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A certain salt has a solubility of 1 mg/1 L. I have 1 mg of this certain salt dissolving in 1 L of water, I have a concentration of 1 ppm.

1st Question: If the solution is left open to evaporation, 0.5 L of water (50 %) evaporated, would I still be getting concentration of 1 ppm while 0.5 mg of this salt became solid again and left at the bottom of the solution?

2nd Question: (if it's a yes to the 1st question) If I then fill the container back to 1 L mark with water, would the concentration be 1 ppm again after the salts are completely dissolved?

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Correct, when the water evaporates you'd have more salt than is soluble in water. The excess would then fall out of solution. Adding the same amount of water that evaporated would leave you with the same kind of solution you started out with, so the concentration of dissolved salt would be the same as when you started (although you might need to stir some to get the salt back in solution).

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  • $\begingroup$ Thank you Derek. Does this only work with solutes that do not evaporate? What if solute is not a salt but some liquid. Like the pesticide Malathion which is liquid at room temp; $\endgroup$
    – pzyxian
    Commented Jun 4, 2015 at 21:46
  • $\begingroup$ I'm not familiar with Malathion but with liquids the situation is a little different. For two liquids to be miscible, i.e. for them to be able to mix together uniformly, they have to have similar polarities. Since water is a polar solvent, Malathion would also have to be polar for the two liquids to mix. If the two liquids are miscible, then the evaporation of some water wouldn't affect whether Malathion remains mixed with the remaining water. $\endgroup$
    – Derek
    Commented Jun 5, 2015 at 10:31

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