Posted by Eric on Tuesday, October 5, 2010 at 9:08pm.
Nowhere in your problem do you say what the volume is for the solubility. In most cases it is quoted as 100 mL; therefore, I ASSUME that 1.5 g @ 20 C MEANS 1.5 grams A will dissolve in 100 mL toluene at 20 C, etc.
20 g sample is
95%A or 20 x 0.95 = 19 grams A.
5% B or 20 x 0.05 = 1 grams B.
How much toluene do we need to dissolve the 19 g A? It is 100 mL x (19/10) = 190 mL toluene needed to dissolve 19 grams A @ 110 C. Will that dissolve all of the B? Yes, because you have only 1 g B and the solubility of B at 110 C is 8 grams.
Now we cool the 190 mL toluene to 20 C. How much A comes out?
1.5 g x (190/100) = 2.85 grams A dissolve; thus, 19-2.85 = ?? grams recrystallize so your percent recovery (although the problem doesn't ask for that) is (16.15/20)*100 = ??
You may want to check to see if it will be contaminated with B. We had the 1 g B dissolved in 190 mL toluene at 110 C. At 20 C, B dissolves to the extent of
0.5 x (190/100) = 0.95. Since we had 1.0 g B, all of it will stay in solution at the lower temperature. Check my thinking.
Your answer is correct.
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