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I found this equation in "Experiments in Physical Chemistry" by Garland, Nibler, Shoemaker. This is on page 174 (Experiment 9).

$$V=\frac{1000+mM_2}{d}$$

where $V$ is the volume of the solution, $m$ is the molality, $M_2$ is the molar mass of the solute and $d$ is the density of the solution.

The equation was in the middle of a derivation and is shown without any background information. I suppose it should be something pretty straightforward but I am not able to obtain this equation from the definition of molality, density and molar mass.

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Recall the definitions of molality and molarity, it is mol solute / kg of solvent and mol solute/ Vol of solution in (L). It is not kg of solution in molality. Instead of memorizing a plug and chug formula, would not it be far better to understand the conversion and do it yourself?

I quote an example from General Chemistry by Ebbing and Gammon (2016). Read the strategy:

Molality to Molarity: An aqueous solution is 0.273 m $\ce{KCl}$. What is the molar concentration of potassium chloride, $\ce{KCl}$ ? The density of the solution is 1.011 x 10$^3$ g/ L.

Problem Strategy You are asked to determine the molarity of the solution, which is the moles of solute $(\ce{KCl})$ per liter of solution (volume of $\ce{H2O}$ and $\ce{KCl}$. The molality tells you that the solution contains 0.273 mol $\ce{KCl}$ in $1 \mathrm{~kg}$ of water. If we determine the volume of solution that contains $0.273 \mathrm{~mol} \ce{KCl}$, we should be able to calculate the molarity. Because the solution is made up of only $\ce{KCl}$ and water, the solution mass is the mass of 0.273 mol $\ce {KCl}$ plus the mass of 1.000 kg of water. Knowing the total mass of solution, we then can use the solution density to convert from the mass of solution to the volume of solution. The molarity is obtained by dividing the moles of solute by the volume of solution in liters.

EDIT: Now that the OP posted the text, we can discuss the hints.

The equation $$V=\frac{1000+mM_2}{d}$$

is derived from mass balance principle. Keeping the above discussion in mind,

$$Total\hspace{0.15cm}mass\hspace{0.15cm}of\hspace{0.15cm}solution, M=mass _{solvent} + mass_{solute}$$

This is in grams. Now recall the mass volume relation from density. Here $V$ is the volume of solution not the solvent.

$$d=\frac{M}{V}$$ or $$M=dV$$

These hints should be enough to derive the desired equation.

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  • $\begingroup$ I don't want to memorize the formula. I want to derive it. It is more general that a simple example from Gen. Chem. The equation is just part of a longer derivation. $\endgroup$
    – ziulfer
    Jan 26 at 2:01
  • $\begingroup$ I edited my answer in case you want to see where the equation appears. $\endgroup$
    – ziulfer
    Jan 26 at 2:05
  • $\begingroup$ Thanks, I will have a look. Please add the full book reference. $\endgroup$
    – AChem
    Jan 26 at 2:06
  • $\begingroup$ "Experiments in Physical Chemistry" Garland, Nibler, Shoemaker. This is experimental 9, page 174 $\endgroup$
    – ziulfer
    Jan 26 at 2:10
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    $\begingroup$ Got it. For the mass of the solvent = 1000 g = 1Kg, the molality = mols. Then mols * Molar mass = mass(g). Thanks! $\endgroup$
    – ziulfer
    Jan 26 at 4:07
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The equation given is $$V = \frac{1000 + m \cdot M}{d} \pu{cm3}$$

It only works if you know to enter the quantities' numerical values after expressing them in the conventional units (g for mass, mol/kg for molality, g/mol for molar mass and e.g. g/mL for density). With these specifications, the expression gives the volume of a solution containing 1000 g of solvent.

Another way of writing this, using the modern abbreviation for molality, $b$, and mass, $m$:

$$V_\mathrm{solution} = \frac{m_\mathrm{solvent} + m_\mathrm{solvent} \cdot b_\mathrm{solute} \cdot M_\mathrm{solute}}{d_\mathrm{solution}}$$

The numerator is the mass of the solution as sum of the mass of solvent and solute. In the cited equation, the 1000 reflects that molarity has units of $\pu{mol kg-1}$ but the molar mass conventionally has units of $\pu{g mol-1}$.

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    $\begingroup$ Good grief! IUPAC has come up with a new symbol for molarity as ($b$)? It is a joke that IUPAC mentions under amount concentration "Also called ...in older literature molarity". Molarity is as current terminology as possible. It is not going anywhere for several decades. What is the need to come up with new terms when there is no problem with molarity and or its symbols? $\endgroup$
    – AChem
    Jan 26 at 3:36
  • $\begingroup$ $$m = \frac{n}{m}$$ is confusing as a definition. $\endgroup$
    – Karsten ♦
    Jan 26 at 3:52
  • $\begingroup$ As is $$M = 3 M$$ $\endgroup$
    – Karsten ♦
    Jan 26 at 3:53
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    $\begingroup$ Sorry typo, I fixed it. $\endgroup$
    – Karsten ♦
    Jan 26 at 9:21
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    $\begingroup$ @Achem Molarity is too short, cannot be confused with its vaque generic meaning (amount) in everyday life and perhaps IUPAC guys are paid by the length of the terms. :-P (deleting older or OT comments) $\endgroup$
    – Poutnik
    Jan 26 at 16:00

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