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I have been given a graph of the amount of solubility of two substances. I am not sure how to complete these questions. Could someone just tell me what I am supposed to do, or show me a formula that I can use to solve them? I know it is hard without the graph. Thanks :)

1) What mass of substance A would you need to add to 325g of water to saturate the solution at 25 C?

2) 27g of substance A is added to 60g of water. At what temperature will the solution be saturated?

3) A saturated solution in 68g of water is cooled from 70 C to 55 C. Waht mass of the solute will crystalize out?

4) If you had a saturated solution in 30g of water each of substance A and substance B, what would be the difference of solute at 45 C?

5) You have a saturated solution in 140g of water of substance A at 20 C. What mass of water must you have to contain a saturated solution of substance B at that temperature, given the same amount of solute?

I really just need formulas, if possible. Thanks! :)

2007-11-22 09:15:46 · 2 answers · asked by Marie 2 in Science & Mathematics Chemistry

2 answers

impossible to give formulas. Help you w/number 1, from there you will have it.

Look at the graph, at the horizontal axis (Temperature) search 25 C, then draw from there a straight perfectly vertical line, where it crosses the curve of solubility, that's the saturation point, from that pont draw a straight horizontal line to find out in the "y" vertical axis the actual solubility of compound A, maybe the units will be in grams of substance per 100 grams of water, (if not, well you must make the proper adjustments to the calculations)
now take the solubility, and multiply it by 325 and divide it by 100 (Rule of three).

Advice: If you come to truly understand the "rule of three" believe me, you will have a masterkey for one third of all problems in the universe.

2007-11-22 09:38:07 · answer #1 · answered by Manuelon 4 · 0 1

impossible without graph really because you need the original masses and stuff

2007-11-22 17:36:44 · answer #2 · answered by Anonymous · 0 0

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