1. 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
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2. 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.
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3. How do you calculate the molarity of a solution?
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4. 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.
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5. 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.
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6. 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.
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7. 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
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8. 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?
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9. 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
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10. 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
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11. 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.
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12. 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?
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13. 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
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14. 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
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15. 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.
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16. 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
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17. 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
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18. What effect does dilution have on the amount of solute?
Making Dilutions
Making Dilutions
What effect does dilution have on the amount of solute?
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19. 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.
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20. 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
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21. 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)
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22. 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.
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23. 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.
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24. 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?
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25. 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
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26. 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.
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27. 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.
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28. 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?
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29. 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
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30. How do percent by volume and percent by mass differ?
Percent Solutions
Percent Solutions
How do percent by volume and percent by mass differ?
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31. Percent Solutions
Percent by volume of a solution is the ratio of the volume of solute to the volume of solution.
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32. 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.
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33. 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.
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34. 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).
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35. 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
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36. 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?
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37. 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)
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38. 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
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39. 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)
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40. 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.
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41. 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
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42. 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.
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43. What are three ways to calculate the concentration of a solution?
CHEMISTRY & YOU
What are three ways to calculate the concentration of a solution?
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44. 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)).
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45. 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?
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46. 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
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47. 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
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48. 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
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49. 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
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50. What is the mass of water in a 2000 g glucose (C6H12O6) solution that is labeled 5.0% (m/m)?
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51. (% (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%
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52. 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.
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53. 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
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54. 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
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55. END OF 16.2
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