Chapter 16 Solutions 16.1 Properties of Solutions

Chapter 16 Solutions 16.1 Properties of Solutions
16.2 Concentrations of Solutions 16.3 Colligative Properties of Solutions 16.4 Calculations Involving Colligative Properties Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How can you grow a tree made out of crystals?
CHEMISTRY & YOU How can you grow a tree made out of crystals? Remember, the crystallization of a solute from solution is a physical change that is different from freezing. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What factors affect how fast a substance dissolves?
Solution Formation Solution Formation What factors affect how fast a substance dissolves? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solution Formation Granulated sugar dissolves faster than sugar cubes, and both granulated sugar and sugar cubes dissolve faster in hot tea or when you stir the tea. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solution Formation The compositions of the solvent and the solute determine whether or not a substance will dissolve. Factors that affect how fast a substance dissolves include: Agitation Temperature Particle size of the solute Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solution Formation Agitation If the contents of the glass are stirred, the crystals dissolve more quickly. The dissolving process occurs at the surface of the sugar crystals. Stirring speeds up the process because fresh solvent (the water) is continually brought in contact with the surface of the solute (sugar). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solution Formation Agitation Agitation (stirring or shaking) affects only the rate at which a solid solute dissolves. It does not influence the amount of solute that will dissolve. An insoluble substance remains undissolved regardless of how vigorously or for how long the solvent/solute system is agitated. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Temperature also influences the rate at which a solute dissolves.
Solution Formation Temperature Temperature also influences the rate at which a solute dissolves. Sugar dissolves much more rapidly in hot tea than in iced tea. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solution Formation Temperature At higher temperatures, the kinetic energy of water molecules is greater than at lower temperatures, so the molecules move faster. The more rapid motion of the solvent molecules leads to an increase in the frequency of the force of the collisions between water molecules and the surfaces of the sugar crystals. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Particle Size of the Solute
Solution Formation Particle Size of the Solute The rate at which a solute dissolves also depends upon the size of the solute particles. The smaller particles in granulated sugar expose a much greater surface area to the colliding water molecules. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Particle Size of the Solute
Solution Formation Particle Size of the Solute The dissolving process is a surface phenomenon. The more surface area of the solute that is exposed, the faster the rate of dissolving. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A. Increasing the temperature B. Stirring the mixture
Which of the following will not speed up the rate at which a solid solute dissolves? A. Increasing the temperature B. Stirring the mixture C. Crushing the solute D. Decreasing the temperature Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How can you describe the equilibrium in a saturated solution?
Solubility Solubility How can you describe the equilibrium in a saturated solution? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What is happening in this figure?
Solubility What is happening in this figure? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What is happening in this figure?
Solubility What is happening in this figure? Particles move from the solid into the solution. Some dissolved particles move from the solution back to the solid. Because these two processes occur at the same rate, no net change occurs in the overall system. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Such a solution is said to be saturated.
Solubility Such a solution is said to be saturated. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Such a solution is said to be saturated.
Solubility Such a solution is said to be saturated. A saturated solution contains the maximum amount of solute for a given quantity of solvent at a constant temperature and pressure. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solubility In a saturated solution, a state of dynamic equilibrium exists between the solution and any undissolved solute, provided that the temperature remains constant. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solubility The solubility of a substance is the amount of solute that dissolves in a given quantity of a solvent at a specified temperature and pressure to produce a saturated solution. Solubility is often expressed in grams of solute per 100 g of solvent (g/100 g H2O). Sometimes the solubility of a gas is expressed in grams per liter of solution (g/L). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solubility A solution that contains less solute than a saturated solution at a given temperature and pressure is an unsaturated solution. If additional solute is added to an unsaturated solution, the solute will dissolve until the solution is saturated. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solubility Some liquids—for example, water and ethanol—are infinitely soluble in each other. Two liquids are miscible if they dissolve in each other in all proportions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Liquids that are insoluble in each other are immiscible.
Solubility Liquids that are insoluble in each other are immiscible. Oil and water are examples of immiscible liquids. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A. grams of solute per 100 liters of solvent
The solubility of a substance is often expressed as which of the following? A. grams of solute per 100 liters of solvent B. grams of solute per 1 cm3 of solvent C. grams of solute per 100 grams of solvent D. grams of solute per 100 grams of solution Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Factors Affecting Solubility
What factors affect the solubility of a substance? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Temperature affects the solubility of solid, liquid, and gaseous solutes in a solvent; both temperature and pressure affect the solubility of gaseous solutes. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs Temperature The solubility of most solid substances increases as the temperature of the solvent increases. Temperature (°C) Solubility (g/100g H2O) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs Temperature The solubility of most solid substances increases as the temperature of the solvent increases. Temperature (°C) Solubility (g/100g H2O) For a few substances, solubility decreases with temperature. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solubilities of Substances in Water at Various Temperatures
Interpret Data Solubilities of Substances in Water at Various Temperatures Solubility (g/100 g H2O) Substance Formula 0°C 20°C 50°C 100°C Barium hydroxide Ba(OH)2 1.67 31.89 — Barium sulfate BaSO4 Calcium hydroxide Ca(OH)2 0.189 0.173 0.07 Potassium chlorate KClO3 4.0 7.4 19.3 56.0 Potassium chloride KCl 27.6 34.0 42.6 57.6 Sodium chloride NaCl 35.7 36.0 37.0 39.2 Sodium nitrate NaNO3 74 88.0 114.0 182 Aluminum chloride AlCl3 30.84 31.03 31.60 33.32 Silver nitrate AgNO3 122 222.0 455.0 733 Sucrose (table sugar) C12H22O11 179 230.9 260.4 487 Hydrogen H2 0.0 Oxygen O2 0.0070 0.0043 0.0026 Carbon dioxide CO2 0.335 0.169 0.076 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Temperature A supersaturated solution contains more solute than it can theoretically hold at a given temperature. The crystallization of a supersaturated solution can be initiated if a very small crystal, called a seed crystal, of the solute is added. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The rate at which excess solute deposits upon the surface of a seed crystal can be very rapid. The solution is clear before a seed crystal is added. Crystals begin to form immediately after the addition of a seed crystal. Excess solute crystallizes rapidly. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU How do you think crystal-growing kits work? Use what you know about solubility and supersaturated solutions to explain your answer. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU How do you think crystal-growing kits work? Use what you know about solubility and supersaturated solutions to explain your answer. Crystal-growing kits usually begin with a supersaturated solution. When a seed crystal is added to the solution, crystals rapidly begin to grow because the supersaturated solution contains more solute than is theoretically possible. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Temperature The effect of temperature on the solubility of gases in liquid solvents is opposite that of solids. The solubilities of most gases are greater in cold water than in hot. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure Changes in pressure have little effect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure Carbonated beverages are a good example. The drinks are bottled under a high pressure of CO2 gas, which forces larger amounts of the gas into solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure Carbonated beverages are a good example. When the container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure How is the partial pressure of carbon dioxide gas related to the solubility of CO2 in a carbonated beverage? The relationship is described by Henry’s law, which states that at a given temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure As the pressure of the gas above the liquid increases, the solubility of the gas increases. As the pressure of the gas decreases, the solubility of the gas decreases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Pressure You can write the relationship in the form of an equation. = S1 P1 S2 P2 S1 is the solubility of a gas at one pressure, P1. S2 is the solubility at another pressure, P2. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.1 Using Henry’s Law If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is its solubility (in g/L) at 1.0 atm of pressure? (The temperature is held constant at 25°C.) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknown.
Sample Problem 16.1 Analyze List the knowns and the unknown. 1 Use Henry’s law to solve for the unknown solubility. KNOWNS UNKNOWN P1 = 3.5 atm S1 = 0.77 g/L P2 = 1.0 atm S2 = ? g/L Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknowns.
Sample Problem 16.1 Calculate Solve for the unknowns. 2 State the equation for Henry’s law. Isolate S2 by multiplying both sides by P2: P2  =  P2 S1 P1 S2 P2 = S1 P1 S2 P2 Solve Henry’s law for S2. Substitute the known values and calculate. S2 = = = 0.22 g/L S1  P2 P1 0.77 g/L  1.0 atm 3.5 atm Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Evaluate Does the result make sense?
Sample Problem 16.1 Evaluate Does the result make sense? 3 The new pressure is approximately one-third of the original pressure. So, the new solubility should be approximately one-third of the original. The answer is correctly expressed to two significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Explain why an opened container of a carbonated beverage is more likely to go flat sitting on the counter than in the refrigerator. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Explain why an opened container of a carbonated beverage is more likely to go flat sitting on the counter than in the refrigerator. The solubility of a gas in a liquid increases with decreasing temperature. More carbon dioxide will remain in solution at the colder temperature found in the refrigerator. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts Factors that determine how fast a substance dissolves are stirring, temperature, and surface area. In a saturated solution, a state of dynamic equilibrium exists between the solution and any undissolved solute, provided that the temperature remains constant. Temperature affects the solubility of solid, liquid, and gaseous solutes in a solvent; both temperature and pressure affect the solubility of gaseous solutes. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Henry’s law: = S1 P1 S2 P2 Key Equation
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms saturated solution: a solution containing the maximum amount of solute for a given amount of solvent at a constant temperature and pressure; an equilibrium exists between undissolved solute and ions in solution solubility: the amount of a substance that dissolves in a given quantity of solvent at specified conditions of temperature and pressure to produce a saturated solution Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms unsaturated solution: a solution that contains less solute than a saturated solution at a given temperature and pressure miscible: describes liquids that dissolve in each other in all proportions immiscible: describes liquids that are insoluble in each other; oil and water are immiscible Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms supersaturated solution: a solution that contains more solute than it can theoretically hold at a given temperature; excess solute precipitates if a seed crystal is added Henry’s law: at a given temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 16 Solutions 16.2 Concentrations of Solutions
16.1 Properties of Solutions 16.2 Concentrations of Solutions 16.3 Colligative Properties of Solutions 16.4 Calculations Involving Colligative Properties Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How can you describe the concentration of a solution?
CHEMISTRY & YOU How can you describe the concentration of a solution? The federal government and state governments set standards limiting the amount of contaminants allowed in drinking water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How do you calculate the molarity of a solution?

Molarity The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. A solution that contains a relatively small amount of solute is a dilute solution. A concentrated solution contains a large amount of solute. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In chemistry, the most important unit of concentration is molarity.
Molarity (M) is the number of moles of solute dissolved in one liter of solution. Molarity is also known as molar concentration. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Molarity The figure below illustrates the procedure for making a 0.5M, or 0.5-molar, solution. Add 0.5 mol of solute to a 1-L volumetric flask half filled with distilled water. Swirl the flask carefully to dissolve the solute. Fill the flask with water exactly to the 1-L mark. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Molarity To calculate the molarity of a solution, divide the number of moles of solute by the volume of the solution in liters. Molarity (M) = moles of solute liters of solution Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.2 Calculating Molarity Intravenous (IV) saline solutions are often administered to patients in the hospital. One saline solution contains 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.2 Analyze List the knowns and the unknown. 1 Convert the concentration from g/100 mL to mol/L. The sequence is: g/100 mL → mol/100 mL → mol/L KNOWNS solution concentration = 0.90 g NaCl/100 mL molar mass NaCl = 58.5 g/mol UNKNOWN solution concentration = ?M Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown.
Sample Problem 16.2 Calculate Solve for the unknown. 2 Use the molar mass to convert g NaCl/100 mL to mol NaCl/100 mL. Then convert the volume units so that your answer is expressed in mol/L. The relationship L = 1000 mL gives you the conversion factor 1000 mL/1 L. Solution concentration =   = 0.15 mol/L = 0.15M 0.90 g NaCl mol NaCl mL 100 mL g NaCl L Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.2 Evaluate Does the result make sense? 3 The answer should be less than 1M because a concentration of 0.90 g/100 mL is the same as 9.0 g/1000 mL (9.0 g/1 L), and 9.0 g is less than 1 mol NaCl. The answer is correctly expressed to two significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating the Moles of Solute in a Solution
Sample Problem 16.3 Calculating the Moles of Solute in a Solution Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70M NaClO? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.3 Analyze List the knowns and the unknown. 1 The conversion is: volume of solution → moles of solute. Molarity has the units mol/L, so you can use it as a conversion factor between moles of solute and volume of solution. KNOWNS volume of solution = 1.5 L solution concentration = 0.70M NaClO UNKNOWN moles solute = ? mol Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.3 Calculate Solve for the unknown. 2 Multiply the given volume by the molarity expressed in mol/L. Make sure that your volume units cancel when you do these problems. If they don’t, then you’re probably missing a conversion factor in your calculations. 1.5 L  = 1.1 mol NaClO 0.70 mol NaCl 1 L Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.3 Evaluate Does the result make sense? 3 The answer should be greater than 1 mol but less than 1.5 mol, because the solution concentration is less than 0.75 mol/L and the volume is less than 2 L. The answer is correctly expressed to two significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

B. enough water to make 1.00 liter of solution
How much water is required to make a 1.00M aqueous solution of NaCl, if 58.4 g of NaCl are dissolved? A liter of water B. enough water to make 1.00 liter of solution C kg of water D. 100 mL of water Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What effect does dilution have on the amount of solute?
Making Dilutions Making Dilutions What effect does dilution have on the amount of solute? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Both of these solutions contain the same amount of solute.
Making Dilutions Both of these solutions contain the same amount of solute. You can tell by the color of solution (a) that it is more concentrated than solution (b). Solution (a) has the greater molarity. The more dilute solution (b) was made from solution (a) by adding more solvent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Moles of solute before dilution = Moles of solute after dilution
Making Dilutions Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change. Moles of solute before dilution = Moles of solute after dilution Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Moles of solute = molarity (M)  liters of solution (V)
Making Dilutions Moles of solute before dilution = Moles of solute after dilution The definition of molarity can be rearranged to solve for moles of solute. Molarity (M) = moles of solute liters of solution (V) Moles of solute = molarity (M)  liters of solution (V) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The total number of moles of solute remains unchanged upon dilution.
Making Dilutions The total number of moles of solute remains unchanged upon dilution. Moles of solute = M1  V1 = M2  V2 M1 and V1 are the molarity and the volume of the initial solution. M2 and V2 are the molarity and volume of the diluted solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Making Dilutions The student is preparing 100 mL of 0.40M MgSO4 from a stock solution of 2.0M MgSO4. She measures 20 mL of the stock solution with a 20-mL pipet. She transfers the 20 mL to a 100-mL volumetric flask. She carefully adds water to the mark to make 100 mL of solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Preparing a Dilute Solution
Sample Problem 16.4 Preparing a Dilute Solution How many milliliters of aqueous 2.00M MgSO4 solution must be diluted with water to prepare mL of aqueous 0.400M MgSO4? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.4 Analyze List the knowns and the unknown. 1 Use the equation M1  V1 = M2  V2 to solve for the unknown initial volume of solution (V1) that is diluted with water. KNOWNS M1 = 2.00M MgSO4 M2 = 0.400M MgSO4 V2 = mL of 0.400M MgSO4 UNKNOWN V1 = ? mL of 2.00M MgSO4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.4 Calculate Solve for the unknown. 2 Solve for V1 and substitute the known values into the equation. V1 = = = 20.0 mL M2  V2 M1 0.400M  mL 2.00M Thus, 20.0 mL of the initial solution must be diluted by adding enough water to increase the volume to mL. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.4 Evaluate Does the result make sense? 3 The initial concentration is five times larger than the dilute concentration. Because the number of moles of solute does not change, the initial volume of solution should be one-fifth the final volume of the diluted solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

100. 0 mL of a 0. 300M CuSO4·5H2O solution is diluted to 500. 0 mL
100.0 mL of a 0.300M CuSO4·5H2O solution is diluted to mL. What is the concentration of the diluted solution? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

100. 0 mL of a 0. 300M CuSO4·5H2O solution is diluted to 500. 0 mL
100.0 mL of a 0.300M CuSO4·5H2O solution is diluted to mL. What is the concentration of the diluted solution? M1  V1 = M2  V2 M2 = M M2 = = M1  V1 V2 0.300M  mL 500.0 mL Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How do percent by volume and percent by mass differ?
Percent Solutions Percent Solutions How do percent by volume and percent by mass differ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Solutions Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. Isopropyl alcohol (2-propanol) is sold as a 91-percent solution by volume. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Solutions Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. Isopropyl alcohol (2-propanol) is sold as a 91-percent solution by volume. You could prepare such a solution by diluting 91 mL of pure isopropyl alcohol with enough water to make 100 mL of solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Solutions Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. Isopropyl alcohol (2-propanol) is sold as a 91-percent solution by volume. The concentration is written as 91 percent by volume, 91 percent (volume/volume), or 91% (v/v). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Solutions Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. Percent by volume (%(v/v)) =  100% volume of solution volume of solute Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating Percent by Volume
Sample Problem 16.5 Calculating Percent by Volume What is the percent by volume of ethanol (C2H6O, or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.5 Analyze List the knowns and the unknown. 1 Use the known values for the volume of solute and volume of solution to calculate percent by volume. KNOWNS volume of solute = 85 mL ethanol volume of solution = 250 mL UNKNOWN Percent by volume = ?% ethanol (v/v) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.5 Calculate Solve for the unknown. 2 State the equation for percent by volume. Percent by volume (%(v/v)) =  100% volume of solution volume of solute Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.5 Calculate Solve for the unknown. 2 Substitute the known values into the equation and solve. Percent by volume (%(v/v)) =  100% 250 mL 85 mL ethanol = 34% ethanol (v/v) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.5 Evaluate Does the result make sense? 3 The volume of the solute is about one-third the volume of the solution, so the answer is reasonable. The answer is correctly expressed to two significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Solutions Another way to express the concentration of a solution is as a percent by mass, or percent (mass/mass). Percent by mass of a solution is the ratio of the mass of the solute to the mass of the solution. Percent by mass (%(m/m)) =  100% mass of solution mass of solute Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Solutions Percent by mass (%(m/m)) =  100% mass of solution mass of solute Percent by mass is sometimes a convenient measure of concentration when the solute is a solid. You have probably seen information on food labels expressed as a percent composition. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What are three ways to calculate the concentration of a solution?
CHEMISTRY & YOU What are three ways to calculate the concentration of a solution? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What are three ways to calculate the concentration of a solution?
CHEMISTRY & YOU What are three ways to calculate the concentration of a solution? The concentration of a solution can be calculated in moles solute per liter of solvent, or molarity (M), percent by volume (%(v/v)), or percent by mass (%(m/m)). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Percent by Mass as a Conversion Factor
Sample Problem 16.6 Using Percent by Mass as a Conversion Factor How many grams of glucose (C6H12O6) are needed to make 2000 g of a 2.8% glucose (m/m) solution? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.6 Analyze List the knowns and the unknown. 1 The conversion is mass of solution → mass of solute. In a 2.8% C6H12O6 (m/m) solution, each 100 g of solution contains 2.8 g of glucose. Used as a conversion factor, the concentration allows you to convert g of solution to g of C6H12O6. KNOWNS mass of solution = 2000 g percent by mass = 2.8% C6H12O6(m/m) UNKNOWN mass of solute = ? g C6H12O6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.6 Calculate Solve for the unknown. 2 Write percent by mass as a conversion factor with g C6H12O6 in the numerator. You can solve this problem by using either dimensional analysis or algebra. 100 g solution 2.8 g C6H12O6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.6 Calculate Solve for the unknown. 2 Multiply the mass of the solution by the conversion factor. 2000 g solution  = 56 g C6H12O6 100 g solution 2.8 g C6H12O6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.6 Evaluate Does the result make sense? 3 The prepared mass of the solution is 20  100 g. Since a 100-g sample of 2.8% (m/m) solution contains 2.8 g of solute, you need 20  2.8 g = 56 g of solute. To make the solution, mix 56 g of C6H12O6 with 1944 g of solvent. 56 g of solute g solvent = 2000 g of solution Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 5.0% (m/m)?

(% (m/m))  mass of solution
What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 5.0% (m/m)? % (m/m) =  100% mass of glucose mass of solution mass of glucose = mass of glucose = 2000 g  = 100 g C6H12O6 mass of water = 2000 g – 100 g = 1900 g H2O (% (m/m))  mass of solution 100% Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts To calculate the molarity of a solution, divide the moles of solute by the volume of the solution in liters. Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change. Percent by volume is the ratio of the volume of solute to the volume of solution. Percent by mass is the ratio of the mass of the solute to the mass of the solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Equations moles of solute Molarity (M) = liters of solution
M1  V1 = M2  V2 Percent by volume =  100% volume of solution volume of solute Percent by mass =  100% mass of solution mass of solute Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

dilute solution: a solution that contains a small amount of solute
Glossary Terms concentration: a measurement of the amount of solute that is dissolved in a given quantity of solvent; usually expressed as mol/L dilute solution: a solution that contains a small amount of solute concentrated solution: a solution containing a large amount of solute molarity (M): the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 liter of solution Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 16 Solutions 16.3 Colligative Properties of Solutions
16.2 Concentrations of Solutions 16.3 Colligative Properties of Solutions 16.4 Calculations Involving Colligative Properties Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Why do you need salt to make ice cream?
CHEMISTRY & YOU Why do you need salt to make ice cream? Ice-cream makers know that if you add rock salt to ice, the mixture freezes at a few degrees below 0°C. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Describing Colligative Properties
What are three colligative properties of solutions? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Tea is not the same as pure water. Some of these differences in properties have little to do with the specific identity of the solute. Instead, they depend upon the mere presence of solute particles in the solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A colligative property is a property of solutions that depends only upon the number of solute particles, not upon their identity. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Three important colligative properties of solutions are vapor-pressure lowering freezing-point depression boiling-point elevation Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Vapor pressure is the pressure exerted by a vapor that is in dynamic equilibrium with its liquid in a closed system. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Vapor pressure is the pressure exerted by a vapor that is in dynamic equilibrium with its liquid in a closed system. A solution that contains a solute that is nonvolatile (not easily vaporized) always has a lower vapor pressure than the pure solvent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Higher vapor pressure Pure solvent Solvent particle Lower vapor pressure Solution containing nonvolatile solute Solute particle Equilibrium is established between the liquid and vapor in a pure solvent. In a solution, solute particles reduce the number of solvent particles able to escape the liquid. Equilibrium is established at a lower vapor pressure. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Ionic solutes that dissociate have greater effects on vapor pressure than does a nondissociating solute. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Ionic solutes that dissociate have greater effects on vapor pressure than does a nondissociating solute. Three moles of sodium chloride dissolved in water produce 6 mol of particles because each formula unit of NaCl dissociates into two ions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Ionic solutes that dissociate have greater effects on vapor pressure than does a nondissociating solute. Three moles of calcium chloride dissolved in water produce 9 mol of particles because each formula unit of CaCl2 dissociates into three ions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Ionic solutes that dissociate have greater effects on vapor pressure than does a nondissociating solute. Three moles of glucose dissolved in water produce 3 mol of particles because glucose does not dissociate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering The decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Which solution has the lowest vapor pressure? The vapor-pressure lowering caused by 0.1 mol of NaCl in 1000 g of water is twice that caused by 0.1 mol of glucose in the same quantity of water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Vapor-Pressure Lowering Which solution has the lowest vapor pressure? In the same way, 0.1 mol of CaCl2 in 1000 g of water produces three times the vapor-pressure lowering as 0.1 mol of glucose in the same quantity of water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Freezing-Point Depression When a substance freezes, the particles of the solid take on an orderly pattern. The presence of a solute in water disrupts the formation of this pattern. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Freezing-Point Depression When a substance freezes, the particles of the solid take on an orderly pattern. The presence of a solute in water disrupts the formation of this pattern. As a result, more kinetic energy must be withdrawn from a solution than from the pure solvent to cause the solution to solidify. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Freezing-Point Depression The freezing point of a solution is lower than the freezing point of the pure solvent. The difference in temperature between the freezing point of a solution and the freezing point of the pure solvent is called the freezing-point depression. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Freezing-Point Depression Freezing-point depression is another colligative property. The magnitude of the freezing-point depression is proportional to the number of solute particles dissolved in the solvent and does not depend upon their identity. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Freezing-Point Depression The freezing-point depression of aqueous solutions plays an important role in helping keep travelers safe in cold, icy weather. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Freezing-Point Depression The freezing-point depression of aqueous solutions plays an important role in helping keep travelers safe in cold, icy weather. The truck spreads a layer of salt on the icy road to make the ice melt. The melted ice forms a solution with a lower freezing point than that of pure water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Boiling-Point Elevation The boiling point of a substance is the temperature at which the vapor pressure of the liquid phase equals atmospheric pressure. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Boiling-Point Elevation The boiling point of a substance is the temperature at which the vapor pressure of the liquid phase equals atmospheric pressure. Adding a nonvolatile solute to a liquid solvent decreases the vapor pressure of the solvent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Boiling-Point Elevation The boiling point of a substance is the temperature at which the vapor pressure of the liquid phase equals the atmospheric pressure. Adding a nonvolatile solute to a liquid solvent decreases the vapor pressure of the solvent. Because of the decrease in vapor pressure, additional kinetic energy must be added to raise the vapor pressure of the liquid phase of the solution to atmospheric pressure and initiate boiling. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Boiling-Point Elevation The boiling point of a solution is higher than the boiling point of the pure solvent. The difference in temperature between the boiling point of a solution and the boiling point of the pure solvent is the boiling-point elevation. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Boiling-Point Elevation The fluid circulating through a car’s cooling system is a solution of water and ethylene glycol, or antifreeze. The antifreeze doesn’t just lower the freezing point of the water in the cooling system. It also elevates the boiling point, which helps protect the engine from overheating in the summer. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Boiling-Point Elevation Boiling-point elevation is a colligative property; it depends on the concentration of particles, not on their identity. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Boiling-Point Elevation Boiling-point elevation is a colligative property; it depends on the concentration of particles, not on their identity. The magnitude of the boiling-point elevation is proportional to the number of solute particles dissolved in the solvent. The boiling point of water increases by 0.512°C for every mole of particles that the solute forms when dissolved in 1000 g of water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU Solutes other than NaCl could be used to produce the same freezing-point depression in an ice-cream machine. What factors do you think make NaCl a good choice? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU Solutes other than NaCl could be used to produce the same freezing-point depression in an ice-cream machine. What factors do you think make NaCl a good choice? NaCl, or rock salt, is readily available, inexpensive, and non-toxic. It is an ionic compound. It produces twice the freezing-point depression of a molecular solid such as sucrose, or table sugar. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

You have 500 mL of 1M solutions of NaCl, Na2SO4, Na3PO4, and Al2(SO4)3
You have 500 mL of 1M solutions of NaCl, Na2SO4, Na3PO4, and Al2(SO4)3. Which solution will have the highest boiling point? A. NaCl(aq) B. Na2SO4(aq) C. Na3PO4(aq) D. Al2(SO4)3(aq) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts Colligative properties of solutions include vapor-pressure lowering, freezing-point depression, and boiling-point elevation. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms colligative property: a property of a solution that depends only upon the number of solute particles, and not upon their identities; boiling- point elevation, freezing-point depression, and vapor-pressure lowering are colligative properties Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms freezing-point depression: the difference in temperature between the freezing point of a solution and the freezing point of the pure solvent boiling-point elevation: the difference in temperature between the boiling point of a solution and the boiling point of the pure solvent Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The Mole and Quantifying Matter
BIG IDEA The Mole and Quantifying Matter Solubility, miscibility, concentration, and colligative properties are used to describe and characterize solutions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 16 Solutions 16.4 Calculations Involving
16.1 Properties of Solutions 16.2 Concentrations of Solutions 16.3 Colligative Properties of Solutions 16.4 Calculations Involving Colligative Properties Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How hot is a pot of boiling pasta?
CHEMISTRY & YOU How hot is a pot of boiling pasta? Recall that dissolved salt elevates the boiling point of water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Recall that colligative properties of solutions depend only on the number of solute particles dissolved in a given amount of solvent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Molality (m) is the number of moles of solute dissolved in 1 kilogram (1000 grams) of solvent. Molality is also known as molal concentration. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Molality (m) is the number of moles of solute dissolved in 1 kilogram (1000 grams) of solvent. Molality is also known as molal concentration. moles of solute kilogram of solvent Molality (m) = Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Molality is not the same as molarity. Molality refers to moles of solute per kilogram of solvent rather than moles of solute per liter of solution. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.7 Using Molality How many grams of potassium iodide must be dissolved in g of water to produce a molal KI solution? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.7 Analyze List the knowns and the unknown. 1 According to the definition of molality, the final solution must contain mol KI per 1000 g H2O. Use the molality as a conversion factor to convert from mass of the solvent (H2O) to moles of the solute (KI). Then use the molar mass of KI to convert from mol KI to g KI. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.7 Analyze List the knowns and the unknown. 1 The steps are as follows: mass of H2O → mol KI → g KI. KNOWNS mass of water = g = kg solution concentration = 0.060m molar mass KI = g/mol UNKNOWN mass of solute = ? g KI Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.7 Calculate Solve for the unknown. 2 Identify the conversion factor based on 0.060m that allows you to convert from g H2O to mol KI. 1.000 kg H2O 0.060 mol KI Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.7 Calculate Solve for the unknown. 2 Identify the conversion factor based on the molar mass of KI that allows you to convert from mol KI to g KI. 1 mol KI 166.0 g KI Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.7 Calculate Solve for the unknown. 2 Multiply the known solvent volume by the conversion factors. To make the molal KI solution, you would dissolve 5.0 g of KI in g of water. kg H2O   = 5.0 g KI 1.000 kg H2O 0.060 mol KI 1 mol KI 166.0 g KI Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.7 Evaluate Does the result make sense? 3 A 1-molal KI solution is one molar mass of KI (166.0 g) dissolved in 1000 g of water. The desired molal concentration (0.060m) is about 1/20 of that value, so the mass of KI should be much less than the molar mass. The answer is correctly expressed to two significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The concentration of a solution can also be expressed as a mole fraction. The mole fraction of a solute in a solution is the ratio of the moles of that solute to the total number of moles of solvent and solute. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In a solution containing nA mol of solute A and nB mol of solvent B, the mole fraction of solute A (XA) and the mole fraction of solvent B (XB) can be expressed as follows: XA = XB = nA + nB nA nB Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Note that mole fraction is a dimensionless quantity. The sum of the mole fractions of all the components in a solution equals unity, or one. XA = XB = nA + nB nA nB Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Ethylene glycol (EG) is added to water as antifreeze. The figure below illustrates how to calculate the mole fractions of the solute and solvent for a solution of EG in water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating Mole Fractions
Sample Problem 16.8 Calculating Mole Fractions Ethylene glycol (EG, or C2H6O2) is added to automobile cooling systems to protect against cold weather. What is the mole fraction of each component in a solution containing 1.25 mol of ethylene glycol and 4.00 mol of water? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknowns.
Sample Problem 16.8 Analyze List the knowns and the unknowns. 1 The given quantities of solute (EG) and solvent (water) are expressed in moles. Use the equations for mole fraction of a solute and mole fraction of a solvent to solve this problem. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknowns.
Sample Problem 16.8 Analyze List the knowns and the unknowns. 1 KNOWNS moles of ethylene glycol (nEG) = 1.25 mol EG moles of water (nH2O) = 4.00 mol H2O UNKNOWNS mole fraction EG (XEG) = ? mole fraction H2O (XH2O) = ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.8 Calculate Solve for the unknowns. 2 Write the equation for the mole fraction of ethylene glycol (XEG) in the solution. Note that the denominator for each mole fraction is the same: the total number of moles of solvent and solute in the solution. XEG = nEG + nH2O nEG Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.8 Calculate Solve for the unknowns. 2 Write the equation for the mole fraction of water (XH2O) in the solution. Note that the denominator for each mole fraction is the same: the total number of moles of solvent and solute in the solution. XH2O = nEG + nH2O nH2O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.8 Calculate Solve for the unknowns. 2 Substitute the known values into each equation. XEG = = = 0.238 nEG + nH2O nEG 1.25 mol 1.25 mol mol XH2O = = = 0.762 nEG + nH2O nH2O 4.00 mol 1.25 mol mol Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.8 Evaluate Does the result make sense? 3 The sum of the mole fractions of all the components in the solution equals 1 (XEG + XH2O = 1.000). Each answer is correctly expressed to three significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What is the mole fraction of He in a gaseous solution containing 4
What is the mole fraction of He in a gaseous solution containing 4.0 g of He, 6.5 g of Ar, and 10.0 g of Ne? A. 0.60 B. 1.5 C. 0.20 D. 0.11 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Freezing-Point Depression and Boiling-Point Elevation
How are freezing-point depression and boiling-point elevation related to molality? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Depressions of freezing points and elevations of boiling points are usually quite small. To measure colligative properties accurately, you would need a thermometer that can measure temperatures to the nearest 0.001°C. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Depressions of freezing points and elevations of boiling points are usually quite small. Another way to determine the magnitudes of colligative properties is by calculating them. You can do this if you know the molality of the solution and some reference data about the solvent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The magnitudes of the freezing-point depression (ΔTf) and the boiling-point elevation (ΔTb) of a solution are directly proportional to the molal concentration (m), assuming the solute is molecular, not ionic. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The magnitudes of the freezing-point depression (ΔTf) and the boiling-point elevation (ΔTb) of a solution are directly proportional to the molal concentration (m), assuming the solute is molecular, not ionic. ΔTf m ΔTb m Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs Temperature (°C) Vapor pressure (kPa) 101 ΔTf is the difference between the freezing point of the solution and the freezing point of the pure solvent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs Temperature (°C) Vapor pressure (kPa) 101 ΔTb is the difference between the boiling point of the solution and the boiling point of the pure solvent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

With the addition of a constant, the proportionality between the freezing-point depression and the molality m can be expressed as an equation. ΔTf = Kf  m Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

With the addition of a constant, the proportionality between the freezing-point depression and the molality m can be expressed as an equation. ΔTf = Kf  m The constant, Kf, is the molal freezing-point depression constant, which is equal to the change in freezing point for a 1-molal solution of a nonvolatile molecular solute. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The boiling-point elevation of a solution can also be expressed as an equation. The proportionality constant is Kb, the molal boiling-point elevation constant, which is equal to the change in boiling point for a 1-molal solution of a nonvolatile molecular solute. ΔTb = Kb  m Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The values of Kf and Kb depend upon the solvent.
Interpret Data The values of Kf and Kb depend upon the solvent. Kf and Kb Values for Some Common Solvents Solvent Kf (°C/m) Kb (°C/m) Acetic Acid 3.90 3.07 Benzene 5.12 2.53 Camphor 37.7 5.95 Cyclohexane 20.2 2.79 Ethanol 1.99 1.19 Nitrobenzene 7.00 5.24 Phenol 7.40 3.56 Water 1.86 0.512 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating the Freezing-Point Depression of a Solution
Sample Problem 16.9 Calculating the Freezing-Point Depression of a Solution Antifreeze protects a car from freezing. It also protects it from overheating. Calculate the freezing-point depression of a solution containing exactly 100 g of ethylene glycol (C2H6O2) antifreeze in kg of water. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.9 Analyze List the knowns and the unknown. 1 Calculate the number of moles of C2H6O2 and the molality of the solution. Then calculate the freezing-point depression using ΔTf = Kf  m. UNKNOWN KNOWNS mass of C2H6O2 = 100 g mass of water = kg Kf for H2O = 1.86°C/m molar mass C2H6O2 = 62.0 g/mol ΔTf = ?°C Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.9 Calculate Solve for the unknown. 2 Use the molar mass of C2H6O2 to convert the mass of solute to moles. 100 g C2H6O2  = 1.61 mol C2H6O2 62.0 g C2H6O2 1 mol C2H6O2 Calculate the molality of the solution. m = = = 3.22m kg solvent mol solute 0.500 kg 1.61 mol Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.9 Calculate Solve for the unknown. 2 Calculate the freezing-point depression. ΔTf = Kf  m = 1.86°C/m  3.22m = 5.99°C The freezing point of the solution is 0.00oC – 5.99°C = – 5.99°C. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.9 Evaluate Does the result make sense? 3 A 1-molal solution reduces the freezing temperature by 1.86°C. So, a decrease of 5.99°C for an approximately 3-molal solution is reasonable. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating the Boiling Point of a Solution
Sample Problem 16.10 Calculating the Boiling Point of a Solution What is the boiling point of a 1.50m NaCl solution? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.10 Analyze List the knowns and the unknown. 1 Each formula unit of NaCl dissociates into two particles, according to the equation NaCl(s) → Na+(aq) + Cl–(aq). Based on the total number of dissociated particles, the effective molality is 2  1.50m = 3.00m. Calculate the boiling-point elevation (using the equation ΔTb = Kb  m) and then add it to 100°C. UNKNOWN KNOWNS Solution concentration = 1.50m NaCl Kb for H2O = 0.512°C/m boiling point = ?°C Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.10 Calculate Solve for the unknown. 2 Calculate the boiling-point elevation, making sure to use the molality of total dissociated particles in solution. ΔTb = Kb  m = 0.512°C/m  3.00m = 1.54°C Calculate the boiling point of the solution. Tb = 100°C °C = °C Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 16.10 Evaluate Does the result make sense? 3 The boiling point increases about 0.5°C for each mole of solute particles. The total change is reasonable. Because the boiling point of water is defined as exactly 100°C, this value does not limit the number of significant figures in the solution of the problem. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU Does pasta cook at 100°C? After you read Sample Problem 16.10, calculate the boiling-point elevation for 2 L of water containing a teaspoon of salt (about 20 g). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU Does pasta cook at 100°C? After you read Sample Problem 16.10, calculate the boiling-point elevation for 2 L of water containing a teaspoon of salt (about 20 g). 20 g NaCl   = 0.7 mol solute 58.5 g NaCl 1 mol NaCl 2 mol solute m = = = 0.4m kg solvent mol solute 2 kg 0.7 mol ΔTb = Kb  m = 0.512°C/m  0.4m = 0.2°C Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How does the freezing-point depression by CaCl2 for a given solvent compare to the freezing-point depression caused by the same molal concentration of ethylene glycol (C2H6O2,)? A. It is exactly the same. B. It is twice as large. C. It is three times as large. D. It is four times as large. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts Chemists use two ways to express the ratio of solute to solvent: in molality and in mole fractions. The magnitudes of freezing-point depression and boiling-point elevation are proportional to molality. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Mole fractions: XA = XB = nA + nB nA nB
Key Equations Molality (m) = moles of solute kilograms of solvent Mole fractions: XA = XB = nA + nB nA nB ΔTf = Kf  m ΔTb = Kb  m Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms molality (m): the concentration of solute in a solution expressed as the number of moles of solute dissolved in 1 kilogram (1000 g) of solvent mole fraction: the ratio of the moles of solute in solution to the total number of moles of both solvent and solute molal freezing-point depression constant (Kf): the change in freezing point for a 1-molal solution of a nonvolatile molecular solute molal boiling-point elevation constant (Kb): the change in boiling point for a 1-molal solution of a nonvolatile molecular solute Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The Mole and Quantifying Matter
BIG IDEA The Mole and Quantifying Matter Solution concentration can be quantified in terms of molarity (moles of solute per liter of solution), molality (moles of solute per kilogram of solvent), percent by volume, and percent by mass. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 16 Solutions 16.1 Properties of Solutions

Similar presentations

Presentation on theme: "Chapter 16 Solutions 16.1 Properties of Solutions"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Chapter 16 Solutions 16.1 Properties of Solutions

Similar presentations

Presentation on theme: "Chapter 16 Solutions 16.1 Properties of Solutions"— Presentation transcript:

Similar presentations

About project

Feedback