What is Raoult's Law of vapor-pressure? Could someone please explain with diagrams?

1 Answer
Jul 30, 2018

Raoult's law just says that the vapor pressure P*A above a pure liquid will decrease to PA<P*A when solvent is added into it.

For ideal mixtures (no change in intermolecular forces after mixing), it is based on the mole fraction χA(l) of solvent in the solution phase:

PA=χA(l)P*A

where A is the solvent.

Since 0<χA(l)<1, it follows that the vapor pressure of the solvent must decrease. It starts out as PA=P*A, and then as χA(l) decreases, PA decreases.

![chemistryonline.guru)

The solute blocks the solvent from vaporizing, so it is harder to boil, and thus the vapor pressure is lower than desired; it is harder to reach the atmospheric pressure, so the boiling point is also higher.