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
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.