know that the liquid and vapour in a container are at dynamic equilibrium,

So the mole fraction which is mentioned in Raoult's Law is measured at equilibrium or initially?


1 Answer 1


Raoults Law says that, for an ideal liquid mixture, the partial pressure of each component in the gas phase is equal to the vapor pressure of the pure component (at the temperature of the system) times the mole fraction of that component in the liquid phase. If the system is at equilibrium, the mole fraction to use in the equation is the current mole fraction.

  • $\begingroup$ So you want to say that if we have to calculate partial pressure at any time then we must use mole fraction of the component in solution at that time in the equation? $\endgroup$
    – Lalit
    Jul 13, 2021 at 18:40
  • $\begingroup$ Yes, that is what I’m saying. $\endgroup$ Jul 13, 2021 at 18:55
  • $\begingroup$ But that creates a problem, Initially partial pressure is zero but the product vapor pressure as a pure component and mole fraction is not equal to zero, leading to a contradiction to your statement $\endgroup$
    – Lalit
    Jul 13, 2021 at 19:00
  • $\begingroup$ Why is the partial pressure zero initially? Do you now have vapor and liquid initially at equilibrium? $\endgroup$ Jul 13, 2021 at 21:08
  • $\begingroup$ Initially partial pressure will be zero as nothing will have been evaporated from solution $\endgroup$
    – Lalit
    Jul 14, 2021 at 2:49

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