However, EVERYTHING we have just said only holds for certain circumstances; solutions for which these statements are true are referred to as ideal solutions.

For a solution to be an ideal solution DHsoln must equal 0. What this means is that the intermolecluar forces that hold solvent A to other solvent A molecules, must be similar to the forces that hold solvent B to other solvent B molecules.

A solution of hexane (CH3CH2CH2CH2CH2CH3) and heptane (CH3CH2CH2CH2CH2CH2CH3) would not deviate from Raoult's law.

Because IMF's for hexane are LDF's, and IMF's for heptane are LDF's too. So, when the solution forms LDF's hold hexane to heptane.

DHsoln = DHhex-hept - DHhept-hept - DHhex-hex = 0

I this solution 1 mol of LDF's from hept are broken, 1 mol LDF's from hexane are broken, and 2 mols of LDF's are formed between hexan and heptane. The solution has simply traded LDF's for LDF's so there is no energy gain or loss.

A solution of n-butanol (CH3CH2CH2CH2OH) and hexane (CH3CH2CH2CH2CH2CH3) would not follow Raoult's law.

By dissolving butanol in hexane the H-bonds between the butanol have been disrupted, so more butanol will be able to evaporate than would be expected based solely on concentration calculations.

DHsoln = DHbutanol-hexane - DHbutanol-butanol - DHhex-hex > 0

Strong butanol-butanol H-bonds must be overcome and replaced with LDF's. So, energy must be put in.

Compare with a butanol-pentanol solution. H-bonding stays the same so partial pressure determined solely be the concentration of butanol. Vapor presure goes down to the expected (Raoult's Law) pressure. The pressure goes down in the butanol pentanol solution more than vapor pressure went down in butanol-hexane solution.

A solution of H2O and acetone (CH3C(O)CH3) would not follow Raoult's law.

By dissolving acetone in water H-bonds between the water and the acetone are created. Because the acetone is attracted to the water not as much acetone would evaporate as would be expected based solely on concentration calculations.

Compare with a methylethylketone-acetone solution. IMF's stay the same, dipole-dipole interactions, so the vapor presure of acetone is determined solely by the concentration of acetone. The vapor pressure of acetone goes down less than the vapor pressure went down in the acetone water mixture.

These observations can be summed up by observing the DHsoln for the different solutions.

hexane-heptane&emdash;the amount of energy required to overcome the london dispersion forces between hexane molecules and between the heptane molecules is about the same as the amount of energy released when the hexane molecules are allowed to interact with the heptane molecules. DHsoln is approximately 0. IMF forces same at beginning and end. Traded London for London.

butanol-hexane&emdash;to make this solution H-bonds must be broken to separate the butanol molecules, and London dispersion forces need to be overcome to separate the hexane molecules. Since butanol cannot H-bond with hexane the amount of energy required to break appart the molecules is larger than the amount of energy released when the molecules, butanol and hexane, interact. DHsoln > 0. IMF stronger at begining than at end. Traded H-bond and London for London

water-acetone&emdash;To make this solution H-bonds must be broken, and dipole-dipole interactions must be overcome. When the solution forms the water molecules undergo H-bonding with the acetone; so, the energy required to break appart water and acetone is less than the energy released when the acetone and water interact. So DHsoln < 0. IMF weaker at begining than at end. Traded H-bond and dipole-dipole for H-bond.

So, if DHsoln = 0 Raoult's law holds

DHsoln < 0 negative deviation; ie, lower vapor pressure than predicted

DHsoln > 0 positive deviation; ie, higher vapor pressure than predicted