1. Ratio of concentration = Ratio of solubility  The added solvent must be more volatile than the desired component.  It must also specifically dissolve the desired component.  The component must have a greater tendency to dissolve in the added solvent than in the solution. K D = C O = S O = C H20 S H20

2. A substance X can be isolated from its plant source by solvent extraction. However, a minor component Y has an appreciable solubility in the solvents that may be used. Given below are the solubilities of X and Y in different solvents: SolventT, ˚CSolubility in 100g solvent at 28˚C XY Ethyl methyl ketone8065 Cyclohexane8182 Benzene8051.8 CCl 4 788.751.25 Water10021

3. a. Which is the best extracting solvent? CCl 4 is the best extracting solvent it dissolves only a little amount of Y relative to the X component needed. It also dissolves the greatest amount of X among the choices. It is immiscible with water, which happens to be the solvent in the solution.

4. ◦ Mathematically speaking, the choice relies on the K D and the best result is the one with the highest K D. The K D values with respect to water as calculated are the following: ◦ From this table, it is evident that CCl 4 is the yields the highest ratio, ergo the best possible choice as the extracting solvent. SolventKDKD Ethyl methyl ketone3 Cyclohexane4 Benzene2.5 CCl 4 4.375

5. b. Given a saturated aqueous solution of X and Y and using 100mL of solvent in (1), determine the percent recovery of X in a single extraction. K D = S CCl4 = 8.75g/100g = 4.375 S H20 2.00g/100g K D = C CCl4 = Xg/100ml = 4.375 C H20 (2.00g-Xg)/100ml X = 1.63g % rec = X pure x 100% = 1.63g x 100% = 81.4% X impure 2.00g

6. c. Repeat (b) using 50mL of solvent in each of the two successive extractions. Determine the percent recovery and compare this with (b). K D = S CCl4 = 8.75g/100g = 4.375 S H20 2.00g/100g First extraction: K D = C CCl4 = Xg/50ml = 4.375 C H20 (2.00g-Xg)/100ml X = 1.37g

7. Second extraction: K D = C CCl4 = Xg/50ml = 4.375 C H20 (0.63g-Xg)/100ml X = 0.43g % rec = X pure x 100%= (1.37+0.43)g x 100%= 90.0% X impure 2.00g In the double extraction, the percent recovery is comparably higher, considering the same amount of the solvent added.

8. d. What is the percent recovery of the minor component in a single extraction using 100mL of the solvent in (a)? K D = S CCl4 = 1.25g/100g = 1.25 S H20 1.00g/100g K D = C CCl4 = Yg/100g = 1.25 C H20 (1.00g-Yg)/100g Y = 0.56 % rec = Y pure x 100% = 0.56g x 100% = 56% Y impure 1.00g