# Confusion in Raoults Law

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.

• The latter.$\;$ – Ivan Neretin Feb 4 at 16:46
• 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. – Devesh Feb 4 at 17:13
• Well, if the air space above the liquid is not very large, the composition change will be negligible. That's the common assumption. – Ivan Neretin Feb 4 at 17:19
• 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 – Devesh Feb 4 at 17:38
• Basically yes, but that's pretty much never done. The assumption is good enough for everyone. – Ivan Neretin Feb 4 at 17:58