20g of water is evaporated from 200g of saturated solution of KNO3(aq) at 60°C by heating and the solution was cooled to 20°C. How many grams of KNO3 were deposited? The solubility of KNO3 in 100g of water is 32 at 20°C and 109 at 60°C.

I have a doubt on "20g water is evaporated from 200g KNO3" statement. I'm assuming that as 20g evaporated that means the initial saturated solution has mass of 220g. This is at 60°C. I also assume the solubility in g/L.

Now, at 60°C the max amount of solution that can dissolve is 109. Thus, if we get 220g this leaves $220-109=111g$ of solvent that is not dissolved.

Then, if the saturated solution is cooled down to 20°C, the solubility will decrease to 32g/L while the amount of precipitation we have remains the same. This leaves $111-32=79 grams$ of KNO3 precipitated. Therefore, the final amount of KNO3 deposited is 79g.

Did I treat the amount of water evaporated in a correct way? I'm not sure whether the water is affecting the total concentration or 'overall' volume of the saturated solution.

  • $\begingroup$ You would serve a good service to yourself and readers, if you evaluate everything via symbolic algebraic expressions and equations. This way you would easier learn the principles and as bonus, both you and readers would easier spot eventual mistakes done. $\endgroup$
    – Poutnik
    Oct 24, 2020 at 14:03
  • $\begingroup$ if 20g is evaporated from 200g, how could be the initial mass 220g ?? I is like saying I have \$200, I give you \$20 and I initially had \$220. Strange, huh? $\endgroup$
    – Poutnik
    Oct 24, 2020 at 14:10
  • $\begingroup$ I'm new here and I was looking for that, I appreciate your info. I'm thinking before the solution get evaporated so initial mass is 220g and I feel this is really wrong because I never cover the topic relating to evaporated water in solubility and I couldn't find the source on internet explaining how it can relate to it's solubility. $\endgroup$
    – Panpaka
    Oct 24, 2020 at 14:13
  • $\begingroup$ Solubility for given T is the constant. No evaporation related operations affect it. $\endgroup$
    – Poutnik
    Oct 24, 2020 at 14:15
  • $\begingroup$ So what is the 20g evaporated water tells us? $\endgroup$
    – Panpaka
    Oct 24, 2020 at 14:16

1 Answer 1


Unfortunately, your calculations are a little bit wrong because of the poor choice of the total initial mass. It is $200 g$, and not $220~ g$. So here are the right calculations.

As $100 g$ water dissolves $109g$ $\ce{KNO3}$ at $60$°C, producing a total mass of $209g$, you may deduce that $200g$ of your initial solution contains $(200/209)·109g = 104.3 g$ $\ce{KNO3}$ and $(200/209)·100g = 95.7g$ water.

Now you loose $20 g$ water by evaporation. The amount of $\ce{KNO3}$ is still $104.3 g$. Remaining water is $95.7g - 20g = 75.7g$ water. At the final temperature $20$°C, $100g$ water can only dissolve $32g$ $\ce{KNO3}$. As a consequence, $75.7$ g water can dissolve $(75.7/100)·32g = 24.2 g$ $\ce{KNO3}$. The rest is deposited in the flask. It is $104.3 g - 24.2 g = 80.1 g$.

Please note that the mass has a unit, the gram. It is incorrect to say "the solubility of $\ce{KNO3}$ in $100g$ of water is $32$ at $20°$C and $109$ at $60°$C". You should have said that "the solubility of $\ce{KNO3}$ in $100$g of water is $32 g$ at $20$°C and $109$ g at $60$°C

  • $\begingroup$ Thank you for confirming this. I literally just solved the question after Poutnik gave me the hint and I was hesitating to confirm my answer. $\endgroup$
    – Panpaka
    Oct 24, 2020 at 16:03
  • $\begingroup$ For the note, that's original from the question. I was wondering why it did not state any unit as well. $\endgroup$
    – Panpaka
    Oct 24, 2020 at 16:11
  • $\begingroup$ Please, both OP and responders, always use as the first step a general algebraic solution, for 3 reasons.: 1/better grasp of principles for OP. 2/ better orientation and spotting eventual mistakes. 3/solution reusability and permanent value. Use eventually MathJax for mathematical/chemical expression/formula formatting. E.g. \$\$\frac{[\ce{SO4^2-}]}{[\ce{HSO4-}]}\$\$ looks like $$\frac{[\ce{SO4^2-}]}{[\ce{HSO4-}]}$$ $\endgroup$
    – Poutnik
    Oct 24, 2020 at 17:20

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