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15 Solutions Section 2 Describing Solution Composition

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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. Molarity 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. 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? Sample Problem 16.2 Calculating Molarity

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

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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. Calculate Solve for the unknown. 2 Sample Problem 16.2 The relationship 1 L = 1000 mL gives you the conversion factor 1000 mL/1 L. Solution concentration =   = 0.15 mol/L = 0.15M 0.90 g NaCl 1 mol NaCl 1000 mL 100 mL 58.5 g NaCl 1 L

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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. Evaluate Does the result make sense? 3 Sample Problem 16.2

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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? Sample Problem 16.3 Calculating the Moles of Solute in a Solution

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. KNOWNS UNKNOWN moles solute = ? mol Analyze List the knowns and the unknown. 1 Sample Problem 16.3 volume of solution = 1.5 L solution concentration = 0.70M NaClO 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.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Multiply the given volume by the molarity expressed in mol/L. Calculate Solve for the unknown. 2 Sample Problem 16.3 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. 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. Evaluate Does the result make sense? 3 Sample Problem 16.3

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

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

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

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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. 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. Making Dilutions 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) Moles of solute before dilution = Moles of solute after dilution

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Making Dilutions The total number of moles of solute remains unchanged upon dilution. Moles of solute = M 1  V 1 = M 2  V 2 M 1 and V 1 are the molarity and the volume of the initial solution. M 2 and V 2 are the molarity and volume of the diluted solution.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Making Dilutions 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. The student is preparing 100 mL of 0.40M MgSO 4 from a stock solution of 2.0M MgSO 4.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. How many milliliters of aqueous 2.00M MgSO 4 solution must be diluted with water to prepare 100.0 mL of aqueous 0.400M MgSO 4 ? Sample Problem 16.4 Preparing a Dilute Solution

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. KNOWNS UNKNOWN V 1 = ? mL of 2.00M MgSO 4 Analyze List the knowns and the unknown. 1 Sample Problem 16.4 M 1 = 2.00M MgSO 4 M 2 = 0.400M MgSO 4 V 2 = 100.0 mL of 0.400M MgSO 4 Use the equation M 1  V 1 = M 2  V 2 to solve for the unknown initial volume of solution (V 1 ) that is diluted with water.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Solve for V 1 and substitute the known values into the equation. Calculate Solve for the unknown. 2 Sample Problem 16.4 V 1 = = = 20.0 mL M2  V2M2  V2 M 1 0.400M  100.0 mL 2.00M Thus, 20.0 mL of the initial solution must be diluted by adding enough water to increase the volume to 100.0 mL.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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. Evaluate Does the result make sense? 3 Sample Problem 16.4

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 100.0 mL of a 0.300M CuSO 4 ·5H 2 O solution is diluted to 500.0 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 CuSO 4 ·5H 2 O solution is diluted to 500.0 mL. What is the concentration of the diluted solution? M 1  V 1 = M 2  V 2 M 2 = 0.0600M M 2 = = M1  V1M1  V1 V2V2 0.300M  100.0 mL 500.0 mL

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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.

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. What is the percent by volume of ethanol (C 2 H 6 O, or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? Sample Problem 16.5 Calculating Percent by Volume

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

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

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Substitute the known values into the equation and solve. Calculate Solve for the unknown. 2 Sample Problem 16.5 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. 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. Evaluate Does the result make sense? 3 Sample Problem 16.5

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

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Solutions 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. Percent by mass (%(m/m)) =  100% mass of solution mass of solute

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

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What are three ways to calculate the concentration of a solution? CHEMISTRY & YOU 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. How many grams of glucose (C 6 H 12 O 6 ) are needed to make 2000 g of a 2.8% glucose (m/m) solution? Sample Problem 16.6 Using Percent by Mass as a Conversion Factor

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. KNOWNS UNKNOWN mass of solute = ? g C 6 H 12 O 6 Analyze List the knowns and the unknown. 1 Sample Problem 16.6 mass of solution = 2000 g percent by mass = 2.8% C 6 H 12 O 6 (m/m) The conversion is mass of solution → mass of solute. In a 2.8% C 6 H 12 O 6 (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 C 6 H 12 O 6.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Write percent by mass as a conversion factor with g C 6 H 12 O 6 in the numerator. Calculate Solve for the unknown. 2 Sample Problem 16.6 100 g solution 2.8 g C 6 H 12 O 6 You can solve this problem by using either dimensional analysis or algebra.

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Multiply the mass of the solution by the conversion factor. Calculate Solve for the unknown. 2 Sample Problem 16.6 2000 g solution  = 56 g C 6 H 12 O 6 100 g solution 2.8 g C 6 H 12 O 6

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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 C 6 H 12 O 6 with 1944 g of solvent. 56 g of solute + 1944 g solvent = 2000 g of solution Evaluate Does the result make sense? 3 Sample Problem 16.6

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What is the mass of water in a 2000 g glucose (C 6 H 12 O 6 ) solution that is labeled 5.0% (m/m)?

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What is the mass of water in a 2000 g glucose (C 6 H 12 O 6 ) 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  0.050 = 100 g C 6 H 12 O 6 mass of water = 2000 g – 100 g = 1900 g H 2 O (% (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 Percent by mass =  100% mass of solution mass of solute Percent by volume =  100% volume of solution volume of solute Molarity (M) = moles of solute liters of solution M 1  V 1 = M 2  V 2

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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

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