I corrected the first post for the Li2SO4 formula but you didn't catch it. Be sure you catch it this time around.
I usually see solubility quoted in grams/100 g soln or grams/100 mL soln. Based on that, and with no density, I don't know how to proceed unless you can clear that up. If I assume the quote is for grams Li2SO4/100 g soln, then that means the solubility is 35g Li2SO4/(35 g Li2SO4+65g H2O). Then the solubility in 250 g H2O is
35.0 x (250 mL/65 mL) = about 134.6 g Li2SO4 for a total mass of soln of 134.6 + 250 =384.6 g. That would make the soln (134.6/384.6)*100 = 35% which makes since quoted that way.
Do the same thing for the 50C water, then subtract the two. If I've interpreted wrong please correct me.
How many more grams of Li SO4 can you dissolve in 250g of water at 10.0 degrees C than at 50 degrees C.
The solubility is 35.0 at 10 degrees and at 50 degrees C it is at 32.5.
Thanks DR Bob for all your help.
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