Deposition of saturated solution and affect of evaporated water

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.

• 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. – Poutnik Oct 24 '20 at 14:03