My textbook informs me that

Molarity is dependent on temperature, whereas molality is not.

I'm aware of the definitions of both molarity and molality, but consider this scenario:

A solution of $\ce{NaCl}$ in $\ce{H2O}$ is heated. Eventually, all the water will escape and I will be left with only $\ce{NaCl}$ crystals. At the end, there is no solvent to even measure concentration of what was once a solution. In this case, how does the argument "molality is independent of temperature" still hold true?

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    $\begingroup$ In your extreme example, obviously it doesn't. There's no solution to speak of, so you can't really define a concentration, either in terms of molarity or molality. Since, molarity is moles per volume of solution, it is sensitive to temperature (since, volume changes with temperature). No such problems with molality since it is defined as moles over mass of solvent (which doesn't change). $\endgroup$ – getafix Aug 28 '16 at 7:55
  • $\begingroup$ Formally, molality does not change if you take in account the vapor. Otherwise you are just cheating by taking away mass of solvent instead of just changing it's temperature. $\endgroup$ – Džuris Aug 28 '16 at 16:46
  • $\begingroup$ You cannot change the composition of a phase and expect that concentrations do not change. Also, applying heat to a solution does not necessarily mean that solvent will evaporate. It does not when the system is closed and consists of the solution phase only. $\endgroup$ – aventurin Aug 28 '16 at 17:23
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    $\begingroup$ Both are quantities of solutions. If you evaporate all the solvent, molarity and molality do not mean anything anymore. $\endgroup$ – Gimelist Sep 2 '16 at 11:43

Recall that molarity $c$ is


whereas for molality $b$ is


While in the extreme case you mentioned, both quantities are not defined because all the solvent is either boiled away (or for heat sensitive species, the solvent can even decompose under heating), molarity depends on volume, and the volume of the solution depends on temperature due to expansion/contraction.


Example: 5 moles of solute in 1 liter of solution.

The Molarity at the beginning is (5moles ÷ 1Liter) = 5 M.

When the solution is warmed up and expanded by 1% say, to 1.01L, the new Molarity = (5moles ÷ 1.01Liter) = 4.95 M

P.S. Don't over think this question! The amount of content should not change, all you're talking about is expansion of content when the temperature rises. All matter expands when temperature rises so if you're calculating the molarity of a solution, you may get different readings on a warm day compared to a freezing cold day!


Molality doesn't change with temperature...


  • $\begingroup$ You are not answering the OP's query about the temperature dependence of molarity vs. molality. $\endgroup$ – Todd Minehardt Jun 6 '17 at 20:15

These two words sound similar but they are not synonyms, even though both of them are used for representing solution concentration. By definition, molarity is the number of moles of solute dissolved per liter of solution. We use capital letter “M” to represent molarity and its formula is M= (# mol SOLUTE)/ (Liters of SOLUTION). Molality is then the number of moles of solute per kilogram of the SOLVENT, NOT solution! We use lower case letter “m” to represent molality and its formula can be represented as: m= (# mole SOLUTE) / (Kilograms of SOLVENT). Most of the time scientists use either molarity or molality to represent solution concentration, but MOLALITY is preferred when the temperature of the solution varies. That is because MOLALITY does not depend on temperature, (Neither number of moles of solute nor mass of solvent will be affected by changes of temperature.) while MOLARITY changes as temperature changes. (Volume of solution in the formula changes as temperature changes, and that is why.)

  • $\begingroup$ Of course, I know the definitions properly! My only doubt was in the fact that even though molality is supposed to be independent of temperature, if I increase the temperature of the solution by heating it, some(or all)of the solvent evaporates and not only does the volume of the solvent change, but also the mass of the same and hence, molality will change. What point am I completely missing? From the other answers, I gathered that this argument will not hold good for extreme cases like the one that I mentioned in my question and must not try to apply it to such situations. Am I correct? $\endgroup$ – user33789 Aug 29 '16 at 0:58

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