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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?

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    $\begingroup$ 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. $\endgroup$
    – MaxW
    Commented Feb 18, 2017 at 16:18
  • $\begingroup$ @MaxW Could you give some examples of situations where each of the units would work best for expressing solubility in binary solvents? $\endgroup$
    – Marcel
    Commented Feb 18, 2017 at 16:19
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    $\begingroup$ 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? $\endgroup$
    – MaxW
    Commented Feb 18, 2017 at 17:12

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For solvent mixtures, especially in mixture or solubility design of experiments, I work in mass for the solvents. It's unaffected by temperature, more accurate (very advantageous if small values are being used) and should get around your problem. You can quote your initial results as %w/w

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