# Chemistry

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At what temperatures can 40mL of water dissolve the follownig quantities of potassium nitrate?
(a) 35.0g (b) 20.0g

can anyone start me off with a formula or a hint?

• Chemistry -

Don't you have somewhere the solubility of KNO3 in water at various temperatures?

• Chemistry -

Would a solubility curve help?

• Chemistry -

• Chemistry -

Yes. What is on the x axis and what is on the y axis?

• Chemistry -

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.

• Chemistry -

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.

• Chemistry -

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.

• Chemistry -

what are the units of 100 and 40?

• Chemistry -

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.

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