[shakuras]

Hi all!
I hope you can help me figure things out. My problem is solution making, for instance: How much sucrose do I need to prepare a 2.5M solution in 100 mL of water?
I go and find the molar mass of sucrose, which is 342.3 g/mol. If find the number of moles needed: n=2.5mol/L×0.1L = 0.25 mol. Then: Mass=0.25mol×342.3g/mol = 85.6 gr.
That is great and all, but then I go and prepare that, and the final volume of the actual solution is like 150 ml, due to sucrose addition. So I didn't prepare a 2.5M solution, but a 1.67 M solution, right? Is that correct?

If I assume ideal conditions, the molar volume of sucrose is V = 342.3 gr/mol/ 1.587g/mL (density of sucrose) = 215.7 ml/mol.
Given my 0.25 mol, I can say that the increase in volume is dV= 215.7 ml/mol x 0.25 mol = 54 ml.

How am I suppose to prepare a solution with a final concentration...I am confused.
Please help me figure it out.

Thanks,
Alex.

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[Babcock_Hall]

I did not notice any flaws in your thinking.  When I make up a solution, I weigh out the correct mass into a beaker, add a stir bar, and then add solvent to about 80% of the final desired volume.  Then I stir until the solid dissolves and transfer into a graduated cylinder, bring the level of solvent up to the appropriate mark, and then mix until the solution is homogeneous. 

[shakuras]

Thank you for replying! So my problem was pouring the entire amount at once? In this case if I need, by calculation, 85gr of solids to be dissolved in 100ml, I might run into an issue with solubility, if I first pour a small amount...
Or maybe in this case, I pour water until the overall volume is 100ml, regardless of if I actually used 100ml of water. That's tricky...
That's my confusion. The approach you suggest works when solid amounts are a lot smaller than the solvent, it would seem. I'm definitely not understanding something here...

[Borek]

So my problem was pouring the entire amount at once?

Yes.

In this case if I need, by calculation, 85gr of solids to be dissolved in 100ml

Beware: 100 mL of solution, not of the solvent. You already know that's the case, but being careful with wording never hurts

I might run into an issue with solubility, if I first pour a small amount...

Using higher volume of water won't help you get past the solubility limit either.

Or maybe in this case, I pour water until the overall volume is 100ml, regardless of if I actually used 100ml of water. That's tricky...

You attempt to dissolve in smaller amount of water, than you "fill up" to the required volume. Otherwise you always risk adding too much water. If there is a solid left you can always attempt to add some more water, but less than the final volume.

Actually sometimes after filling up and mixing you will find out you need to add even a bit more water, due to the solution contraction.

Some numbers: according to density tables 2.5 M solution of sucrose has a density of 1.31 g/mL. That means 100 mL of the solution weights 131 g. 85 g of that is sucrose, so you need 46 g of water (which is 46 mL). But: mixing according to density tables is not worth it, filling up to the known volume is much easier and much more accurate.

That's my confusion. The approach you suggest works when solid amounts are a lot smaller than the solvent, it would seem. I'm definitely not understanding something here...

Sucrose is highly soluble, saturated solution is close to 90%, or 3.8 M.

[shakuras]

Thank you very much! That was very helpful