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As I understand it, the boiling point elevation in a particular solvent (in this case water) is a function of a particular constant of that solvent, and the molality of a dissolved substance. In turn, the maximum molality is a function of the solubility of the compound, as well as the van 't' Hoff factor. Thus it would follow that the highest possible elevation of water's boiling point would be the compound with the greatest ratio of solubility to van 't' Hoff factor. So the question remains: which compound is that?

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  • $\begingroup$ There are melts of salts with some % of water (eg nitrates used for blueing of steel) with a boiling point about 250 degrees. But all Your reasoning is nonsense, because boiling point elevation equations are limited to DILUTE systems. $\endgroup$ – Georg May 26 '13 at 10:47
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The "particular constant" $k$ you refer to in the question isn't an arbitrary constant. This "law" is usually written as

$\Delta T = km = \frac{M_1 RT_b^2}{1000 \Delta H_1} m$

where k is the "constant you refer to; $M_1$ is the molecular weight of the solvent, $R$ is the gas constant, $T_b$ is the boiling point of the pure solvent, the constant 1000 makes the formula applicable only to water, and $\Delta H_1$ is the heat of vaporization of the solvent, here assumed to be water. $m$ is the molality of the solute.

This equation is derived in courses in physical chemistry and is itself a simplification to allow slide rule simplification (this is true for a lot of formulas in chemistry).

So what you want to do to make $\Delta T$ as large as possible is to find a solvent with a large molecular weight and a high boiling point and a low heat of vaporization. The constant 1000, which is the number of grams of solvent in a liter of solvent will also have to be adjusted. And of course $m$ can be played with.

I don't think anyone has ever worked out what works best for a large $\Delta T$ since boiling point elevation, once used for molecular weight approximation, isn't used for that purpose any more -- and thermometers are a lot more sensitive and accurate than they used to be.

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  • $\begingroup$ If the 1000 is just the density of the solvent, then why not write it in that way to make the equation more generally applicable? Or are other assumptions leading to the equation also water specific? $\endgroup$ – Michiel May 22 '13 at 5:51

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