FWIW - in the chemmy lab we usually talk in terms of 'molar' solutions - a 'mole' being the mass (or weight) of a chemical equal in grams to it's molecular weight. So - if we take a chemical with a molecular weight of (say) 100, then 200 grams of that chemical in 1 litre of water would then produce a 2.0 molar solution.

But - you don't just add 200 grams of it to 1 litre of water because chemicals continue to occupy volume even when dissolved in water, and so more than 1 litre would be produced as a result (at a slightly lower concentration than that desired).

So - the drill is to add 200 grams of the chemical to (say) 750 mls or thereabouts of water, then - when fully dissolved - top up the volume to exactly 1 litre, which then produces the desired 2.0 molar solution.

There are some examples of how to achieve fairly complex dilutions from 'stock' molar solutions at:

http://dilutions.quansysbio.com/dilutions-explanations-and-examples/ from where I've pinched the following:

Using C1V1 = C2V2

To make a fixed amount of a dilute solution from a stock solution, you can use the formula: C1V1 = C2V2 where:

V1 = Volume of stock solution needed to make the new solution

C1 = Concentration of stock solution

V2 = Final volume of new solution

C2 = Final concentration of new solution

So - to turn our strength_2 solution into a strength_1 solution, we simply double the amount of solvent (water), thus:

2M x 1 litre = 1M x 2 litres, i.e. to give double the volume at half the concentration (strength)

Hope this helps (sorry about the chemmy lecture - chemistry has kinda conditioned how I think about things ...)

'best,

LJ