2. 16.2 Concentrations of Solutions > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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

3. 16.2 Concentrations of Solutions > 2 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.

4. 16.2 Concentrations of Solutions > 3 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.

5. 16.2 Concentrations of Solutions > 4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.Molarity Procedure for making a 0.5M 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.

6. 16.2 Concentrations of Solutions > 5 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.Molarity Molarity (M) = moles of solute liters of solution

7. 16.2 Concentrations of Solutions > 6 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

8. 16.2 Concentrations of Solutions > 7 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 g/100 mL → mol/100 mL → mol/L

9. 16.2 Concentrations of Solutions > 8 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate Solve for the unknown. 2 Sample Problem 16.2 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

10. 16.2 Concentrations of Solutions > 9 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

11. 16.2 Concentrations of Solutions > 10 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

12. 16.2 Concentrations of Solutions > 11 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate Solve for the unknown. 2 Sample Problem 16.3 1.5 L  = 1.1 mol NaClO 0.70 mol NaCl 1 L

13. 16.2 Concentrations of Solutions > 12 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

14. 16.2 Concentrations of Solutions > 13 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

15. 16.2 Concentrations of Solutions > 14 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Making Dilutions What effect does dilution have on the amount of solute?

16. 16.2 Concentrations of Solutions > 15 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

17. 16.2 Concentrations of Solutions > 16 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Making Dilutions So: 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.

18. 16.2 Concentrations of Solutions > 17 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

19. 16.2 Concentrations of Solutions > 18 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

20. 16.2 Concentrations of Solutions > 19 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.

21. 16.2 Concentrations of Solutions > 20 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Solutions How do percent by volume and percent by mass differ?

22. 16.2 Concentrations of Solutions > 21 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.

23. 16.2 Concentrations of Solutions > 22 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.

24. 16.2 Concentrations of Solutions > 23 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.

25. 16.2 Concentrations of Solutions > 24 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).

26. 16.2 Concentrations of Solutions > 25 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

27. 16.2 Concentrations of Solutions > 26 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

28. 16.2 Concentrations of Solutions > 27 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

29. 16.2 Concentrations of Solutions > 28 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

30. 16.2 Concentrations of Solutions > 29 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)

31. 16.2 Concentrations of Solutions > 30 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Solutions 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).

32. 16.2 Concentrations of Solutions > 31 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Solutions A convenient measure of concentration when the solute is a solid. Information on food labels is expressed as a percent composition. Percent by mass (%(m/m)) =  100% mass of solution mass of solute

33. 16.2 Concentrations of Solutions > 32 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What are three ways to calculate the concentration of a solution? CHEMISTRY & YOU

34. 16.2 Concentrations of Solutions > 33 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What are three ways to calculate the concentration of a solution? CHEMISTRY & YOU 1. Molarity (M), 2. Percent by volume (%(v/v)), 3. Percent by mass (%(m/m)).

35. 16.2 Concentrations of Solutions > 34 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

36. 16.2 Concentrations of Solutions > 35 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)

37. 16.2 Concentrations of Solutions > 36 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.

38. 16.2 Concentrations of Solutions > 37 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

39. 16.2 Concentrations of Solutions > 38 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)?

40. 16.2 Concentrations of Solutions > 39 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 = 0.050  2000 g = 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%