Knowing the moles of substance per mass of solution would be useful. Pour some solution and you'd know how much substance you have. Why is molality defined not this way, but as moles of substance per mass solute instead? It seems like this unit was engineered to have a value similar to molarity rather than to be maximally useful.

Why was molality defined this way? Is there an advantage I'm missing? This measurement seems easy to prepare but more cumbersome to use, and I can't see why that would be an advantage.

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    $\begingroup$ "Why is molality defined […] as moles of substance per solute instead?" — no, molality is the amount of solute per mass of the solvent. And if I understand the question correctly, your proposition to use the amount of solution would make preparation of such solution unnecessarily complex. Besides, what you describe as "moles of substance per amount of solution" is already known as mole fraction. $\endgroup$
    – andselisk
    Dec 14, 2019 at 8:33
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    $\begingroup$ @andselisk I edited to clarify—I meant mass in every case of unspecified unit. Mole fraction is not per kg but per moles of solution. $\endgroup$
    – piojo
    Dec 14, 2019 at 8:42
  • $\begingroup$ To clarify, you seem to be asking about usefulness similar to that of molarity where the solvent is generally considered an inert carrier that is only used for ease of measuring a dose. If molality was defined per unit of solution it would basically be a temperature insensitive molarity that could also be used with solid solutions. Is this rewording accurate? $\endgroup$
    – Max Power
    Apr 18, 2020 at 16:21
  • $\begingroup$ Now I wonder if there is a mass based version of molarity, or if the need for that level of temperature compensation, or solid solutions, is rare enough that it is simply handled as a custom procedure/unit on an as needed bases. It is easy enough to say each Kg of solution contained 5 moles of solute. $\endgroup$
    – Max Power
    Apr 18, 2020 at 16:27
  • $\begingroup$ This all probably dates back before digital balances when volumes were quick and masses were cumbersome. $\endgroup$
    – Max Power
    Apr 18, 2020 at 16:29

2 Answers 2


Molality $b$ is defined its way because being proportional for the given solvent to the solute/solvent molecular ratio.

Molarity or molar/mass fractions or molar_amount/solution_mass are not proportional to it.

The boiling and melting point shifts are proportional via ebullioscopic/kryoscopic solvent constants to molality.

$$\Delta T_\mathrm{b}=K_\mathrm{e} \cdot b$$

$$\Delta T_\mathrm{m}=K_\mathrm{k} \cdot b$$

The number of solvent (s) molecules per 1 solute molecules

$$\frac{n_\mathrm{s}}{n}=\frac{m_\mathrm{s}}{M_\mathrm{s}\cdot n}=\frac 1{ M_\mathrm{s}\cdot b}$$

If we provide as the input the solute mass and solvent volume and density, the above molecule ratio is

$$\frac{n_\mathrm{s}}{n}= \frac{m_\mathrm{s}\cdot M}{M_\mathrm{s}\cdot V \cdot \rho}$$


The molality is defined by moles of solutes per mass of the solvent, There is a practical reason for this choice. This concentration unit is mainly used in cryoscopy. In cryoscopy, you start with a given mass of solvent, and you don't want to know about the volume. Then you dissolve a known amount of solute. The final volume does not matter. You just want to know at which temperature the solution begins to solidify or the solid begins to melt. This temperature allows you to determine the molar mass of the solute. And this is why cryoscopy is useful.

For example, adding n moles of a solute to 1 kg solvant makes the freezing point of water decrease by n · 1.86°C. So if you dissolve m grams of a solute in 1 kg water, and if you observe that the freezing point of water is -3.72°C, you know that there was 2 moles of your solute in the m grams used. This allows you to determine the molar mass of the solute. The total volume is of no interest in this determination.

The same measurements can be made in other solvents. In naphtalène, melting at 80°C, the freezing point constant is not $1.86$ , but $6.8$ K L $mol^{-1}$, if the unit molality is used.

  • $\begingroup$ The question as it is currently written is not asking about volume. $\endgroup$
    – Max Power
    Apr 18, 2020 at 16:17

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