# Expressing solubility in binary solvents with non-additive volumes

I have experimentally obtained data that shows the molarity of a saturated solution in mixed solvents, with different proportions of solvents.

The issue here is that the two solvents (water/ethanol) have non-additive volumes, and expressing the solubility as $\mathrm{mol}$ $\mathrm{L^{-1}}$ thus seems a bit odd, especially when comparing solubilities between different solvent ratios. Would mole fraction be a better alternative? If so, why? And is it possible to convert from the "old" amount-per-volume answer to the new, better one?

Also, I include a graphical representation of my results, with solvent composition on the x-axis and solubility on the y-axis. Would it be appropriate to use mole fraction (or whatever other unit is better suited) on both axes: the mole fraction of solvents on the x-axis and that of the solute in the mixed solvent on the y-axis?

• Not sure where you're trying to go with this. Using molar concentrations works for some calculations and mole fraction works best for others. Neither works well for all calculations. – MaxW Feb 18 '17 at 16:18
• @MaxW Could you give some examples of situations where each of the units would work best for expressing solubility in binary solvents? – Marcel Feb 18 '17 at 16:19
• I'll start the list for you. For mixing sulfuric acid with water molarity works best. For density of a binary mixtures of two miscible liquids mole fraction works best // Sorry, I'm confused. I am not driving this question you are. What are you trying to prove? What don't you understand? – MaxW Feb 18 '17 at 17:12