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The molality $b$ of a solute dissolved in a solvent is given by:

$$b_\mathrm{solute} = \dfrac{n_\mathrm{solute}}{m_\mathrm{solvent}}$$

where $n_\mathrm{solute}$ is the amount of solute (e.g. in mol) and $m_\mathrm{solvent}$ is the mass of solvent (e.g. in kg).

Why is molality measured in terms of the mass of the solvent instead of the whole solution? Is there any specific advantage to taking the ratio this way?

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2 Answers 2

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We already have a unit to measure with respect to the full solution, molarity (because for a solution, the volume matters and atoms/molecules/ions per litre), as you may have seen in problems of ionic equilibrium. And we do know the density and other properties of the solution to convert the volume into mass of solution. But with respect to mass of solute, it is used in industrial areas, where volume is not used, because the per gram of solute intake matters for the body (food stuff) and makeup cream and other.

I don't know of any other significance for the two : molarity, and molality, units of concentration.

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  • $\begingroup$ The choice of molality in cryoscopie measurements is due to practical reasons. It is coming from the question : Which temperature do I obtain when adding $m$ grams (or $n$ moles) of a salt to $1$ kg ice ? The final volume of the obtained solution and its final mass are of no importance. The final temperature depends on the ratio $n/m$, i.e. on the molality of the future solution. $\endgroup$
    – Maurice
    Apr 4, 2023 at 19:16
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The choice of molality is due to practical reasons. It is coming from the question : Which temperature do I obtain when adding 𝑚 grams (or 𝑛 moles) of a substance to $m_o = 1$ kg ice at $0°C$, if the substance becomes a solute later on ? The final volume of the obtained solution and its final mass are of no importance to the scientist. The final temperature $T' < 0°C$ depends on the ratio $n/m_o$, i.e. on the molality of the future solution. The law of cryoscopy states that the temperature of fusion of ice + salt is $T'$ < $0°C$ and that this value does not depend on the nature of the dissolved substance : $$T' = - 1.86 n/m_o$$ This result is valid for all sorts of solutes, provided that they get dissolved in water without dissociation. If they get dissociated, the parameter $1.86$ must be multiplied by the number of ions produced by the dissociation. As a consequence, the fall of temperature $T'$ depends only on the molality $n/m_o$ of the substance to be added to the ice. Here is the origin of the use of the term "molality" in this sort of experiment.

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  • $\begingroup$ You have already answered this question over three years ago: chemistry.stackexchange.com/a/125105. $\endgroup$
    – andselisk
    Apr 4, 2023 at 20:05
  • $\begingroup$ You are right. But I was and still am unable to find the "old" report I wrote three years ago. $\endgroup$
    – Maurice
    Apr 5, 2023 at 9:57

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