can anyone help me set this up? I think the first is simple a matter
of setting up the ratios to be equal, i.e., 18/100 = 80/x * 100. But
the 2nd part is confusing me. I know that is .54g is the solubility,
then .46g/100 mL must precipitate out. But I am confused about how to
account for the 80g that were dissolved at the higher temp. I used
.54/100 = y *80g/444.44 = .03 = y, therefore .03 * 80 gives 2.4h that
are soluble meaning that 77.6 grams precipitate out. Is this correct?

The solubility of a compound in water is: 0.54g/100 mL at 14°C,
18g/100 mL at 100°C.
How many mL of boiling water are required to dissolve 80 g of the
compound? If solution were cooled to 14°C, how many grams of the
compound would crystallize out?

I came up with the same answer, though by a different (and less
algebraic) method.  I started by saying, how much water is needed to
dissolve 80 g if 18 g can be dissolved in 100ml.  80/18=4.44 (to the
nearest hundredth), so 444 ml (or 444.44....) is needed.  Next I
figured how much would be dissolved at 14C, which is .54g/100ml or
.54*4.44=2.40(rounded to nearest hundredth again).  Subtracting the
2.4 that would still be dissolved from the 80 to start leaves 77.6g
precipitating out - assuming you are not concerned with the
possibility of supersaturation.  I'm not sure where the .46g/100ml
figure came from.  Another way to look at it would be that
(18-.54)g/100 ml must precipitate out, or 17.46g/100ml =
17.46*(444.44/100)= 77.6g.