0
$\begingroup$

I'm trying to make stock solution of 3M sucrose in water. In 25ml, this is 25.672g sucrose (FW 342.3). However, that mass of sucrose, in the powder, occupies a pretty substantial volume, such that if I add 25ml of water to it, I end up with a dissolved solution ~42ml in volume (it is fully dissolved). Puzzled by this, I've also prepared a second solution in which I am only adding water up to the 25ml line in my 50ml conical (so far only 9ml). This is obviously struggling to dissolve as the saturation solubility for sucrose is like 5M or something like that.

So what's the deal? Is my 40ml solution with the correct mass of solute and volume of solvent really 3M? My head says yes but my gut is uneasy.

$\endgroup$
  • 1
    $\begingroup$ I've never made that particular 3.0 molar solution, but the correct method as you noted is to dissolve the 25.6 grams of sucrose in some water, and then dilute the whole solution to a final volume of 25.0 ml. $\endgroup$ – MaxW Feb 5 '16 at 21:11
1
$\begingroup$

The English wikipedia page for sucrose notes a solubility of 2000 g/L at 25 degrees Celsius, corresponding to an approximately 5.8 M solution. 3 M shouldn't be a problem in that respect, but to make dissolving easier you could warm up your solution slightly if that is appropriate in your protocol.

Now, molar is a concentration equivalent to mole per liter and this volume is the total volume including solvent and the dissolved material. So 25 ml of a 3 M solution of anything, say x, should have a final volume of 25 ml, including the solvent and 75 millimole of x.

This means your second approach is correct (being 25.6725 g of sucrose, to which you add water until the total volume is 25 ml).

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.