Posted by Marty on Wednesday, November 14, 2007 at 4:18pm.
Don't you have somewhere the solubility of KNO3 in water at various temperatures?
Would a solubility curve help?
gave answer my question
Yes. What is on the x axis and what is on the y axis?
on the x axis is temperature (degrees celsius) and on the y axis is solubility (g/100g H20).
Is there any mathematical way to figure this out instead of using the solubility curve.
Yes. Generally the question is how much water will it take to dissolve so much KNO3 at a particular T (and you are given the solubility at those Ts); however, this problem isn't that. You will need a table of solubility vs T or a curve.
You want to dissolve 20 g in 40 mL water. We will assume the density of water is 1.00 g/mL; therefore, 40 mL = 40 g water.
How much is that per 100?
20 g x (100/40) = 50 g KNO3
Now look on your curve and read the T that dissolves 50 g KNO3.
Same kind of procedure for 35 g KNO3.
what are the units of 100 and 40?
grams.
20 g KNO3 is what you want to dissolve.
Your curve shows solubility per 100 g H2O. So we convert 20 g KNO3/40 g H2O to xx g KNO3/100 g H2O.
20 g KNO3 x (100 g H2O/40 g H2O) = 50 g KNO3/100 g H2O. And that is how the solubility curve is plotted. So on the x axis find 50 g KNO3, move straight up to the curve, at the intersection with the curve, move horizontally to the left and read T on the Y axis.