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1 FOCUS Connecting to Your World Cooking instructions for a wide Guide for Reading variety of foods, from dried pasta to packaged beans to frozen fruits to Objectives fresh vegetables, often call for the addition of a small amount of salt to the Key Concepts • What are two ways of 16.4.1 Solve problems related to the cooking water. Most people like the flavor of expressing the concentration food cooked with salt. But adding salt can of a solution? molality and mole fraction of a have another effect on the cooking pro- • How are freezing-point solution cess. Recall that dissolved salt elevates depression and boiling-point elevation related to molality? 16.4.2 Describe how freezing-point the boiling point of water. Suppose you Vocabulary depression and boiling-point added a teaspoon of salt to two liters of molality (m) elevation are related to water. A teaspoon of salt has a mass of mole fraction molality. about 20 g. Would the resulting boiling molal freezing-point depression K point increase be enough to shorten constant ( f) the time required for cooking? In this molal boiling-point elevation Guide to Reading constant (K ) section, you will learn how to calculate the b amount the boiling point of the cooking Reading Strategy Build Vocabulary L2 water would rise. Before you read, make a list of the vocabulary terms above. As you Graphic Organizers Use a chart to read, write the symbols or formulas that apply to each term and organize the definitions and the math- Molality and Mole Fraction describe the symbols or formulas ematical formulas associated with each using words. Recall that colligative properties depend only upon solute concentration. vocabulary term. The unit molality and mole fractions are two additional ways in which chemists express the concentration of a solution. The unit molality (m) is the Reading Strategy L2 number of moles of solute dissolved in 1 kilogram (1000 g) of solvent. Preview Have students preview this Molality is also known as molal concentration. section by reading the key concepts and skimming the headings, visuals, Molality ϭ moles of solute kilogram of solvent and boldfaced materials. Note that 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 solu- 2 tion. In the case of water as the solvent, 1 kg or 1000 g equals a volume of INSTRUCT 0.500 mol 1000 mL, or 1 L. (29.3 g) You can prepare a solution that is 1.00 molal (1m) in glucose, for exam- NaCl ple, by adding 1.00 mol (180 g) of glucose to 1000 g of water. Figure 16.17 shows how a 0.500 molal (0.500m) solution in sodium chloride is prepared Ask, What effect does adding salt by dissolving 0.500 mol (29.3 g) of NaCl in 1.000 kg (1000 g) of water. have on the cooking process? (Add- ing salt increases the boiling point of the cooking water.) Ask, Do you think the resulting boiling point increase m Figure 16.17 To make a 0.500 solution 1.000 kg would be enough to significantly of NaCl, use a balance to measure 1.000 kg H2O of water and add 0.500 mol (29.3 g) NaCl. shorten the time required for Calculating What would be the molality cooking? (Students are likely to predict, if only 0.500 kg of water were used? correctly, that the change would be neg- ligible.) Section 16.4 Calculations Involving Colligative Properties 491 Molality and Mole Fraction Use Visuals L1 Section Resources Figure 16.17 Display the figure on an Print Technology overhead projector. Ask students to •Guided Reading and Study Workbook, •Interactive Textbook with ChemASAP, write the definition of molality in their Section 16.4 Problem-Solving 16.29, 16.31, 16.33, notebooks. Show the step-by-step pro- •Core Teaching Resources, Section 16.3 16.36; Simulation 21; Assessment 16.4 cedure a chemist would use to prepare Review a 0.500m solution of NaCl. Have stu- •Transparencies, T177–T179 dents confirm your calculations and the •Laboratory Manual, Lab 33 data given in the figure.

Answers to... Figure 16.17 1.00m

Solutions 491 4.00 mol Total moles A B 5.25 mol H O Section 16.4 (continued) Figure 16.18 Ethylene 1.25 mol 2 EG A 1.25 glycol (EG) is added to B Mole fraction EG water as antifreeze in the A B 5.25 A proportions shown. A mole Sample Problem 16.6 B fraction is the ratio of the 4.00 Mole fraction H2O Answers number of moles of one A B 5.25 substance to the total 29. 750 g water × 0.400 mol NaF / number of moles of all 1000 g water × 42.0 g NaF / substances in the solution. SAMPLE PROBLEM 16.6 × 1 Inferring What is the sum 1 mol NaF = 1.26 10 g NaF of all mole fractions in a Using Solution Molality × solution? 30. 10.0 g NaCl / 600 g water 1 mol How many grams of potassium iodide must be dissolved in 500.0 g of NaCl / 58.5 g NaCl × 1000 g water/ water to produce a 0.060 molal KI solution? 1 kg water = 2.85 × 10−1m NaCl Analyze List the knowns and the unknown. Practice Problems Plus L2 Knowns Unknown How many grams of lithium bromide • mass of water 500.0g 0.5000 kg • mass of solute ? g KI • solution concentration 0.060m must be dissolved in 444 g of water • molar mass KI 166.0 g/mol to prepare a 0.140m LiBr solution? According to the definition of molal, the final solution must contain (5.40 g LiBr) 0.060 mol KI per 1000 g H2O. Use the molality as a conversion factor to convert from mass of water to moles of the solute (KI). Then use the Discuss L2 molar mass of KI to convert from mol KI to g KI. The steps are ¡ ¡ Write the expressions defining molar- mass of H2O mol KI g KI. ity and molality on the board. Compare Calculate Solve for the unknown. the chemical quantities in each expres- Math Handbook 0.060 mol KI 166.0 g KI sion. Point out that molarity is denoted 0.5000 kg H2O 5.0 g KI For help with dimensional 1.000 kg H2O 1 mol KI by M and molality by m. Explain that analysis, go to page R66. the molality of a solution does not vary Evaluate Does the result make sense? with temperature because the mass of 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 the solvent does not change. In con- 20 of that value, so the mass of KI should be much less than the molar trast, the molarity of a solution does mass. The answer is correctly expressed to two significant figures. vary with temperature because the liquid can expand and contract. When Practice Problems studying colligative properties such as 29. How many grams of sodium 30. Calculate the molality of a boiling-point elevation and freezing- Problem-Solving 16.29 Solve fluoride are needed to prepare solution prepared by dissolv- point depression, it is preferable to use Problem 29 with the help of an a 0.400m NaF solution that ing 10.0 g NaCl in 600 g of interactive guided tutorial. a concentration that does not depend contains 750 g of water? water. withChemASAP on temperature. The concentration of a solution also can be expressed as a mole fraction, as shown in Figure 16.18. The mole fraction of a solute in a solution is Math Handbook the ratio of the moles of that solute to the total number of moles of solvent n n and solute. In a solution containing A mol of solute A and B mol of solvent For a math refresher and practice, X X B, the mole fraction of solute A ( A) and the mole fraction of solvent B ( B) direct students to dimensional can be expressed as follows. analysis, page R66. n n X A X B A n + n B n + n A B A B

Checkpoint What is a mole fraction?

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Differentiated Instruction Facts and Figures Less Proﬁcient Readers L1 Mole Fraction Uses Have students write sentences using the Explain that the mole fraction compares the words molarity and molality, and have number of moles of a solute to the total them circle the letter in each word that number of moles in the solution. Organic makes them distinct. The mnemonic, ‘r’ for chemists, who frequently work with non- molarity and liter may help. aqueous solvent systems, often use this method of expressing concentration. The mole fraction is also used when calculating the vapor pressure of a solution.

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SAMPLE PROBLEM 16.7 Figure 16.19 The concentration of Sample Problem 16.7 antifreeze used in an automobile cooling system can be described Calculating Mole Fractions by mole fractions. Answers Ethylene glycol (C H O ) is added to automobile cooling systems to 31. X = 0.190; X = 0.810 2 6 2 C2H5OH H2O protect against cold weather. What is the mole fraction of each compo- 32. X = 0.437; X = 0.563 nent in a solution containing 1.25 mol of ethylene glycol (EG) and CCl4 CHCl3 4.00 mol of water? Practice Problems Plus L2 Analyze List the knowns and the unknowns. Calculate the mole fraction of each Knowns component in a solution of 42 g • moles of ethylene glycol (nEG) 1.25 mol EG • moles of water 1n 2 4.00 mol H O CH3OH, 35 g C2H5OH, and 50 g H2O 2 Unknowns C3H7OH. (CH3OH = 0.45; C2H5OH = • mole fraction EG (X ) ? EG 0.26; C3H7OH = 0.29) • mole fraction H O 1X 2 ? 2 H2O

The mole fraction of ethylene glycol (XEG) in the solution is the number of moles of ethylene glycol divided by the total number of moles in Math Handbook the solution: n For a math refresher and practice, EG XEG n + n EG H2O direct students to using a calculator, Similarly, the mole fraction of water 1X 2 in the solution is the number page R62. H2O of moles of water divided by the total number of moles in the solution:

nH O X 2 H2O n + n EG H2O

Calculate Solve for the unknowns. CLASS Activity Math Handbook n X EG 1.25 mol 0.238 EG n + n + For help with using a Diagramming Methods of EG H2O 1.25 mol 4.00 mol Practice Problems calculator, go to page R62. L3 nH O Concentration Calculation X 2 4.00 mol 0.762 H2O n + n + EG H2O 1.25 mol 4.00 mol Purpose Students organize various methods used to calculate the Evaluate Does the result make sense? concentration of a solution. The mole fraction is a dimensionless quantity. The sum of the mole Procedure Have students design a fractions of all the components in a solution must equal 1. Note that diagram that includes deﬁnition of XEG X 1.000. Each answer is correctly expressed to three signif- H2O icant ﬁgures. each type of concentration unit. More than one design is possible, but each Practice Problems should include the terms concentration 31. What is the mole fraction 32. A solution contains 50.0 g of of a solution, percent by mass, percent by

of each component in a carbon tetrachloride (CCl4) volume, molarity, molality, and mole solution made by mixing and 50.0 g of chloroform fraction. Students may wish to include 300 g of ethanol (C H OH) (CHCl ). Calculate the mole Problem-Solving 16.32 Solve 2 5 3 additional detail by showing intercon- and 500 g of water? fraction of each component Problem 32 with the help of an interactive guided tutorial. mass of solute in the solution. versions between and withChemASAP moles of solute. Underneath each concentration unit, students should provide examples of when the unit would be used in the laboratory and in everyday life.

Section 16.4 Calculations Involving Colligative Properties 493

Answers to... Figure 16.18 one

Checkpoint Mole fraction is the ratio of the number of moles of one component of a mixture to the total number of moles of all the components of the mixture.

Solutions 493 Section 16.4 (continued) Table 16.2 Freezing-Point Depression Kf Values for Some and Boiling-Point Elevation Freezing-Point Common Solvents The graph in Figure 16.20 shows that the freezing point of a solvent is low- Solvent K (C/m) Depression and Boiling- f ered and its boiling point is raised by the addition of a nonvolatile solute. Water 1.86 ∆ The magnitudes of the freezing-point depression ( Tf) and the boiling- Point Elevation Acetic acid 3.90 ∆ point elevation ( Tb) of a solution are directly proportional to the molal con- Benzene 5.12 centration (m), when the solute is molecular, not ionic. Interpreting Graphs L2 Nitrobenzene 7.00 ∆T ∝ m ∆T ∝ m a. 0°C; 100°C Phenol 7.40 f b The change in the freezing temperature (∆T ) is the difference between the b. The freezing point of the solution is Cyclohexane 20.2 f Camphor 37.7 freezing point of the solution and the freezing point of the pure solvent. Sim- lower than that of pure water and the ∆ ilarly, the change in the boiling temperature ( Tb) is the difference between boiling point is higher than that of the boiling point of the solution and the boiling point of the pure solvent. pure water. The term m is the molal concentration of the solution. c. Adding a solute to water allows it to With the addition of a constant, the proportionality between the freezing point depression (∆T ) and the molality m can be expressed as an equation. remain as a liquid over a longer tem- f ∆T K m perature range because the solution f f molal freezing-point depression constant, changes to a solid at a lower tempera- The constant, Kf, is the which is ture and changes to a vapor at a higher equal to the change in freezing point for a 1-molal solution of a nonvolatile molecular solute. The value of Kf depends upon the solvent. Its units are temperature. Simulation 21 Discover ° the principle underlying the C/m. Table 16.2 lists the Kf values for water and some other solvents. Enrichment Question L3 colligative properties of The boiling-point elevation of a solution can also be expressed as an solutions. equation. Which aqueous solution would have withChemASAP ∆T K m the larger boiling point elevation b b The constant, K , is the molal boiling-point elevation constant, which is and freezing point depression: b equal to the change in boiling point for a 1-molal solution of a nonvolatile 1m KCl or 1m CaBr ? (CaBr ) 2 2 molecular solute. The value of Kb depends upon the solvent. Its units are °C/m. Table 16.3 lists the K values for water and some other solvents. Use Visuals L2 b Figure 16.20 Display Figure 16.20 on Figure 16.20 The graph Vapor Pressure versus Temperature an overhead projector. Choose an arbi- shows the relationship between vapor pressure and T T trary concentration for an aqueous temperature for pure water 101 f b solution of NaCl or ethylene glycol. Cal- and aqueous solutions. Liquid culate the boiling-point elevation and freezing-point depression. Then write INTERPRETING GRAPHS Vapor pressure the temperatures on the horizontal a. Identify What is the of pure water axis. Show students how to read the freezing point of water? What is the boiling point? data for the boiling points of the pure b. Describe How do the solvent and of the solution. Then have freezing and boiling points students read the corresponding val- of the solution compare to Solid Vapor pressure those of pure water? of solution ues for the freezing points. Ask, What c. Apply Concepts Does Vapor Boiling point ∆ ∆ adding a solute to water are Tb and Tf for this solution? Pressure (kPa) Vapor of water allow it to remain as a liquid (Point out that the solute affects both the Freezing point of solution Boiling point over a longer or shorter Freezing point of water of solution freezing point and boiling point of a liq- temperature range? Explain. uid.) Why would knowing the boil- 0 100 ing-point elevation and freezing- Temperature (C) point depression be important when choosing antifreeze for car radia- 494 Chapter 16 tors? (Despite its name, antifreeze protects against both freezing and overheating.)

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Sample Problem 16.8 shows how these equations can be used for cal- Table 16.3 Sample Problem 16.8 ∆T ∆T culating f and b of solutions if the solute is a molecular compound. For ionic compounds, both the freezing-point depression and the boiling- Kb Values for Some Answers Common Solvents point elevation depend upon the number of ions produced by each for- 33. mol C H O = 10.0 g glucose Solvent K (C/m) 6 12 6 × mula unit. This number is used to calculate an effective molality, as shown b m in Sample Problem 16.9. Water 0.512 1 mol / 180.2 g = 0.0555 mol; = Ethanol 1.19 0.0555 mol / 50.0 g H2O × 1000g / Benzene 2.53 1 kg = 1.11m; T = K m = SAMPLE PROBLEM 16.8 ∆ f f × Cyclohexane 2.79 1.86°C/m × 1.11m = 2.06°C Acetic acid 3.07 Calculating the Freezing-Point Depression of a Solution 34. mol C H O = 200 g C H O × 1 mol/ Phenol 3.56 3 6 3 6 58.0 g C H O = 3.45 mol C H O; Antifreeze protects a car from freezing. It also protects it from over- Nitrobenzene 5.24 3 6 3 6 heating. Calculate the freezing-point depression and the freezing point Camphor 5.95 m = 3.45 mol C3H6O/400 g × of a solution containing 100 g of ethylene glycol (C H O ) antifreeze in 2 6 2 1000 g/1 kg = 8.63m; Kf × m = 0.500 kg of water. 5.12°C/m × 8.63m = 44.2°C Analyze List the knowns and the unknowns. Knowns Unknown ∆T ° Math • mass of solute 100 g C2H6O2 • f ? C Handbook ° • mass of solvent 0.500 kg H2O •freezing point ? C K ° m For a math refresher and practice, • f for H2O 1.86 C/ ∆T K m direct students to algebraic • f f Calculate the number of moles of solute and the molality. Then calcu- equations, page R69. late the freezing-point depression and freezing point.

Calculate Solve for the unknown. L2 Math Handbook Discuss moles C H O 100 g C H O 1 mol 1.61 mol Some students may think that freezing 2 6 2 2 6 2 62.0 g C H O For help with algebraic 2 6 2 equations, go to page R69. points and boiling points can be depressed or elevated without end. m mol solute 1.61 mol 3.22m kg solvent 0.500 kg Explain that as the concentration of a solute, such as ethylene glycol, increases, ∆T K m 1.86°C/m 3.22m 5.99°C f f there comes a point when the quantity ° ° ° The freezing point of the solution is 0.00 C 5.99 C 5.99 C. of solute exceeds the quantity of solvent. Evaluate Does the result make sense? Ethylene glycol becomes the solvent and APr 1-molalactice Problemssolution reduces the freezing temperature by 1.86°C, so a water becomes the solute. The trends in decrease of 5.99°C for an approximately 3-molal solution is reason- colligative properties begin to reflect able. The answer is correctly expressed with three significant figures. ethylene glycol instead of water. If the Practice Problems solute is a solid, such as NaCl, eventually the solution becomes saturated. But 33. What is the freezing point 34. Calculate the freezing-point even before this saturation point, the depression of an aqueous depression of a benzene solution of 10.0 g of glucose solution containing 400 g of magnitude of certain colligative proper-

(C6H12O6) in 50.0 g H2O? benzene and 200 g of the ties may reach a maximum. molecular compound acetone Problem–Solving 16.33 Solve (C H O). K for benzene is Problem 33 with the help of an 3 6 f interactive guided tutorial. 3 5.12°C/m. ASSESS withChemASAP Evaluate Understanding L2 Ask, Which solution has a higher boiling point, 1 mol of Al(NO ) in Section 16.4 Calculations Involving Colligative Properties 495 3 3 1000 g of water or 1.5 mol of KCl in 1000 g of water? Have students explain their answers. (The solution of Al(NO3)3 has a higher boiling point because Al(NO3)3 dissociates into a larger number of particles.) Why is it important to distinguish between nonvolatile and volatile compounds when discussing certain colligative properties? (Volatile solutes would quickly evaporate at higher temperatures, which would change the molal concentration of the solution.)

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Section 16.4 (continued) SAMPLE PROBLEM 16.9

Calculating the Boiling Point of a Solution Sample Problem 16.9 What is the boiling point of a 1.50m NaCl solution? Answers Analyze List the knowns and the unknown. × Knowns Unknown 35. m = 1.25 mol CaCl2 / 1400 g • concentration 1.50m NaCl • boiling point ?°C m K ° m 1000 g / 1 kg = 0.893 CaCl2; Each • b for H2O 0.512 C/ ΔT K × m formula unit of CaCl2 dissociates • b b + − into three particles; 0.893m CaCl2 Each formula unit of NaCl dissociates into two particles, Na and Cl . × Δ × Based on the total number of dissociated particles, the effective molal- 3 = 2.68m ; Tb = Kb m = 0.512°C /m × 2.68m = 1.37°C; ity is 2 × 1.50m 3.00m. Calculate the boiling-point elevation and ° 100°C + 1.37°C = 101.37°C then add it to 100 C. ° ° Calculate Solve for the unknown. 36. 2.00 C / 0.512 C/m = 3.91m; Each Math Handbook formula unit of NaCl dissociates 0.512C For help with significant DT K m 3.00m 1.54C into two particles; 3.91m / 2 = b b m figures, go to page R59. ° ° ° 1.96m; In 1 kg of water, 1.96 mol The boiling point of the solution is 100 C 1.54 C 101.54 C NaCl × 58.5 g / 1 mol = 115 g NaCl Evaluate Does the result make sense? ° Practice Problems Plus L2 The boiling point increases about 0.5 C for each mole of solute particles, so the total change is reasonable. Because the boiling point of What is the boiling point of a solu- water is defined as exactly 100°C, this value does not limit the number tion containing 96.7 g of sucrose of significant figures in the solution of the problem. (C H O ) in 250.0 g water at 12 22 11 Practice Problems 1 atm? (100.579°C) 35. What is the boiling point of 36. What mass of NaCl would Problem-Solving 16.36 Solve Practicea solution Problems that contains have to be dissolved in Problem 36 with the help of an 1.25 mol CaCl in 1400 g of 1.000 kg of water to raise the Math interactive guided tutorial. 2 Handbook water? boiling point by 2.00°C? For a math refresher and practice, withChemASAP direct students to significant figures, page R59. 16.4 Section Assessment

37. Key Concept What are two ways of expressing Reteach L1 the ratio of solute particles to solvent particles? Handbook Remind students that when they do cal- 38. Key Concept How are freezing-point depres- Element Distribution Look at the table on page culations involving ionic solids, they sion and boiling-point elevation related to R4 of the Elements Handbook showing the distribu- molality? tion of elements in the oceans. What generalization must ﬁnd the molality of the solution in can you make about the temperature at which 39. How many grams of sodium bromide must be ocean water will freeze? What effect does the pres- terms of the total number of particles. dissolved in 400.0 g of water to produce a Work through several examples to be ence of dissolved elements in the ocean have on 0.500 molal solution? the rate of evaporation of ocean water? sure they understand this important 40. Calculate the mole fraction of each component in point. a solution of 2.50 mol ethanoic acid (CH3COOH) in 10.00 mol of water. 41. What is the freezing point of a solution of 12.0 g of Elements Handbook Assessment 16.4 Test yourself CCl4 dissolved in 750.0 g of benzene? The freezing on the concepts in Section 16.4. point of benzene is 5.48°C; K is 5.12°C/m. Given the list of dissolved elements f withChemASAP and their ppm, students should infer 496 Chapter 16 that the freezing point of ocean water is less than 0°C at 1 atm and that the rate of evaporation of ocean waters is slower because of lowered vapor pressure. Section 16.4 Assessment 37. Molality and mole fractions are two con- proportional to the molal concentration venient ways of expressing the ratio of (m), when the solute is molecular, not solute particles to solvent particles. ionic. 38. The magnitudes of the freezing-point 39. 20.6 g NaBr depression (ΔT ) and the boiling-point 40. X = 0.200 X = 0.800 If your class subscribes to the f CH3COOH H2O elevation (ΔT ) of a solution are directly 41. 4.95°C Interactive Textbook, use it to b review key concepts in Section 16.4.

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Small-Scale LAB Small-Scale LAB Making a Solution

Purpose Making a Solution L2 To make a solution and use carefully measured data to Objective calculate the solution's concentration. After completing this activity, students Materials will be able to: solid NaCl • •make a solution water • •calculate the solution’s molarity, • 50-mL volumetric flask molality, percent by mass, and mole • balance fraction Procedure Measure the mass of a clean, dry, volumetric flask. Add 4. Molarity (M) tells how many moles of solute are Prep Time 20 minutes enough solid NaCl to approximately fill one-tenth of the dissolved in 1 L of solution. Class Time 20 minutes volume of the flask. Measure the mass of the flask again. M ϭ mol NaCl Half fill the flask with water and shake it gently until all the L solution Safety Wear safety goggles and a lab NaCl dissolves. Fill the flask with water to the 50-mL mark apron. a. Calculate the liters of solution. 1000 mL = 1 L and measure the mass again. b. Calculate the molarity of the NaCl solution. Teaching Tips Analyze 5. Density tells how many grams of solution are present in Household table salt contains a small Using your experimental data, record the answers to the 1 mL of solution. following questions below your data table. amount of desiccant, such as sodium g solution Density ϭ silicate, which gives the salt solution an 1. Percent by mass tells how many grams of solute are mL solution present in 100 g of solution. opaque appearance. Use lab grade Calculate the density of the solution. NaCl instead. Remind students about %%by mass ϭ mass of solute ϫ 100 You’re The Chemist mass of solute + solvent significant figures. Point out that the The following small-scale activities allow you to develop balance is probably more accurate your own procedures and analyze the results. a. Calculate the mass of the solute (NaCl). than the volumetric flask. Calculations b. Calculate the mass of the solvent (water). 1. Analyze It Use a small-scale pipet to extract a sample of your NaCl solution and deliver it to a massed empty based only on measurements using c. Calculate the percent by mass of NaCl in the volumetric flask. Measure the mass of the flask and fill the balance will have a larger number solution. it with water to the 50-mL line. Measure the mass of the of significant figures. 2. Mole fraction tells how many moles of solute are flask again. Calculate the concentration of this dilute present for every 1 mol of total solution. solution using the same units you used to calculate the Expected Outcome Students make a solution and calculate the solution’s Mole fraction ϭ mol NaCl concentration of the NaCl solution. Are the results you mol NaCl + mol H O 2 obtained reasonable? concentration in various units. a. Calculate the moles of NaCl solute. 2. Design It! Design and carry out an experiment to Molar mass NaCl = 58.5 g/mol make a solution of table sugar quantitatively. Calculate Analyze b. Calculate the moles of water. the concentration of the table sugar solution using the Sample data: same units you used to calculate the concentration of Molar mass H2O = 18.0 g/mol dry bottle = 15.98 g the NaCl solution. Is the effective molality of the table c. Calculate the mole fraction of your solution. flask + NaCl = 22.88 g sugar solution the same as the effective molality of a 3. Molality (m) tells how many moles of solute are present sodium chloride solution of the same concentration? flask + NaCl + water = 69.09 g in 1 kg of solvent. Recall that effective molality is the concentration value 1. a. 6.90 g b. 46.21 g c. 13.0% m ϭ mol NaCl used to calculate boiling-point elevation and freezing- kg H O 2. a. 0.118 mol NaCl b. 2.57 mol H2O 2 point depression. Calculate the molality of your solution. c. 0.0439 3. 2.55m 4. a. 0.050 L b. 2.4M 5. 1.1 g/mL Small-Scale Lab 497 You’re the Chemist 1. Sample data: dry flask = 15.98 g 2. Sample data: molarity = 0.242M flask + NaCl solution = 22.88 g mass of dry flask = 16.72 g density = 1.1 g/mL flask + NaCl solution + water = 69.09 g mass of flask + sugar = 20.85 g No, the effective molaliy of the NaCl solution mass of NaCl solution = 2.09 g mass of flask + sugar + water = 69.53 g is twice the effective molality of the sugar mass of water = 47.77 g mass of sugar = 4.13 g solution. mass of NaCl = 0.271 g mass of solvent = 48.68 g percent mass of NaCl = 0.544% L3 –3 percent mass of sugar = 7.82% For Enrichment moles NaCl = 4.64 × 10 mol moles of sugar = 0.0121 mol Students could compare the freezing point mass of water = 49.59 g moles of water = 2.70 mol depressions of the sugar and salt solutions. moles of water = 2.76 mol –3 mole fraction = 4.46 × 10–3 (One way to do this is to partially freeze the solu- mole fraction = 1.68 × 10 molality = 0.249m tions and determine the temperature of the mix- molality = 0.0936m tures of liquid and solid.) density = 1.0 g mL Solutions 497