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Question:

Determine whether liquid hexane ($\ce{C6H14}$) or liquid methanol ($\ce{CH3OH}$) is the more appropriate solvent for the substances grease ($\ce{C20H42}$) and potassium iodide ($\ce{KI}$).

It seems logical that $\Delta H$ to dissociate $\ce{C6H14}$ would be small, since it's non-polar, that $\Delta H$ to dissociate $\ce{C20H42}$ would also be small, since it's non-polar, which would be conducive to a solution forming between those two.

The question that I'm stuck on is if the $\Delta H$ of formation of a solution between $\ce{K+}$, $\ce{I-}$, and the mildly polar $\ce{CH3OH}$ would be greater than the amount of energy it would take to dissociate potassium iodide. How can I estimate how much energy will be released by the attraction of the $\ce{I-}$ to the slightly positive hydrogen on the hydroxide group is greater than energy required to dissociate $\ce{KI}$?

Am I missing something?

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You don't need to do compare the said enthalpies. Maybe the process of dissolution of $\ce{KI}$ in $\ce{CH3OH}$ is energetically favorable, maybe it isn't (the latter seems more likely to me); that's irrelevant. Thanks to entropy, everything dissolves in everything, if only in minuscule amounts.

You just need to compare solubilities of $\ce{KI}$ in methanol and hexane, which basically boils down to the comparison of solvation enthalpy in methanol vs. that in hexane, and here the difference in their polarity provides a clear and obvious answer.

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