In a solution with two volatile components, the more volatile component will have higher vapor pressure. That makes sense since more volatile component will be present in a greater quantity in vapor. However, as per Raoult's law------ Pvapor = (mole fraction in solution) * Pvapor(pure)----- since the more volatile component has lesser mole fraction in solution, will not the Pvapor value decrease i.e., the more volatile component has lesser vapor pressure? I am a bit confused about this concept. Any clarifications??? For reference u can read page 84 (Raoult's law), TBR GChem book 2.
Raoults law and Volatility....
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Let's say you add 1 mole of 2 different liquids together. One of them is super volatile, the other isn't. The moment you add them together, you can use Raoult's law to determine the partial vapor pressures they will both contribute. Once they actually vaporize and are at equilibrium, then yes, your logic is correct and their mole fraction in solution will change. That is, the more volatile liquid will have a lower mole fraction and the less volatile liquid will have a higher mole fraction. This sort of segues into a discussion of distillation, but to clarify your confusion, you were attempting to apply Raoult's law after the solution had reached equilibrium.
