In Raoults law for a solution of two volatile liquids do we measure the mole fraction of components at the time when we just add the two liquids and there is no vapour pressure or we measure their mole fractions at equilibrium condition that too in the liquid which is left after evaporation.

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    $\begingroup$ The latter.$\;$ $\endgroup$ – Ivan Neretin Feb 4 at 16:46
  • $\begingroup$ Im currently in class 12 and in my textbook in all questions they have taken the mole fraction of components when they are mixed initially when there is no vapour pressure. So is this an assumption? Maybe taking the mole fraction of the components at equilibrium condition in liquid state is out of the scope of my syllabus. $\endgroup$ – Devesh Feb 4 at 17:13
  • $\begingroup$ Well, if the air space above the liquid is not very large, the composition change will be negligible. That's the common assumption. $\endgroup$ – Ivan Neretin Feb 4 at 17:19
  • $\begingroup$ So basically if there are no assumptions involved then the mole fraction Xa or Xb of the components will be calculated in the liquid phase which is left in the vessel after evaporation and reaching equilibrium.And we will not consider the moles that are present in gaseous phase. Am i right $\endgroup$ – Devesh Feb 4 at 17:38
  • $\begingroup$ Basically yes, but that's pretty much never done. The assumption is good enough for everyone. $\endgroup$ – Ivan Neretin Feb 4 at 17:58

Total mole fractions of a system are known, constant and can be measured or calculated at/for any time ( no reaction is assumed ).

Mole fractions within a gaseous or liquid phase are measured or calculated in context of Raoult law for the state of equilibrium, unless they are defined as an initial state for a particular scenario.

  • $\begingroup$ Ok so in raoults law when we write partial pressure of any component as suppose Pa Xa then Pa represents the pure vapour pressure of the component and Xa represents the mole fraction of component in the liquid which is left after evaporation at equilibrium. It might sound like im asking the same thing again but i just need assurity because my textbook is taking the mole fraction of components at initial condition when we just mix them and there is no vapour pressure. So can you please tell what assumption are they using $\endgroup$ – Devesh Feb 5 at 7:38
  • $\begingroup$ Yes, it is like you say pa.xa is partial vapour p of a, which if pure has p.p. pa, , if xa is its molar fraction in liquid at equilibrium conditions. Said by other words, equilibrium saturated partial vapour pressure is proportional to molar fraction of volatile liquid in mixture. $\endgroup$ – Poutnik Feb 5 at 8:51
  • $\begingroup$ Alright, thank you for helping me. $\endgroup$ – Devesh Feb 5 at 9:09

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