Upon being asked to prepare a solution of a given molarity from a solution with high concentration I was confused whether or not I should factor in the volume taken from the high concentration substance into the volume of the solvent. I think that an example could illustrate what I am asking more appropriately. Lets say I have to prepare a 5 M substance in 50 ml of water. The substance is in a container that is 10 M so I must calculate the number of moles and then the amount of solute to extract from the 10 M container. $$M = {m\over{L}} $$ $$5 M = {m\over{.05}} $$ $$moles = .25$$ given that the solution's concentration is 10 M $$10M={.25\over{L}}$$ $$L = .025$$ Now in order to achieve the 5 M in the 50 ml of water, do I pour 25 ml of water and then the 25 ml, which then adds to 50 ml? or do I pour the 50 ml of water and then the 25ml resulting in a total volume of 75 ml?


2 Answers 2


The only volume that is important in the denominator of your equation is the total volume of solution; neither the volume of solute or solvent are explicitly relevant.

Pragmatically, there is not likely to be much of a volume change when taking a concentrated solution and diluting it.

So with reference to your question, you should add enough solvent to get the proper final volume. Explicitly, add the 25 mL of stock solution and 'dilute to mark'-- that is add enough solvent so the total solution is 50 mL. The added solvent might not be exactly 25 mL to make this final volume.


You only add the $\ce{25 mL}$ water. After all you want to dilute to a certain molarity. You start from a 10 M solution and want to go to a 5M solution. Neglecting the volume of the dissolved compound you simply add another equivalent of solvent to dilute by a factor 2, i.e. $\ce{25 mL}$ water in this case.

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    $\begingroup$ This answer is explicitly untrue, but may be pragmatically true. The volume of solute may be negligible, but if it isn't this approach will lead to improper concentration. $\endgroup$
    – Lighthart
    Commented Jan 7, 2015 at 23:50
  • $\begingroup$ If you want the final concentration to be any more precise than "about 5M," this method is definitely incorrect. This is why you fill to volume using a volumetric flask (as @Lighthart wrote in their answer). The total volume is all that matters here, not the volume of solvent. Never assume the volume of solute is negligible. $\endgroup$
    – Kyle
    Commented Jan 11, 2015 at 16:19

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