 # Chapter 16 Solutions 16.2 Concentrations of Solutions

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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.

Analyze List the knowns and the unknown.
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.

Evaluate Does the result make sense?
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.

Analyze List the knowns and the unknown.
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.

Calculate Solve for the unknown.
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.

Evaluate Does the result make sense?
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.

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?

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.

Analyze List the knowns and the unknown.
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.

Calculate Solve for the unknown.
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.

Evaluate Does the result make sense?
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 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.

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.

Analyze List the knowns and the unknown.
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.

Calculate Solve for the unknown.
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.

Calculate Solve for the unknown.
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.

Evaluate Does the result make sense?
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?

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.

Analyze List the knowns and the unknown.
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.

Calculate Solve for the unknown.
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.

Calculate Solve for the unknown.
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.

Evaluate Does the result make sense?
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.