Solutions The federal government and state governments set standards limiting the amount of contaminants allowed in drinking water.

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1 Solutions The federal government and state governments set standards limiting the amount of contaminants allowed in drinking water Properties of Solutions 16. Concentrations of Solutions 16. Colligative Properties of Solutions 16.4 Calculations Involving Colligative Properties Molarity & YOU How can you describe the concentration of a solution? Chapter 16 1 CHEMISTRY Molarity Molarity The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. How do you calculate the molarity of a solution? A solution that contains a relatively small amount of solute is a dilute solution. A concentrated solution contains a large amount of solute. Molarity 4 Molarity The figure below illustrates the procedure for making a 0.5M, or 0.5-molar, solution. 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. Add 0.5 mol of solute to a 1-L volumetric flask half filled with distilled water. 5 6 Swirl the flask carefully to dissolve the solute. Fill the flask with water exactly to the 1-L mark. 1

2 Molarity Sample Problem 16. 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 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? 7 8 Sample Problem 16. Sample Problem Analyze List the knowns and the unknown. 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 9 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 1 L = 1000 ml gives you the conversion factor 1000 ml/1 L. Solution 0.90 g NaCl 1 mol NaCl 1000 ml concentration = 100 ml 58.5 g NaCl 1 L = 0.15 mol/l = 0.15M 10 Sample Problem 16. Sample Problem 16. Evaluate Does the result make sense? 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. 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? 11 1

3 Sample Problem 16. Sample Problem Analyze List the knowns and the unknown. 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 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 mol NaCl 1.5 L = 1.1 mol NaClO 1 L 1 14 Sample Problem 16. Evaluate Does the result make sense? 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 L. The answer is correctly expressed to two significant figures. 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 Making Dilutions 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 Making Dilutions What effect does dilution have on the amount of solute? 17 18

4 Making Dilutions Making Dilutions Both of these solutions contain the same amount of solute. 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. 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. 19 Moles of solute Moles of solute before dilution = after dilution Making Dilutions 0 Making Dilutions Moles of solute Moles of solute before dilution = after dilution The total number of moles of solute remains unchanged upon dilution. The definition of molarity can be rearranged to solve for moles of solute. Moles of solute = M1 V1 = M V moles of solute Molarity (M) = liters of solution (V) M1 and V1 are the molarity and the volume of the initial solution. Moles of solute = molarity (M) liters of solution (V) 1 Making Dilutions M and V are the molarity and volume of the diluted solution. The student is preparing 100 ml of 0.40M MgSO4 from a stock solution of.0m MgSO4. Sample Problem 16.4 Preparing a Dilute Solution How many milliliters of aqueous.00m MgSO4 solution must be diluted with water to prepare ml of aqueous 0.400M MgSO4? She measures 0 ml of the stock solution with a 0mL pipet. She transfers the 0 ml to a 100mL volumetric flask. She carefully adds water to the mark to make 100 ml of solution. 4 4

5 Sample Problem 16.4 Sample Problem Analyze List the knowns and the unknown. Use the equation M 1 V 1 = M V to solve for the unknown initial volume of solution (V 1 ) that is diluted with water. KNOWNS M 1 =.00M MgSO 4 M = 0.400M MgSO 4 V = ml of 0.400M MgSO 4 UNKNOWN V 1 =? ml of.00m MgSO 4 5 Solve for V 1 and substitute the known values into the equation. M V 0.400M ml V 1 = = = 0.0 ml.00m M 1 Thus, 0.0 ml of the initial solution must be diluted by adding enough water to increase the volume to ml. 6 Sample Problem 16.4 Evaluate Does the result make sense? 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 ml of a 0.00M CuSO 4 5H O solution is diluted to ml. What is the concentration of the diluted solution? ml of a 0.00M CuSO 4 5H O solution is diluted to ml. What is the concentration of the diluted solution? How do percent by volume and percent by mass differ? M 1 V 1 = M V M 1 V 1 M = = V M = M 0.00M ml ml 9 0 5

6 Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. Isopropyl alcohol (-propanol) is sold as a 91-percent solution by volume. 1 Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. Isopropyl alcohol (-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. Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. Isopropyl alcohol (-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). 4 Sample Problem 16.5 Calculating Percent by Volume Percent by volume of a solution is the ratio of the volume of solute to the volume of solution. volume of solute Percent by volume (%(v/v)) = 100% volume of solution What is the percent by volume of ethanol (C H 6 O, or ethyl alcohol) in the final solution when 85 ml of ethanol is diluted to a volume of 50 ml with water? 5 6 6

7 Sample Problem 16.5 Sample Problem Analyze List the knowns and the unknown. 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 = 50 ml State the equation for percent by volume. volume of solute Percent by volume (%(v/v)) = 100% volume of solution UNKNOWN Percent by volume =?% ethanol (v/v) 7 8 Sample Problem 16.5 Sample Problem 16.5 Substitute the known values into the equation and solve. 85 ml ethanol Percent by volume (%(v/v)) = 100% 50 ml = 4% ethanol (v/v) Evaluate Does the result make sense? 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 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. mass of solute Percent by mass (%(m/m)) = 100% mass of solution mass of solute Percent by mass (%(m/m)) = 100% mass of solution 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

8 CHEMISTRY & YOU CHEMISTRY & YOU What are three ways to calculate the concentration of a solution? 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)) Sample Problem 16.6 Sample Problem 16.6 Using Percent by Mass as a Conversion Factor How many grams of glucose (C 6 H 1 O 6 ) are needed to make 000 g of a.8% glucose (m/m) solution? 45 1 Analyze List the knowns and the unknown. The conversion is mass of solution mass of solute. In a.8% C 6 H 1 O 6 (m/m) solution, each 100 g of solution contains.8 g of glucose. Used as a conversion factor, the concentration allows you to convert g of solution to g of C 6 H 1 O 6. KNOWNS mass of solution = 000 g percent by mass =.8% C 6 H 1 O 6 (m/m) UNKNOWN mass of solute =? g C 6 H 1 O 6 46 Sample Problem 16.6 Sample Problem 16.6 Write percent by mass as a conversion factor with g C 6 H 1 O 6 in the numerator. Multiply the mass of the solution by the conversion factor. You can solve this.8 g C 6 H 1 O 6 problem by using 100 g solution either dimensional analysis or algebra..8 g C 6 H 1 O g solution = 56 g C 6 H 1 O 100 g solution

9 Sample Problem 16.6 Evaluate Does the result make sense? The prepared mass of the solution is g. Since a 100-g sample of.8% (m/m) solution contains.8 g of solute, you need 0.8 g = 56 g of solute. To make the solution, mix 56 g of C 6 H 1 O 6 with 1944 g of solvent. 56 g of solute g solvent = 000 g of solution What is the mass of water in a 000 g glucose (C 6 H 1 O 6 ) solution that is labeled 5.0% (m/m)? Key Concepts What is the mass of water in a 000 g glucose (C 6 H 1 O 6 ) solution that is labeled 5.0% (m/m)? mass of glucose % (m/m) = 100% mass of solution mass of glucose = (% (m/m)) mass of solution 100% mass of glucose = 000 g = 100 g C 6 H 1 O 6 mass of water = 000 g 100 g = 1900 g H O 51 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. 5 Key Equations Glossary Terms Molarity (M) = moles of solute liters of solution concentration: a measurement of the amount of solute that is dissolved in a given quantity of solvent; usually expressed as mol/l M 1 V 1 = M V volume of solute Percent by volume = 100% volume of solution mass of solute Percent by mass = 100% mass of solution 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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