I'm hoping that this isn't too obvious of a question (it's been awhile since I've taken chemistry), but I was looking at the solubility table on Wikipedia, which gives the solubility as grams per 100 mL. My understanding is that 100 mL of water weighs 100 grams.

So here's what I was having trouble wrapping my head around: it lists the solubility as 108 (at least at 20 °C). If I understand that right, that would mean that I can absorb 108 grams of urea in only 100 grams of water, right? Can someone explain why it's possible to have more urea than water in this case?

  • 1
    $\begingroup$ Why would it be a problem to have more of urea than of water by mass? $\endgroup$
    – Jan
    Jan 25, 2017 at 21:25
  • $\begingroup$ @Jan It just seems really weird to me that that would be the case, I'm trying to wrap my mind around why that's possible. $\endgroup$ Jan 25, 2017 at 21:35

1 Answer 1


Well, that is just the way that it is… You should also look at this a bit differently. Urea has molar mass of $\pu{60 g mol^-1}$ and water has molar mass of $\pu{18 g mol^-1}:$

$$n(\ce{CO(NH2)2}) = \frac{m(\ce{CO(NH2)2})}{M(\ce{CO(NH2)2}) } = \frac{\pu{108 g}}{\pu{60 g mol^-1}} = \pu{1.8 mol}\tag{1}$$

$$n(\ce{H2O}) = \frac{m(\ce{H2O})}{M(\ce{H2O})} = \frac{\pu{100 g}}{\pu{18 g mol^-1}} = \pu{5.6 mol}\tag{2}$$

There are $\pu{5.6 mol}/\pu{1.8 mol} = 3.1$ molecules of water per every molecule of urea.

  • $\begingroup$ Thanks, that makes a lot of sense, so the thing that makes that possible is the fact that there are strictly more water molecules than urea molecules (i.e. the total number of molecules in each case is more relevant than the total weight)? $\endgroup$ Jan 25, 2017 at 21:35
  • $\begingroup$ In general the species with the most molecules is the solvent and the species with the fewer molecules is the solute. However solvent can also mean whatever is the liquid and solute something that is a gas (e.g. oxygen in water) or solid (a solid salt that dissolves in water). So truthfully it is hard to come up with absolute definitions. You have to look at the context in which the words are being used. $\endgroup$
    – MaxW
    Jan 25, 2017 at 21:46

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