can someone please tell me what formula's to use for the 2 following chemistry problems.

Question

1. Calculate the solubility of potassium bromide at 23 degrees Celsius. hint. Assume that the solubility increases by an equal amount for each degrees between 20 degrees C and 30 degrees C. at 20 degrees Celsius it is 65.2.
at 30 degrees Celsius it is 70.6.

2. a saturated solution of barium chloride at 30 degrees Celsius contains 150 g of water. How much additional barium chloride can be dissolved by heating this solution to 60 degrees Celsius.
F.Y.I. these are the entire questions I think I could figure it out if I knew what formula to use.

thanking you in advance

Answer 1

#1.
Take the solubility at 30 and subtract solubility at 20 and divide by 10. That gives you an average solubility per degree C. You have it at 20, then add 3*solubility/degree and you will have it at 23 c.

#2. Surely you have a solubility curve or a graph that will tell you the solubility at 60 C. There is nothing in the problem that tells you what the solubility is in the saturated solution.

Answer 2

I will make some assumptions for #2.
That the 85.5 g solubility you quote later is 85.5g BaCl2/100 g water @ 60 C. If that's so then you will be able to dissolve 85.5g x (150gH2O/100 gH2O) = approx 128 g BaCl2 that can be dissolved at 60 C. The only piece of information needed from the 30 C solution is the 150 g H2O. Basically, this is the question you are being asked although it is masquerading with that other information. .
If the solubility of BaCl2 at 60 C is 85.5 g BaCl2/100 g H2O, how much BaCl2 can be dissolved in 150 g H2O at 60 C? Stating it that way makes it simpler I think.

Answer 3

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