1. Solutions Preview Understanding Concepts Reading Skills
Interpreting Graphics

2. Understanding Concepts
1. An industrial chemist stirs some crystals into water. The resulting liquid appears completely uniform under a microscope. What is the liquid?
A. a colloid
B. a solution
C. a suspension
D. a heterogeneous mixture

3. Understanding Concepts, continued
1. An industrial chemist stirs some crystals into water. The resulting liquid appears completely uniform under a microscope. What is the liquid?
A. a colloid
B. a solution
C. a suspension
D. a heterogeneous mixture

4. Understanding Concepts, continued
2. After the lungs take in oxygen during respiration, oxygen gas is dissolved in the bloodstream. Under what conditions can the most oxygen be dissolved in the blood?
F. high blood pressure and high body temperature
G. low blood pressure and low body temperature
H. low blood pressure and high body temperature
I. high blood pressure and low body temperature

5. Understanding Concepts, continued
2. After the lungs take in oxygen during respiration, oxygen gas is dissolved in the bloodstream. Under what conditions can the most oxygen be dissolved in the blood?
F. high blood pressure and high body temperature
G. low blood pressure and low body temperature
H. low blood pressure and high body temperature
I. high blood pressure and low body temperature

6. Understanding Concepts, continued
3. Potassium sulfate, K2SO4, has a molar mass of 174 g. If potassium sulfate is the solute in 2 L of a solution that has a concentration of 0.25 M, how many grams of potassium sulfate are in the solution?
A. 1,392 g
B. 348 g
C. 87 g
D g

7. Understanding Concepts, continued
3. Potassium sulfate, K2SO4, has a molar mass of 174 g. If potassium sulfate is the solute in 2 L of a solution that has a concentration of 0.25 M, how many grams of potassium sulfate are in the solution?
A. 1,392 g
B. 348 g
C. 87 g
D g

8. Understanding Concepts, continued
4. The molecules of compound A have unevenly distributed electric charge. The molecules of compound B have evenly distributed electric charge. Which compound is more likely to dissolve easily in water? Explain your reasoning.

9. Understanding Concepts, continued
4. The molecules of compound A have unevenly distributed electric charge. The molecules of compound B have evenly distributed electric charge. Which compound is more likely to dissolve easily in water? Explain your reasoning.
Answer: Compound A is polar, so it will be most likely to dissolve in water because water is polar.

10. Reading Skills SUPERCRITICAL DECAFFEINATION
One of the most common ways to remove caffeine from coffee and to preserve the flavor of the beverage is to use a solvent that dissolves the caffeine but leaves the rest of the plant material undissolved. One of the main difficulties with removing caffeine is
that caffeine is a nonpolar compound. Therefore, a nonpolar solvent is required to dissolve caffeine. But most effective nonpolar solvents are poisonous to humans. Although carbon dioxide, CO2, is a safe nonpolar compound, it is a gas under normal conditions
and cannot act as a solvent for caffeine.

11. Reading Skills, continued
When both the pressure and temperature
of a fluid are increased beyond a specific (or critical) point, a fluid has some properties of liquids and some properties of gases. Fluids under these conditions are called supercritical fluids. In the 1960s, coffee companies began using supercritical CO2 to extract caffeine. The gaslike behavior of CO2 allows its molecules to penetrate into the plant material. The liquid
aspects of CO2 allow it to dissolve caffeine molecules.

12. Reading Skills, continued
5. Once the caffeine is removed from the plant materials, how might the caffeine be recovered from the supercritical CO2?
F. by forcing the CO2 back through the plant materials
G. by lowering the temperature and pressure of the CO2
H. by changing the polarity of the caffeine molecules
I. by using a harsher solvent

13. Reading Skills, continued
5. Once the caffeine is removed from the plant materials, how might the caffeine be recovered from the supercritical CO2?
F. by forcing the CO2 back through the plant materials
G. by lowering the temperature and pressure of the CO2
H. by changing the polarity of the caffeine molecules
I. by using a harsher solvent

14. Reading Skills, continued
6. Why can’t water be used to dissolve the caffeine found in coffee and tea?

15. Reading Skills, continued
6. Why can’t water be used to dissolve the caffeine found in coffee and tea?
Answer: Caffeine is nonpolar and water is polar.

16. Interpreting Graphics
The graphic below represents three beakers that contain 500 g of
water with different amounts of sugar. One is unsaturated, one is
saturated, and one is supersaturated. Use this graphic to answer
questions 7–8.

17. Interpreting Graphics, continued
7. How many grams of sugar are there altogether in the three cylinders?
A. 2,670 g C. 534 g
B. 1,602 g D. 267 g

18. Interpreting Graphics, continued
7. How many grams of sugar are there altogether in the three cylinders?
A. 2,670 g C. 534 g
B. 1,602 g D. 267 g

19. Interpreting Graphics, continued
8. Which solution would show the most dramatic results if an additional crystal of sugar were dropped into it? What would those results be?

20. Interpreting Graphics, continued
8. Which solution would show the most dramatic results if an additional crystal of sugar were dropped into it? What would those results be?
Answer: The solution that contains 220 g of sugar per 100 g of water is the most concentrated solution, so it is the supersaturated solution. If a sugar crystal were added, all of the excess sugar would crystallize out of the solution until the solution was saturated again.

21. Interpreting Graphics, continued
The graph below shows how the solubility of a mystery solid in water depends on temperature. Use this graph to answer questions 9–10.

22. Interpreting Graphics, continued
9. The solubility of the solid is 88 g/100 g H2O at 20 ˚C. What is the solubility of the substances at 60 ˚C?
F. 88 g/100 g H2O H. 122 g/100 g H2O
G. 100 g/100 g H2O I. 264 g/100 g H2O

23. Interpreting Graphics, continued
9. The solubility of the solid is 88 g/100 g H2O at 20 ˚C. What is the solubility of the substances at 60 ˚C?
F. 88 g/100 g H2O H. 122 g/100 g H2O
G. 100 g/100 g H2O I. 264 g/100 g H2O

24. Interpreting Graphics, continued
10. Water normally boils at 100 ˚C. The scientists performing this experiment were able to measure the solubility at 100 ˚C because the solution was not boiling. Why was the solution not boiling at 100 ˚C ?

25. Interpreting Graphics, continued
10. Water normally boils at 100 ˚C. The scientists performing this experiment were able to measure the solubility at 100 ˚C because the solution was not boiling. Why was the solution not boiling at 100 ˚C ?
Answer: The dissolved solute raised the boiling point of the solution.