This is the boiling-point elevation equation:

$$ \Delta\mathrm{T}_{b}=ib\mathrm{K}_{eb}. $$

If we do the limit of the above, for $b\rightarrow\infty$ we find that the boiling-point elevation is infinite. How this could be explained under a physical view?

$$ \lim_{b~\rightarrow~\infty}ib\mathrm{K}_{eb}=\infty,~\text{if}~i\wedge b > 0. $$

For example, if we use a solute molality that is so big that the solution would be something like the pure solute, the final boiling-point of the solution, in the case of water as solvent for example:

$$ \mathrm{T}_{b}=ib\mathrm{K}_{eb} + 100\mathrm{°C}. $$

wouldn't be like the boiling-point of the pure solute? I mean, something like this:

$$ \lim_{b~\rightarrow~\infty}\mathrm{T}_{solution}=\mathrm{T}_{solute}. $$

Now, please forget what I've said above about $b\rightarrow\infty$. And take a look at this:

$$ T(b) = 0.86b + 64.7. $$

enter image description here

As you can see is, the boiling-point of a solution of $\mathrm{CH_3OH}$ with $\mathrm{H_2O}$ as solute would be $322.7\mathrm{°C}$ for $b = 300$. This it's weird to me 'cause that temperature would be about three times higher then the boiling-point of the pure water!

So the following:

$$ \mathrm{T}_{b}=ib\mathrm{K}_{eb} + \mathrm{T_{solvent}}. $$

holds true if

$$ b = \frac{x~\mathrm{g}_{solute}}{1000~\mathrm{g}_{solvent}} \le 1 $$

right? If yes why (in the above case):

$$ 0.86\frac{1000~\mathrm{g}}{18.01528~\mathrm{g}/\mathrm{mol}}\frac{1~\mathrm{mol}^2}{1~\mathrm{kg}} + 64.7~\mathrm{°C} = 112.4~\mathrm{°C} $$


  • $\begingroup$ After your edit... To use that equation you're imposing the constraint that the solute won't boil. You're looking at the boiling point of the solution, under the assumption that the methanol is the only component that will boil. $\endgroup$
    – user1160
    Mar 2, 2014 at 3:07
  • $\begingroup$ Could you please add units to $b$ and your $T(b)$ equation? If $b = 300$ means something like $b=300 \, \text{mol}_{\ce{H2O}}/\text{kg}_{\ce{CH3OH}}$ then $300$ is not a realistic value as per the arguments I provided in my answer. If you considered realistic molalities ($< 50 \, \text{mol}_{\ce{H2O}}/\text{kg}_{\ce{CH3OH}}$) you would be well in the reasonable temperature regime below 100 °C in your plot. $\endgroup$
    – Philipp
    Mar 2, 2014 at 3:22
  • $\begingroup$ @Brian take a look at my updates $\endgroup$
    – Aurelius
    Mar 2, 2014 at 4:05
  • $\begingroup$ You have inserted the wrong values for the molality in your equation. The molality is $b = n(\text{solute})/m(\text{solvent})$. So, say, you want to know the boiling point of a methanol/water mixture where you have $1 \, \text{mol}$ of water in $1 \, \text{kg}$ of methanol. Then you'd get $T(b) = 0.86 \frac{\text{K} \, \text{kg}}{\text{mol}} \cdot \frac{1 \, \text{mol}}{1 \, \text{kg}} + 64.7~\mathrm{°C} = 65.56~\mathrm{°C}$. $\endgroup$
    – Philipp
    Mar 2, 2014 at 4:29
  • $\begingroup$ It's not so simple as saying it holds true if b <= 1. Rather it holds true if the solute is non-volatile and perfectly miscible. It will hold true for water with a soluble amount of NaCl for example. $\endgroup$
    – user1160
    Mar 2, 2014 at 7:14

2 Answers 2


In reality, when you're saying that the molality is infinite, you are indeed basically asking for solute with infinitesimal amounts of solvent, and it will be true that $T_\text{solution} \approx T_\text{solute}$.

This equation describes the boiling point of the solvent under the effects of the non-volatile solute. Ie. you're making the assumption that the solute does not enter the gas phase at the temperatures you're considering.

A substance boils when the chemical potential of the liquid phase equals the chemical potential of the gas phase: $\mu_\text{liquid} = \mu_\text{gas}$, or when the vapor pressure equals the surrounding pressure.

So to elevate the boiling point, we're making the claim that for a given temperature, the chemical potential of the liquid phase is lowered (it's made more stable), or equivalently, that the vapor pressure is lowered. The effect is entropy based.

When particles are interspersed between the solvent molecules, the solvent molecules become less ordered, and are thus at a state of higher entropy than the pure solvent. This translates into a lower chemical potential. The gas phase will have only the solvent molecules, because we made the assumption that the solute is non-volatile. Therefore, the gas-phase is unaffected by this effect, and the chemical potential of the gas phase remains the same.

It's easy to visualise from the vapour pressure perspective. Imagine the liquid-gas interface. At a given temperature, a certain portion of the liquid solvent molecules at the liquid-gas-interface will enter the gas phase. The gaseous solvent molecules at the liquid-gas-interface will likewise enter the liquid phase when they come in contact with it.

The amount of solvent molecules going from gas->liquid depends on the atmospheric pressure. The amount of solvent molecules going from liquid->gas depends on the kinetic energy of the molecules (the temperature). A molecule may have enough energy to break free of the solvent-solvent interactions, or it may not. The higher the temperature, the more likely.

Now imagine that we add solute molecules. There will be less liquid solvent-molecules present at the liquid-gas interface, and thus a lower rate of liquid -> gas (meaning a lower vapour pressure), simply because the probability of a liquid molecule hitting the phase interface is lower... So the temperature has to be higher to compensate, if we want to counter the unchanged rate solvent molecules going from gas -> liquid.

Why boiling point goes infinite: At the limit where molality goes towards infinity, there will be an infinitesimal amount of liquid solvent molecules at the phase interface... And if we want that infinitesimal amount of molecules to sustain a vapour pressure that equals the atmospheric pressure, we need a temperature that approaches infinity as well.

  • 1
    $\begingroup$ I've heard the explanation about solute molecules "pushing" the liquid molecules out of the liquid-gas phase boundary, but something doesn't feel quite right to me. Would this explanation still be valid if I added a surfactant? Even at low concentrations, surfactants quickly cover a large part of the liquid-gas phase boundary. Does that mean surfactant solutions have a higher effect of the boiling point of a liquid compared to a normal solute, on a per mole basis? $\endgroup$ Mar 2, 2014 at 2:18
  • $\begingroup$ That's an interesting thought. I don't think I can come up with anything right now. THe surfactant will form its own extremely thin liquid phase, that the liquid solvent will interact with. And I suppose it impedes both entry from liquid into the gaseous solvent phase but from the gas into the liquid phase. Does that make any sense? $\endgroup$
    – user1160
    Mar 2, 2014 at 3:24
  • $\begingroup$ I'm not sure. Honestly I feel like the whole idea of blocking the surface is sort of an over-explanation. The correct answer would simply be the change in entropy of the solution (as you mention), and the idea of covering the surface would be trying to go too far in order to provide further physical insight. I'm searching for some articles on water boiling point elevation by surfactants, but I haven't found anything enlightening yet. $\endgroup$ Mar 2, 2014 at 11:41

The flaw in your reasoning is that the molality $b = n(\text{solute})/m(\text{solvent})$ cannot be infinite even if you use a solute molality that is so big that the solution would be something like the pure solute. Take water as an example: How concentrated is water? Very concentrated you may say - but the concentration is limited. We know that one mole of pure water has a mass of 18 g and occupies 18 cm³. So, in one dm³, there are 1000/18 = 55.56 mol which gives you a water concentration of 55.56 mol/dm³. You cannot get more concentrated water than this (unless you did something drastic like taking it into a black hole!). For the molality it is quite similar: If you dissolve something in water the solute concentration cannot exceed the concentration of water, so $n(\text{solute})$ (and therefore $b$) has an upper bound.

  • 1
    $\begingroup$ What if I took 1 kg of my solvent and decided to add as many moles of solute as I like. Then I can make b as large as I like, since I would have as many moles of solute in my kg of solvent as I like. :-) $\endgroup$
    – user1160
    Mar 2, 2014 at 2:05
  • $\begingroup$ @Brian That is the nice part: At some point before reaching the 55.56 mol limit the solute would either stop dissolving and precipitate or there will be some kind of phase inverion such that water will then be dissolved in the solute. In that case the maximum concentration the pure solute could attain (analogous to the little calculation I did for water) would be the upper bound for $n(\text{solute})$. $\endgroup$
    – Philipp
    Mar 2, 2014 at 2:11
  • $\begingroup$ @Brian Of course, my answer is more some kind of rationale to show that the premise of an infinite molality is impossible. Your answer goes much more to the heart of the problem (+1 for that). I think it's nice that you took the time to write out that point of view. That way FormlessCloud gets one answer that really explains what boiling point elevation is about and one that shows him a quick way to rationalize why the limit $b \to \infty$ is not possible. $\endgroup$
    – Philipp
    Mar 2, 2014 at 2:20
  • $\begingroup$ We can readily agree that having a non-volatile, perfectly miscible solute can probably not be, in reality :-) and as you say that puts an upper bound on b $\endgroup$
    – user1160
    Mar 2, 2014 at 3:29

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