# Should the volume of a solute be factored into the volume of a solvent when preparing a substance of a certain molarity?

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?

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.