# Raoult's Law confusion

I have this confusion regarding the statement of Raoult's Law. It states that

The partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Does mole fraction in the mixture mean the combined liquid-vapour mixture or only the liquid which is left after evaporation on attaining equilibrium?

I had this problem in my textbook, one mole of two volatile substances each is mixed. Pure pressures of both of them are given. Should I just use mole fraction as one-half? Aren't some of the moles gonna go into vapour phase?

• For the liquid-vapour mixture to be at equilibrium, the concentrations in the vapor phase and the liquid phase would have to be stable.
– MaxW
Aug 18 '20 at 8:43
• Let me put it this way: does the vapour pressure of a mixture depend on how much amount (preferably in moles) of volatile components you put in a closed vessel, or on how much amount of that, which you put, is left in the liquid phase after equilibrium vapour pressure is attained.
– Rew
Aug 18 '20 at 9:28
• You must have heard about the mass conservation law. Gas cannot contain more than you have put there. Aug 18 '20 at 9:43

It should be like:

The saturated partial pressure of each component of an ideal mixture of liquids is equal to the saturated vapour pressure of the pure component multiplied by its mole fraction in the liquid mixture.

These data are independent on the actual composition of the gaseous phase, that may, or may not be in the equilibrium with the liquid.

OTOH, composition of the liquid phase does depend on the amount and composition of the original liquid mixture put into a container, as evaporation changes liquid composition in favour of less volatile component.

If the pure liquid $$\mathrm{A}$$ has saturated vapour presure $$p_\mathrm{0,A}$$ and the pure liquid $$\mathrm{B}$$ has saturated vapour presure $$p_\mathrm{0,B}$$ and if the $$\mathrm{A}$$ in the liquid has molar fraction $$x_\mathrm{A,l}$$, then the gaseous molar fraction of $$\mathrm{A}$$ ( in the saturated gaseous phase, with assumption of ideal behaviour given by the Raoult law ) is:

$$x_\mathrm{A,g} = \frac{x_\mathrm{A,l} \cdot p_\mathrm{0,A}}{ x_\mathrm{A,l} \cdot p_\mathrm{0,A} + (1-x_\mathrm{A,l}) \cdot p_\mathrm{0,B}}$$

For calculation for particular volumes of the container and the initial mixture of volumes, you would need more data, like these volumes and densities, and assumptions how to treat gases, like. e.g. as the ideal gas, but it goes behind the Raoult law. In all the complexity, it is solving a set of equations for set of variables, like molar amounts in both phases and phase volumes.

If the fraction of components in the gaseous volume is negligible, e.g. if we consider the total vapour pressure in a full or almost full container, we can consider total mixture composition is the same as liquid composition.

• So, how would you answer the question I mentioned I have in my textbook?
– Rew
Aug 18 '20 at 9:30
• I intuitively thought that the mole fraction in liquid will be favoured for the component which is less volatile. Thus, the one-half mole fraction would be changed. But the solution uses the mole fraction of the amount initially put in, which confused me in the first place. This is the very reason I asked the question. Could you tell me if I'm correct in my intuition? Could you summarize it all in one sentence? (Solution = textbook answer, no pun intended)
– Rew
Aug 18 '20 at 9:36
• It was Benzene and Toulene. How am I supposed to know if the gas phase volume is comparable to the liquid or not?
– Rew
Aug 18 '20 at 9:43
• As a complete side note, if they volumes were comparable, how would I calculate the amount that evaporated?
– Rew
Aug 18 '20 at 9:44
• The saturated partial pressure of each component of an ideal mixture of liquids is equal to the saturated vapour pressure of the pure component multiplied by its mole fraction in the liquid mixture. What is this mole fraction? The initial one or when liquid-vapour equilibrium is achieved
– Lllt
Jul 18 at 14:06