Posted by John on Monday, November 21, 2011 at 1:14pm.
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.
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