Volume of a solution in terms of molality, molar mass and density [duplicate]








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[Can you find the mass of solvent with mass of solute, volume of solution, and solution density?](/questions/24823/can-you-find-the-mass-of-solvent-with-mass-of-solute-volume-of-solution-and-so)

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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.




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.


 3

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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