# Name Date Class. SECTION 16.1 PROPERTIES OF SOLUTIONS (pages )

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1 16 SOLUTIONS SECTION 16.1 PROPERTIES OF SOLUTIONS (pages ) This section identifies the factors that affect the solubility of a substance and determine the rate at which a solute dissolves. Solution Formation (pages ) Look at Figure 16.1 on page 471 to help you answer Questions 1 and Underline the condition that causes sugar to dissolve faster in water. a. as a whole cube or in granulated form? b. when allowed to stand or when stirred? c. at a higher temperature or a lower temperature? 2. Name three factors that influence the rate at which a solute dissolves in a solvent. a. b. c. agitation (stirring or shaking) temperature particle size 3. Is the following sentence true or false? Finely ground particles dissolve more rapidly than larger particles because finer particles expose a greater surface area to the colliding solvent molecules. true Solubility (pages ) 4. Complete the following table showing the steps in a procedure to determine the total amount of sodium chloride that will dissolve in 100 g of water at 25 C. Procedure Amount Dissolved Amount Not Dissolved Add 36.0 g of sodium chloride to the water 36.0 g 0.0 g Add an additional 1.0 g of sodium chloride Determine the total amount that dissolves 0.2 g 0.8 g 36.2 g 5. The amount of a substance that dissolves in a given quantity of solvent at a constant temperature is called the substance s solubility at that temperature. 6. If a solution contains the maximum amount of solute for a given quantity of solvent at a constant temperature, it is called a(n) saturated solution. Chapter 16 Solutions 167

2 CHAPTER 16, Solutions (continued) 7. Look at Figure 16.2 on page 472. Circle the letter of each sentence that is true about a saturated solution. a. The total amount of dissolved solute remains constant. b. The total mass of undissolved crystals remains constant. c. When the rate of solvation equals the rate of crystallization, a state of dynamic equilibrium exists. d. If more solute were added to the container, the total amount of dissolved solute would increase. 8. If two liquids dissolve each other, they are said to be miscible. 9. Look at Figure 16.3 on page 473. Why does the oil float on the vinegar? Vinegar is water-based, and oil and water are immiscible. The less dense oil floats on the water-based solution. Factors Affecting Solubility (pages ) 10. Is the following sentence true or false? The solubility of sodium chloride in water increases to 39.2 g per 100 g of water at 100 C from 36.2 g per 100 g of water at 25 C. true 11. Circle the letter of the sentence that best answers the following question. How does the solubility of solid substances change as the temperature of the solvent increases? a. The solubility increases for all solids. b. The solubility increases for most solids. c. The solubility remains constant. 12. Look at Table 16.1 on page 474. Which solid substance listed in the table is nearly insoluble at any temperature? barium sulfate 13. How does the solubility of a gas change with an increase in temperature? As the temperature increases, the solubility of a gas decreases. 14. The directly proportional relationship between the solubility of a gas in a liquid and the pressure of the gas above the liquid is known as Henry s law. 168 Guided Reading and Study Workbook

3 15. Describe the two diagrams of a bottled carbonated beverage below as greater pressure or lower pressure, and then as greater solubility or lower solubility. How do these two examples illustrate the relationship between the solubility of a gas and its vapor pressure? In the closed bottle, the pressure is high and lots of CO 2 is dissolved in the carbonated beverage. Once the cap is removed, the vapor pressure decreases, allowing more CO 2 in the carbonated beverage to escape. greater pressure and greater solubility lower pressure and lower solubility 16. How does a solution become supersaturated? If the temperature of a solution having a small excess of solid solute is raised, the solute dissolves. If the solution is then allowed to cool slowly, the excess solute may stay dissolved at a temperature below the temperature at which it would ordinarily crystallize. SECTION 16.2 CONCENTRATIONS OF SOLUTIONS (pages ) This section explains how to solve problems involving molarity of a solution, how to prepare dilute solutions from more concentrated solutions, and what is meant by percent by volume and percent by mass. Molarity (pages ) 1. A measure of the amount of solute dissolved in a given quantity of solvent is the concentration of a solution. 2. The most important unit of concentration in chemistry is molarity. 3. Is the following sentence true or false? Molarity is the number of moles of dissolved solute per liter of solvent. false Chapter 16 Solutions 169

5 Questions 8 and 9 refer to the following situation. Solvent is added to a solution until the total volume of the solution doubles. 8. What happens to the number of moles of solute present in the solution when the volume doubles? The number of moles of solute remains constant. 9. Circle the letter of the correct description of the change in molarity of a solution when the volume doubles. a. The molarity of the solution is cut in half. b. The molarity of the solution doubles. c. The molarity of the solution remains constant. d. The molarity of the solution increases slightly. 10. List the information you need to find how many milliliters of a stock solution of 2.00M MgSO 4 you would need to prepare ml of 1.00M MgSO 4. Known Unknown M M V 1? ml of 2.00M MgSO 4 M M V ml M 1 V 1 M 2 V 2 Percent Solutions (pages ) 11. List the information needed to find the percent by volume of ethanol in a solution when 50 ml of pure ethanol is diluted with water to a volume of 250 ml. Known Unknown Volume of ethanol 50 ml % ethanol by volume? % Volume of solution 250 ml volume of solute 100% % (v/v) volume of solution Reading Skill Practice Writing a summary can help you remember the information you have read. When you write a summary, include only the most important points. Write a summary of the information about percent solutions on pages Your summary should be shorter than the text on which it is based. Do your work on a separate sheet of paper. Student summaries should explain that if both the solvent and the solute are liquids, volumes of the solvent and the solution can be measured, and the concentration can be expressed as a percent by volume. If solids are dissolved in liquids, the masses of both the solvent and the solute must be obtained to express the concentration as a percent by mass. Chapter 16 Solutions 171

6 CHAPTER 16, Solutions (continued) SECTION 16.3 COLLIGATIVE PROPERTIES OF SOLUTIONS (pages ) This section explains why a solution has a lower vapor pressure, an elevated boiling point, and a depressed freezing point compared with the pure solvent of that solution. Vapor Pressure Lowering (pages ) 1. Properties of a solution that depend only on the number of particles dissolved, but not the identity of solute particles in the solution are called colligative properties. 2. Is the following sentence true or false? A nonvolatile substance is one that does not vaporize easily. true 3. Look at Figure on page 487. What happens to the vapor pressure equilibrium when a nonvolatile solute is added to a pure solvent? The equilibrium is disturbed as solvent particles form shells around solute particles. Equilibrium is eventually re-established at a lower vapor pressure. 4. How is the decrease in vapor pressure of a solution with a nonvolatile solute related to the number of particles per formula unit of solute? The decrease in vapor pressure is proportional to the number of particles the solute produces in solution. Solutes with more particles per formula unit produce a larger decrease in vapor pressure. 5. Assume 3 mol each of three different solutes have been added to three identical beakers of water as shown below. If the beakers are covered to form closed systems at constant temperature, rank the vapor pressures in each container from 1 (lowest) to 3 (highest) Glucose Na Cl Ca Guided Reading and Study Workbook

7 Freezing-Point Depression (pages ) 6. Circle the letter of each sentence that is true about the freezing point of a solution formed by a liquid solvent and nonvolatile solute. a. When a substance freezes, the arrangement of its particles becomes less orderly. b. The presence of a solute in water disrupts the formation of orderly patterns as the solution is cooled to the freezing point of pure water. c. More kinetic energy must be withdrawn from a solution than from a pure solvent in order for the solution to solidify. d. The freezing point of the solution is lower than the freezing point of the pure solvent. 7. One mole of which substance, glucose or sodium chloride, will produce more freezing-point depression when added to equal amounts of water? Why? Sodium chloride because it produces twice as many particles per formula unit in solution than glucose. Boiling-Point Elevation (page 490) 8. Circle the letter next to each sentence that is true concerning the boiling point of a solution formed by a liquid solvent and a nonvolatile solute. a. The boiling point is the temperature at which the vapor pressure equals atmospheric pressure. b. Adding a nonvolatile solute decreases the vapor pressure. c. Because of the decrease in vapor pressure, additional kinetic energy must be added to raise the vapor pressure of the liquid phase to atmospheric pressure. d. The boiling point of the solution is higher than the boiling point of the pure solvent. 9. The difference between the boiling point of a solution and that of the pure solvent is called the boiling-point elevation. SECTION 16.4 CALCULATIONS INVOLVING COLLIGATIVE PROPERTIES (pages ) This section explains how to calculate the molality and mole fraction of a solution, and how to use molality to calculate the freezing-point depression or boiling-point elevation of a solution. Molality and Mole Fraction (pages ) 1. For a solution, the ratio of moles of solute to mass of solvent in kilograms, represented by m, is the solution s molality. Chapter 16 Solutions 173

8 CHAPTER 16, Solutions (continued) 2. Is the following sentence true or false? Molarity and molality are always the same for a solution. false 3. What is the molality of a solution prepared by adding 1.0 mol of sodium chloride to 2.0 kg of water? 0.50m 4. The circle graph below shows the ratio of ethylene glycol (EG) to water in one antifreeze solution. Write the mole fractions for each substance. Mole fraction EG Mole fraction H 2 O mol EG 4.80 mol H 2 O Freezing-Point Depression and Boiling-Point Elevation (pages ) 5. Assuming a solute is molecular and not ionic, the magnitude of the boiling-point elevation of the solution, T b, is directly proportional to the molal concentration of the solute (m) 6. Look at Table 16.3 on page 495. What is the molal boiling-point elevation constant, K b, for water? C/m 7. You need to find the freezing point of a 1.50m aqueous NaCl solution. You calculate T f to be 1.86 C/m 3.00m or 5.86 C. What is the temperature at which the solution freezes? 5.86 C. 174 Guided Reading and Study Workbook

9 GUIDED PRACTICE PROBLEMS GUIDED PRACTICE PROBLEM 1 (page 477) 1. The solubility of a gas in water is 0.16 g/l at 104 kpa of pressure. What is the solubility when the pressure of the gas is increased to 288 kpa? Assume the temperature remains constant. Analyze Step 1. What is the equation for the relationship between solubility and S S 1 2 P1 pressure? P2 Step 2. What is this law called? Henry s law Step 3. What are the known values in this problem? P 1 S kpa 0.16 g/l P kpa Step 4. What is the unknown in this problem? S 2 Solve Step 5. Rearrange the equation to solve for the unknown. S 2 Step 6. Substitute the known values into the equation and solve. S g/l 288 kpa 104 kpa Evaluate S 1 P 2 P1 Step 7. How do you know that your answer is correct? Step 8. Are the units correct? Explain g/l The pressure increased nearly threefold, so the solubility should increase nearly threefold, which it does. Yes, because solubility is expressed as grams of gas dissolved per liter of solution. Chapter 16 Solutions 175

10 CHAPTER 16, Solutions (continued) GUIDED PRACTICE PROBLEM 8 (page 481) 8. A solution has a volume of 2.0 L and contains 36.0 g of glucose. If the molar mass of glucose is 180 g/mol, what is the molarity of the solution? Step 1. What is the equation for molarity of a solution? Molarity (M) moles of solute liters of solution Step 2. How many moles of g 1 mol mol glucose glucose are in the solution? 180 g Step 3. Substitute the known values into the equation for molarity. M 0.20 mol 2.0 L Step 4. Solve. M 0.10 mol/l GUIDED PRACTICE PROBLEM 9 (page 481) 9. A solution has a volume of 250 ml and contains 0.70 mol NaCl. What is its molarity? Analyze Step 1. List the knowns and the unknown. Knowns number of moles of solute 0.70 mol NaCl solution volume 250 ml L Unknown solution concentration (molarity)? M The units of molarity, M, is mol solute/l solution. Calculate Step 2. Solve for the unknown. As long as the units are correct, division gives the result. molarity 2.8 M mol solute solution volume 0.70 mol NaCl L 176 Guided Reading and Study Workbook

11 Evaluate Step 3. Does the result make sense? The numerical answer should be 4 times greater than the number of moles of solute, because 250 ml is one-fourth of 1 L. EXTRA PRACTICE (similar to Practice Problem 10, page 482) 10. How many moles of ammonium nitrate are in 375 ml of 0.40M NH 4 NO 3? 0.40 mol 1 L 375 ml 0.15 mol 1 L 1000 ml GUIDED PRACTICE PROBLEM 12 (page 484) 12. How many milliliters of a solution of 4.00M KI is needed to prepare ml of 0.760M KI? Analyze Step 1. List the knowns and the unknown. Knowns M M KI M M KI V ml M 1 V 1 M 2 V 2 Unknown V 1? ml of 4.00M KI Calculate Step 2. Solve for the unknown. Rearranging the equation above will give the result V 1 M 2 V 2 M M KI ml 4.00M KI 47.5 ml of 4.00M KI Chapter 16 Solutions 177

12 CHAPTER 16, Solutions (continued) Step 3. Evaluate Does the result make sense? The concentration of the initial solution is more than 5 times larger than that of the final solution, so the volume of initial solution should be about one-fifth of the total volume of the diluted solution. GUIDED PRACTICE PROBLEM 14 (page 485) 14. If 10 ml of pure propanone (or acetone) is diluted with water to a total solution volume of 200 ml, what is the percent by volume of acetone in the solution? Step 1. What is the equation for calculating percent by volume? Step 2. What are the knowns in this problem? volume of solute % (v/v) volume of solution 100% volume of acetone 10 ml volume of solution 200 ml Step 3. Substitute the known values into the equation and solve. % (v/v) 10 ml 100% 5 % 200 ml GUIDED PRACTICE PROBLEM 29 (page 492) 29. How many grams of sodium fluoride are needed to prepare a 0.400m NaF solution that contains 750 g of water? Analyze Step 1. List the knowns and the unknown. Knowns mass of water 750 g solution concentration 0.400m molar mass NaF 42.0 g/mol Unknown mass of solute? g NaF 178 Guided Reading and Study Workbook

13 The final solution must contain mol of NaF per 1000 g of water. This information will provide a conversion factor. The process of conversion will be: grams of water mol NaF grams NaF Calculate Step 2. Solve for the unknown. Multiply by the appropriate conversion factors: mol NaF 42.0 g NaF 750 g H 2 O 100g H 2 O 1 mol NaF 12.6 g NaF Evaluate Step 3. Does the result make sense? A one molal NaF solution contains 42.0 g NaF. Given only three-fourths of 1000 g of H 2 O and less than half concentration, the result should be less than three-eighths the molar mass. EXTRA PRACTICE (similar to Practice Problem 31, page 493) 31. What is the mole fraction of each component in a solution made by mixing 230 g of ethanol (C 2 H 5 OH) and 450 g of water? n ETH n WAT X ETH X WAT 1 mol 230 g 5.0 mol 46.0 g 1 mol 450 g 25 mol 18.0 g 5.0 mol mol 25 mol 25 mol mol 25 mol Chapter 16 Solutions 179

14 CHAPTER 16, Solutions (continued) GUIDED PRACTICE PROBLEM 33 (page 495) 33. What is the freezing point depression of an aqueous solution of 10.0 g glucose (C 6 H 12 O 6 ) in 50.0 g H 2 O? Analyze Step 1. List the knowns and the unknown. Knowns mass of solute 10.0 g C 6 H 12 O 6 mass of solvent 50.0 g H 2 O kg H 2 O K f for water 1.86 C/m T f K f m Unknown T f? C To use the given equation, first convert the mass of solute to the number of moles, then calculate the molality, m. Calculate Step 2. Solve for the unknown. Calculate the molar mass of C 6 H 12 O 6 : 1mol C 6 H 12 O g Calculate the number of moles of solute using this conversion: mol C 6 H 12 O g C 6 H 12 O mol C 6 H 12 O 6 Calculate the molality: m mol solute kg solvent 1.11m 1 mol C 6 H 12 O g C 6 H 12 O mol C 6 H 12 O kg H 2 O Calculate the freezing point depression using the known formula: T f K f m 1.86 C/m 1.11 m 2.06 C 180 Guided Reading and Study Workbook

15 Evaluate Step 3. Does the result make sense? Because a one molal solution decreases the freezing point of water by 1.86 C, a solution with concentration about one-tenth larger should decrease the freezing point further by about one-tenth of K f. GUIDED PRACTICE PROBLEM 35 (page 496) 35. What is the boiling point of a solution that contains 1.25 mol CaCl 2 in 1400 g of water? Step 1. What is the concentration 1.25 mol 1000 g 0.89 m of the CaCl 2 solution? 1400 g 1 kg Step 2. How many particles are CaCl 2 (s) Ca + 2 Cl, JKL produced by the ionization of each therefore 3 particles are produced. formula unit of CaCl 2? Step 3. What is the total molality of m 2.7m the solution? Step 4. What is the molal boiling point elevation constant (K b ) for water? K b (water) C/m Step 5. Calculate the boiling point T b C/m 2.7 m elevation. 1.4 C Step 6. Add T b to 100 C to find the 1.4 C 100 C C new boiling point. GUIDED PRACTICE PROBLEM 36 (page 496) 33. What mass of NaCl would have to be dissolved in kg of water to raise the boiling point by 2.00 C? Analyze Step 1. List the knowns and the unknown. Chapter 16 Solutions 181

16 CHAPTER 16, Solutions (continued) Knowns mass of solvent kg H 2 O K f for water 1.86 C/m T b 2.00 C T b K b m Unknown mass of solute 10.0 g NaCl First calculate the molality, m, using the given equation. Then use a molar mass conversion to determine the mass of solute. Calculate Step 2. Solve for the unknown. Calculate the molality by rearranging T b K b m: m T b /K b 2.00 C 1.86 C/m = 1.08m Calculate the molar mass of NaCl: 1 mol NaCl = 58.5 g = kg NaCl Determine the number of moles of solute using the molality and the amount of solvent: 1.08 mol NaCl moles NaCl mass of solvent molality kg H 2 O 1 kg H2 O = 1.08 mol NaCl Finally, convert the number of moles of NaCl to mass: kg NaCl 1.08 mol NaCl 1 mol NaCl = kg NaCl Evaluate Step 3. Does the result make sense? Because the solution s molality is slightly more than one, the mass of solute should be slightly more than one times the numerical value of molar mass. 182 Guided Reading and Study Workbook

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