# Why does ethanol and water mixture show positive deviation from Raoult's Law?

So, my teacher said, "For a solution to show positive deviation from Raoult's Law, it must have a compound which is lacking hydrogen bond and a compound which has hydrogen bond."

This is because the compound not having a hydrogen bond breaks the hydrogen bond of the other compound.

Now, why does ethanol and water mixture show positive deviation? Both of them have hydrogen bonds.

In more generality, Raoult's law is stated as $$P_i = x_iP_\text{vap}$$ and read "the partial vapor pressure of species $i$ is its (pure) vapor pressure multiplied by its mole fraction in the liquid phase." It assumes an ideal-gas gas phase and an ideal-solution liquid phase; an ideal solution is one in which all intermolecular attractions are of equal strength. More explicitly, the $a{-}b$ interactions are of equal strength to the $a{-}a$ and $b{-}b$ interactions, where $a$ and $b$ represent different chemical species in solution.

A positive deviation from Raoult's law occurs when the partial vapor pressure is greater than would be expected from Raoult's law. From a microscopic perspective, this implies that the $a{-}b$ interactions are weaker than the $a{-}a$ and $b{-}b$ interactions.

You should see now that your teacher's statement is a general rule-of-thumb following from the concepts outlined above, but does not necessarily hold true for all situations. Here, water forms hydrogen bonds of greater strength and magnitude than does methanol, and one expects that a water-methanol mixture will therefore possess weaker intermolecular interactions than those in a solution of pure water.

• To OP. The statement of the teacher can be made more general replacing hydrogen bonds with attractive interactions. S/he was referring to cases where the deviation is particularly strong, and thus thumbnail rules particularly useful. Plus one to this answer, too. Mar 5 '20 at 10:26