Principles of Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro

Principles of Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 12 Solutions Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

Solutions homogeneous mixtures
Composition may vary from one sample to another. appears to be one substance, though really contains multiple materials Most homogeneous materials we encounter are actually solutions. e.g., air and seawater Nature has a tendency toward spontaneous mixing. Generally, uniform mixing is more energetically favorable.

Solutions When table salt is mixed with water, it seems to disappear, or become a liquid—the mixture is homogeneous. The salt is still there, as you can tell from the taste, or simply boiling away the water. Homogeneous mixtures are called solutions. The component of the solution that changes state is called the solute. The component that keeps its state is called the solvent. If both components start in the same state, the major component is the solvent.

Seawater Drinking seawater will dehydrate you and give you diarrhea.
The cell wall acts as a barrier to solute moving. The only way for the seawater and the cell solution to have uniform mixing is for water to flow out of the cells of your intestine and into your digestive tract.

Common Types of Solution
Solution Phase Solute Phase Solvent Phase Example gaseous solutions gas air (mostly N2 & O2) liquid solutions liquid solid soda (CO2 in H2O) vodka (C2H5OH in H2O) seawater (NaCl in H2O) solid solutions brass (Zn in Cu) Solutions that contain Hg and some other metal are called amalgams. Solutions that contain metal solutes and a metal solvent are called alloys.

Brass Type Color % Cu % Zn Density g/cm3 MP °C Tensile Strength psi
Uses Gilding reddish 95 5 8.86 1066 50K pre-83 pennies, munitions, plaques Commercial bronze 90 10 8.80 1043 61K doorknobs, grillwork Jewelry 87.5 12.5 8.78 1035 66K costume jewelry Red golden 85 15 8.75 1027 70K electrical sockets, fasteners & eyelets Low deep yellow 80 20 8.67 999 74K musical instruments, clock dials Cartridge yellow 70 30 8.47 954 76K car radiator cores Common 67 33 8.42 940 lamp fixtures, bead chain Muntz metal 60 40 8.39 904 nuts & bolts, brazing rods

Solubility When one substance (solute) dissolves in another (solvent) it is said to be soluble. Salt is soluble in water. Bromine is soluble in methylene chloride. When one substance does not dissolve in another it is said to be insoluble. Oil is insoluble in water. The solubility of one substance in another depends on two factors—nature’s tendency towards mixing, and the types of intermolecular attractive forces.

Mixing and the Solution Process Entropy
Most processes occur because the end result has less potential energy. However, formation of a solution does not necessarily lower the potential energy of the system. When two ideal gases are put into the same container, they spontaneously mix. even though the difference in attractive forces is negligible The gases mix because the energy of the system is lowered through the release of entropy.

Mixing and the Solution Process Entropy
Entropy is the measure of energy dispersal throughout the system. Energy has a spontaneous drive to spread out over as large a volume as it is allowed. By each gas expanding to fill the container, it spreads its energy out and lowers its entropy.

Intermolecular Forces and the Solution Process
Energy changes in the formation of most solutions also involve differences in attractive forces between the particles. In order for the solvent and solute to mix, you must overcome all of the solute–solute attractive forces. some of the solvent–solvent attractive forces. Both processes are endothermic. At least some of the energy to do this comes from making new solute–solvent attractions. which is exothermic

Relative Interactions and Solution Formation
When the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy.

Solubility There is usually a limit to the solubility of one substance in another. Gases are always soluble in each other. Two liquids that are mutually soluble are said to be miscible. Alcohol and water are miscible. Oil and water are immiscible. The maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility. The solubility of one substance in another varies with temperature and pressure.

Will It Dissolve? Like Dissolves Like Chemist’s Rule of Thumb—
A chemical will dissolve in a solvent if it has a similar structure to the solvent. When the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other.

Classifying Solvents

Example 12.1a Predict whether the following vitamin is soluble in fat or water.
The 4 OH groups make the molecule highly polar and it will also H-bond to water. Vitamin C is water soluble. Vitamin C

Example 12.1b Predict whether the following vitamin is soluble in fat or water.
The 2 C=O groups are polar, but their geometric symmetry suggests their pulls will cancel and the molecule will be nonpolar. Vitamin K3 is fat soluble. Vitamin K3

Practice—Decide if the following are more soluble in hexane, C6H14,or water.
naphthalene formaldehyde stearic acid nonpolar molecule more soluble in C6H14 polar molecule more soluble in H2O nonpolar part dominant more soluble in C6H14

Energetics of Solution Formation: The Enthalpy of Solution
In order to make a solution you must: overcome all attractions between the solute particles—endothermic. overcome some attractions between solvent molecules—endothermic. form new attractions between solute particles and solvent molecules—exothermic. The overall DH for making a solution depends on the relative sizes of the DH for these 3 processes. DHsol’n = DHsolute + DHsolvent + DHmix

Solution Process 2. add energy in to overcome some solvent–solvent attractions 1. add energy in to overcome all solute–solute attractions 3. form new solute–solvent attractions, releasing energy

Energetics of Solution Formation
If the total energy cost for breaking attractions between particles in the pure solute and pure solvent is greater than the energy released in making the new attractions between the solute and solvent, the overall process will be endothermic. If the total energy cost for breaking attractions between particles in the pure solute and pure solvent is less than the energy released in making the new attractions between the solute and solvent, the overall process will be exothermic.

Heats of Hydration For aqueous solutions of ionic compounds, the energy added to overcome the attractions between water molecules and the energy released in forming attractions between the water molecules and ions is combined into a term called the heat of hydration. attractive forces between ions = lattice energy DHsolute = –DHlattice energy attractive forces in water = H-bonds attractive forces between ion and water = ion–dipole DHhydration = heat released when 1 mole of gaseous ions dissolves in water = DHsolvent + DHmix

Heats of Solution for Ionic Compounds
For an aqueous solution of an ionic compound, the DHsolution is the difference between the heat of hydration and the lattice energy. DHsolution = DHsolute+ DHsolvent + DHmix DHsolution = −DHlattice energy+ DHsolvent + DHmix DHsolution = DHhydration− DHlattice energy DHsolution = −DHlattice energy + DHhydration

Comparing Heat of Solution to Heat of Hydration
Since the lattice energy is always exothermic, the size and sign on the DHsol’n tells us something about DHhydration. If the heat of solution is large and endothermic, then the amount of energy it costs to separate the ions is more than the energy released from hydrating the ions. DHhydration < DHlattice when DHsol’n is (+) If the heat of solution is large and exothermic, then the amount of energy it costs to separate the ions is less than the energy released from hydrating the ions. DHhydration > DHlattice when DHsol’n is (−)

Ion–Dipole Interactions
When ions dissolve in water, they become hydrated. Each ion is surrounded by water molecules.

Practice—What is the lattice energy of KI if DHsol’n = +21
Practice—What is the lattice energy of KI if DHsol’n = kJ/mol and the DHhydration = −583 kJ/mol?

Practice—What is the lattice energy of KI if DHsol’n = +21
Practice—What is the lattice energy of KI if DHsol’n = kJ/mol and the DHhydration = −583 kJ/mol? Given: Find: DHsol’n = kJ/mol, DHhydration = −583 kJ/mol DHlattice, kJ/mol Conceptual Plan: Relationships: DHsol’n = DHhydration− DHlattice DHsol’n, DHhydration DHlattice Solution: Check: The unit is correct, and the lattice energy being exothermic is correct.

Solution Equilibrium The dissolution of a solute in a solvent is an equilibrium process. Initially, when there is no dissolved solute, the only process possible is dissolution. Shortly, solute particles can start to recombine to reform solute molecules, but the rate of dissolution >> rate of deposition and the solute continues to dissolve. Eventually, the rate of dissolution = the rate of deposition—the solution is saturated with solute and no more solute will dissolve.

Solubility Limit A solution that has the maximum amount of solute dissolved in it is said to be saturated. depends on the amount of solvent depends on the temperature and pressure of gases A solution that has less solute than saturation is said to be unsaturated. A solution that has more solute than saturation is said to be supersaturated.

How Can You Make a Solvent Hold More Solute Than It Is Able To?
Solutions can be made saturated at non-room conditions, then allowed to come to room conditions slowly. For some solutes, instead of coming out of solution when the conditions change, they get stuck in-between the solvent molecules and the solution becomes supersaturated. Supersaturated solutions are unstable and lose all the solute above saturation when disturbed. e.g., shaking a carbonated beverage

Adding Solute to a Supersaturated Solution of NaC2H3O2

Temperature Dependence of Solubility of Solids in Water
Solubility is generally given in grams of solute that will dissolve in 100 g of water. For most solids, the solubility of the solid increases as the temperature increases. when DHsolution is endothermic Solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line).

Solubility Curves

Temperature Dependence of Solid Solubility in Water (g/100 g H2O)

Purification by Recrystallization
One of the common operations performed by a chemist is purifying a material. One method of purification involves dissolving a solid in a hot solvent until the solution is saturated. As the solution slowly cools, the solid crystallizes out leaving impurities behind.

Recrystallization of KNO3
KNO3 can be purified by dissolving a little less than 106 g in 100 g of water at 60 ºC, then allowing it to cool slowly. When it cools to 0 ºC, only 13.9 g will remain in solution; the rest will precipitate out.

Practice—Decide if each of the following solutions is saturated, unsaturated, or supersaturated.
50 g KNO3 in 100 g 34 ºC saturated 50 g KNO3 in 100 g 50 ºC unsaturated 50 g KNO3 in 50 g 50 ºC supersaturated 100 g NH4Cl in 200 g 70 ºC unsaturated 100 g NH4Cl in 150 g 50 ºC supersaturated

Temperature Dependence of Solubility of Gases in Water
Solubility is generally given in moles of solute that will dissolve in 1 liter of solution. generally lower solubility than ionic or polar covalent solids because most are nonpolar molecules For all gases, the solubility of the gas decreases as the temperature increases. The DHsolution is exothermic because you do not need to overcome solute–solute attractions.

Temperature Dependence of Gas Solubility in Water (g/100 g H2O)

Pressure Dependence of Solubility of Gases in Water
The larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the liquid.

Henry’s Law The solubility of a gas (Sgas) is directly proportional to its partial pressure, (Pgas). Sgas = kHPgas kH is called Henry’s Law Constant.

Relationship between Partial Pressure and Solubility of a Gas

Pressure

Example 12. 2 What pressure of CO2 is required to keep the [CO2] = 0
Example 12.2 What pressure of CO2 is required to keep the [CO2] = 0.12 M in soda at 25 °C? Given: Find: S = [CO2] = 0.12 M P of CO2, atm Conceptual Plan: Relationships: S = kHP, kH = 3.4 × 10–2 M/atm [CO2] P Solution: Check: The unit is correct, and the pressure higher than 1 atm meets our expectation from general experience.

Practice—How many grams of NH3 will dissolve in 0
Practice—How many grams of NH3 will dissolve in 0.10 L of solution when its partial pressure is 7.6 torr? (kH = 58 M/atm)

Practice—How many grams of NH3 will dissolve in 0
Practice—How many grams of NH3 will dissolve in 0.10 L of solution when its partial pressure is 7.6 torr? Given: Find: P of NH3 = 7.6 torr; 0.10 L mass of NH3, g Conceptual Plan: Relationships: S = kHP, kH = 58 M/atm, 1 atm = 760 torr, 1 mol = g Solution:

Concentrations Solutions have variable composition.
To describe a solution, you need to describe the components and their relative amounts. The terms dilute and concentrated can be used as qualitative descriptions of the amount of solute in solution. concentration = amount of solute in a given amount of solution occasionally amount of solvent

Solution Concentration Molarity
moles of solute per 1 liter of solution used because it describes how many molecules of solute in each liter of solution if a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar

Molarity and Dissociation
The molarity of the ionic compound allows you to determine the molarity of the dissolved ions. CaCl2(aq) = Ca2+(aq) + 2 Cl−(aq) A 1.0 M CaCl2(aq) solution contains 1.0 mole of CaCl2 in each liter of solution. 1 L = 1.0 mole CaCl2, 2 L = 2.0 moles CaCl2 Because each CaCl2 dissociates to give one Ca2+, a 1.0 M CaCl2 solution is 1.0 M Ca2+. 1 L = 1.0 mole Ca+2, 2 L = 2.0 moles Ca2+ Because each CaCl2 dissociates to give 2 Cl−, a 1.0 M CaCl2 solution is 2.0 M Cl−. 1 L = 2.0 moles Cl−, 2 L = 4.0 moles Cl−

Solution Concentration Molality, m
moles of solute per 1 kilogram of solvent defined in terms of amount of solvent, not solution like the others does not vary with temperature because based on masses, not volumes

Parts Solute in Parts Solution
Parts can be measured by mass or volume. Parts are generally measured in same units. by mass in grams, kilograms, lbs. etc. by volume in mL, L, gallons, etc. mass and volume combined in grams and mL percentage = parts of solute in every 100 parts solution If a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution. or 0.9 kg solute in every 100 kg solution parts per million = parts of solute in every 1 million parts solution If a solution is 36 ppm by volume, then there are 36 mL of solute in 1 million mL of solution.

Percent Concentration

Parts per Million Concentration
60

PPM grams of solute per 1,000,000 g of solution
mg of solute per 1 kg of solution 1 liter of water = 1 kg of water For aqueous solutions we often approximate the kg of the solution as the kg or L of water. For dilute solutions, the difference in density between the solution and pure water is usually negligible.

Parts per Billion Concentration
62

Using Concentrations as Conversion Factors
Concentrations show the relationship between the amount of solute and the amount of solvent. 12%(m/m) sugar(aq) means 12 g sugar  100 g solution or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution 22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution The concentration can then be used to convert the amount of solute into the amount of solution, or vice versa.

Example 12. 3 What volume of 10. 5% by mass soda contains 78
Example 12.3 What volume of 10.5% by mass soda contains 78.5 g of sugar? Given: Find: 78.5 g sugar volume, mL Conceptual Plan: Relationships: g solute g sol’n mL sol’n 100 g sol’n = 10.5 g sugar, 1 mL sol’n = 1.04 g Solution: Check: The unit is correct; the magnitude seems reasonable as the mass of sugar  10% the volume of solution.

Preparing a Solution need to know amount of solution and concentration of solution Calculate the mass of solute needed. Start with amount of solution. Use concentration as a conversion factor. 5% by mass Þ 5 g solute  100 g solution “Dissolve the grams of solute in enough solvent to total the total amount of solution.”

Practice—How would you prepare 250. 0 mL of 19. 5% by mass CaCl2
Practice—How would you prepare mL of 19.5% by mass CaCl2? (d = 1.18 g/mL)

Dissolve 57.5 g of CaCl2 in enough water to total 250.0 mL.
Practice—How would you prepare mL of 19.5% by mass CaCl2? (d = 1.18 g/mL) Given: Find: 250.0 mL solution mass CaCl2, g Conceptual Plan: Relationships: mL sol’n g sol’n g solute 100 g sol’n = 19.5 g CaCl2, 1 mL sol’n = 1.18 g Solution: Answer: Dissolve 57.5 g of CaCl2 in enough water to total mL.

Solution Concentrations Mole Fraction, XA
The mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution. total of all the mole fractions in a solution = 1 unitless The mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution. = mole fraction × 100%

The unit is correct; the magnitude is reasonable.
Example 12.4a What is the molarity of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n M mol C2H6O2, kg H2O, L M Conceptual Plan: Relationships: g C2H6O2 mol C2H6O2 mL sol’n L sol’n M M = mol/L, 1 mol C2H6O2 = g, 1 mL = L Solution: Check: The unit is correct; the magnitude is reasonable.

Practice—Calculate the molarity of a solution made by dissolving 34
Practice—Calculate the molarity of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 mL of solution. (MMNH3 = g/mol)

Practice—Calculate the molarity of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 mL of solution. Given: Find: 34.0 g NH3, 2000 mL sol’n M 2.00 mol NH3, 2.00 L sol’n M Conceptual Plan: Relationships: g NH3 mol NH3 mL sol’n L sol’n M M = mol/L, 1 mol NH3 = g, 1 mL = L Solution: Check: The unit is correct; the magnitude is reasonable. 71

Example 12.4b What is the molality of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: mol C2H6O2, kg H2O, 515 mL sol’n m 17.2 g C2H6O2, kg H2O, 515 mL sol’n m Conceptual Plan: Relationships: m = mol/kg, 1 mol C2H6O2 = g g C2H6O2 mol C2H6O2 kg H2O m Solution: Check: The unit is correct; the magnitude is reasonable.

Practice—Calculate the molality of a solution made by dissolving 34
Practice—Calculate the molality of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 mL of water. (MMNH3 = g/mol, dH2O = 1.00 g/mL)

Practice—Calculate the molality of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 mL of water. Given: Find: 34.0 g NH3, 2000 mL H2O m 2.00 mol NH3, 2.00 kg H2O m Conceptual Plan: Relationships: g NH3 mol NH3 mL H2O g H2O m kg H2O m = mol/kg, 1 mol NH3 = g, 1 kg = 1000 g, 1.00 g = 1 mL Solution: Check: The unit is correct; the magnitude is reasonable. 74

Practice—Calculate the molality of a solution made by dissolving 34
Practice—Calculate the molality of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 g of solution. (MMNH3 = g/mol)

Practice—Calculate the molality of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 g of solution. Given: Find: 2.00 mol NH3, 1.97 kg H2O m 34.0 g NH3, 2000 g solution m Conceptual Plan: Relationships: g NH3 mol NH3 g sol’n g H2O m kg H2O m = mol/kg, 1 mol NH3 = g, 1 kg = 1000 g Solution: Check: The unit is correct; the magnitude is reasonable. 76

Example 12.4c What is the percent by mass of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n %(m/m) Conceptual Plan: Relationships: 1 kg = 1000 g g C2H6O2 g solvent g sol’n % Solution: Check: The unit is correct; the magnitude is reasonable.

Practice—Calculate the percent by mass of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 mL of water. (MMNH3 = g/mol, dH2O = 1.00 g/mL)

Practice—Calculate the percent by mass of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 mL of water. Given: Find: 34.0 g NH3, 2000 g H2O, 2034 g sol’n %(m/m) 34.0 g NH3, 2000 mL H2O %(m/m) Conceptual Plan: Relationships: g NH3 mL H2O % g sol’n g H2O % = g/g × 100%, 1.00 g = 1 mL Solution: Check: The unit is correct; the magnitude is reasonable. 79

Practice—Calculate the parts per million of a solution made by dissolving g of NH3 in 2.00 × 103 mL of water. (MMNH3 = g/mol, dH2O = 1.00 g/mL)

Practice—Calculate the parts per million of a solution made by dissolving g of NH3 in 2.00 × 103 mL of water. Given: Find: 0.340 g NH3, 2000 g H2O, 2000 g sol’n ppm 0.340 g NH3, 2000 mL H2O ppm Conceptual Plan: Relationships: g NH3 mL H2O ppm g sol’n g H2O ppm = g/g × 106, 1.00 g = 1 mL Solution: Check: The unit is correct; the magnitude is reasonable. 81

Example 12.4d What is the mole fraction of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n C Conceptual Plan: Relationships: C = molA/moltot, 1 mol C2H6O2 = g, 1 mol H2O = g g C2H6O2 mol C2H6O2 g H2O mol H2O C Solution: Check: The unit is correct; the magnitude is reasonable.

Practice—Calculate the mole fraction of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 mL of water. (MMNH3 = g/mol, dH2O = 1.00 g/mL)

Practice—Calculate the mole fraction of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 mL of water. Given: Find: 2.00 mol NH3, mol H2O, tot mol C 34.0 g NH3, 2000 mL H2O C Conceptual Plan: Relationships: g NH3 mol NH3 mL H2O g H2O C mol H2O C = mol/mol, 1 mol NH3 = g, 1 mol H2O = g, 1.00 g = 1 mL Solution: Check: The unit is correct; the magnitude is reasonable. 84

Example 12.4d What is the mole percent of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n C% Conceptual Plan: Relationships: C = molA/moltot, 1 mol C2H6O2 = g, 1 mol H2O = g g C2H6O2 mol C2H6O2 g H2O mol H2O C% Solution: Check: The unit is correct; the magnitude is reasonable.

Converting Concentration Units
Write the given concentration as a ratio. Separate the numerator and denominator. Separate into the solute part and solution part. Convert the solute part into the required unit. Convert the solution part into the required unit. Use the definitions to calculate the new concentration units.

Example 12.5 What is the molarity of 6.55% by mass glucose (C6H12O6) solution? Given: Find: 6.55 g C6H12O6, 100 g sol’n M mol C2H6O2, L M 6.55%(m/m) C6H12O6 M Conceptual Plan: Relationships: g C6H12O6 mol C6H12O6 mL L sol’n M g sol’n M = mol/L, 1 mol C6H12O6 = g, 1 mL = L, 1 mL = 1.03 g Solution: Check: The unit is correct; the magnitude is reasonable. 87

Practice—Calculate the molality of a 16. 2 M H2SO4 solution
Practice—Calculate the molality of a 16.2 M H2SO4 solution. (MMH2SO4 = g/mol, dsol’n = 1.80 g/mL)

Practice—Calculate the molality of a 16.2 M H2SO4 solution.
Given: Find: 16.2 M H2SO4 m 16.2 mol H2SO4, kg H2O m 16.2 mol H2SO4, 1.00 L sol’n m Conceptual Plan: Relationships: L mol H2SO4 g sol’n g H2O m kg H2O mL g H2SO4 m = mol/kg, 1 mol H2SO4 = g, 1 kg = 1000 g, 1.80 g = 1 mL Solution: Check: The unit is correct; the magnitude is reasonable. 89

Colligative Properties
Colligative properties are properties whose value depends only on the number of solute particles, and not on what they are. The value of the property depends on the concentration of the solution. The difference in the value of the property between the solution and the pure substance is generally related to the different attractive forces and solute particles occupying solvent molecules positions. 90

Vapor Pressure of Solutions
The vapor pressure of a solvent above a solution is lower than the vapor pressure of the pure solvent. The solute particles replace some of the solvent molecules at the surface. Eventually, equilibrium is re-established, but with a smaller number of vapor molecules; therefore, the vapor pressure will be lower. The pure solvent establishes a liquid  vapor equilibrium. Addition of a nonvolatile solute reduces the rate of vaporization, decreasing the amount of vapor.

Thirsty Solutions A concentrated solution will draw solvent molecules toward it due to the natural drive for materials in nature to mix. Similarly, a concentrated solution will draw pure solvent vapor into it due to this tendency to mix. The result is reduction in vapor pressure.

Thirsty Solutions When equilibrium is established, the liquid level in the solution beaker is higher than the solution level in the pure solvent beaker—the thirsty solution grabs and holds solvent vapor more effectively. Beakers with equal liquid levels of pure solvent and a solution are placed in a bell jar. Solvent molecules evaporate from each one and fill the bell jar, establishing an equilibrium with the liquids in the beakers.

Psolvent in solution = csolvent∙P°
Raoult’s Law The vapor pressure of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P°. Psolvent in solution = csolvent∙P° Since the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent.

Example 12.6 Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of C12H22O11 with mL of H2O. Given: Find: 99.5 g C12H22O11, mL H2O PH2O Conceptual Plan: Relationships: g C12H22O11 mol C12H22O11 g H2O mol H2O CH2O mL H2O PH2O P°H2O = 23.8 torr, 1 mol C12H22O11 = g, 1 mol H2O = g Solution:

Practice—Calculate the total vapor pressure of a solution made by dissolving 25.0 g of glucose (C6H12O6) in 215 g of water at 50 °C. (MMC6H12O6 = g/mol, MMH2O = g/mol Vapor Pressure of 50 °C = 92.5 torr)

Practice—Calculate the total vapor pressure of a solution made by dissolving 25.0 g of glucose (C6H12O6) in 215 g of water at 50 °C. Given: Find: 25.0 g C6H12O6, 215 g H2O PH2O Conceptual Plan: Relationships: P°H2O = 92.5 torr, 1 mol C6H12O6 = g, 1 mol H2O = g g C6H12O6 mol C6H12O6 g H2O mol H2O CH2O PH2O Solution: Since glucose is nonvolatile, the total vapor pressure = vapor pressure H2O. 97

Raoult’s Law for Volatile Solute
When both the solvent and the solute can evaporate, both molecules will be found in the vapor phase. The total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent. for an ideal solution Ptotal = Psolute + Psolvent The solvent decreases the solute vapor pressure in the same way the solute decreased the solvent’s. Psolute = csolute∙P°solute and Psolvent = csolvent∙P°solvent

Example 12.7 Calculate the component and total vapor pressure in a solution prepared by mixing 3.95 g of CS2 with 2.43 g of C3H6O. Given: Find: mol CS2, PCS2 = 285 torr, mol C3H6O, PC3H6O = 148 torr PCS2, PC3H6O, Ptotal mol CS2, PºCS2 = 515 torr, mol C3H6O, PºC3H6O = 332 torr PCS2, PC3H6O, Ptotal 3.95 g CS2, PºCS2 = 515 torr, 2.43 g C3H6O, PºC3H6O = 332 torr PCS2, PC3H6O, Ptotal Conceptual Plan: Relationships: 1 mol CS2 = g, 1 mol C3H6O = 58.0 g g CS2 mol CS2 g C3H6O mol C3H6O C P Solution: 99

Practice—Calculate the total vapor pressure of a solution made by mixing mol of ether (C4H10O) with mol of ethanol (C2H6O) at 20 °C. (Vapor Pressure of 20 °C = 440 torr Vapor Pressure of 20 °C = 44.6 torr)

Practice—Calculate the total vapor pressure of a solution made by mixing mol of ether (C4H10O) with mol of ethanol (C2H6O) at 20 °C. Given: Find: 0.500 mol C4H10O, PC4H10O = 293 torr, mol C2H6O, PC2H6O = 14.9 torr Ptotal 0.500 mol C4H10O, Pºether = 440 torr, mol C2H6O, Pºethanol = 332 torr Ptotal Conceptual Plan: Relationships: mol C4H10O mol C2H6O C P Solution: 101

Freezing Point Depression
The freezing point of a solution is lower than the freezing point of the pure solvent. for a nonvolatile solute Therefore, the melting point of the solid solution is lower. The difference between the freezing point of the solution and freezing point of the pure solvent is directly proportional to the molal concentration of solute particles. (FPsolvent – FPsolution) = DTf = m∙Kf The proportionality constant is called the Freezing Point Depression Constant, Kf. The value of Kf depends on the solvent. The units of Kf are °C/m.

Example 12. 8 What is the freezing point of a 1
Example 12.8 What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C2H6O2? Given: Find: 1.7 m C2H6O2(aq) Tf, °C Conceptual Plan: Relationships: DTf = m∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00 °C m DTf Solution: Check: The unit is correct; the freezing point lower than the normal freezing point makes sense.

Practice—Calculate the molar mass of a compound if a solution of 12
Practice—Calculate the molar mass of a compound if a solution of 12.0 g dissolved in 80.0 g of water freezes at −1.94 °C. (Kf water = 1.86 °C/m)

Practice—Calculate the molar mass of a compound if a solution of 12
Practice—Calculate the molar mass of a compound if a solution of 12.0 g dissolved in 80.0 g of water freezes at −1.94 °C. Given: Find: masssolute = 12.0 g, massH2O = 80.0 g, FPsol’n = −1.94 °C Tf, °C Conceptual Plan: Relationships: m DTf FPsol’n mol MM DTf = FPH2O − FPsol’n m  kgH2O = mol MM = g/mol DTf = m Kf, m = mol/kg, MM = g/mol, Kf = 1.86 °C/m, FPH2O = 0.00 °C Solution: 109

Boiling Point Elevation
The boiling point of a solution is higher than the boiling point of the pure solvent. for a nonvolatile solute The difference between the boiling point of the solution and boiling point of the pure solvent is directly proportional to the molal concentration of solute particles. (BPsolution – BPsolvent) = DTb = m∙Kb The proportionality constant is called the Boiling Point Elevation Constant, Kb. The value of Kb depends on the solvent. The units of Kb are °C/m.

Example 12.9 How many grams of ethylene glycol, C2H6O2, must be added to 1.0 kg H2O to give a solution that boils at 105 °C? Given: Find: 1.0 kg H2O, Tb = 105 °C mass C2H6O2, g Conceptual Plan: Relationships: DTb m kg H2O mol C2H6O2 g C2H6O2 DTb = m∙Kb, Kb H2O = °C/m, BPH2O = °C MMC2H6O2 = g/mol, 1 kg = 1000 g Solution:

Practice—Calculate the boiling point of a solution made by dissolving 1.00 g of glycerin, C3H8O3, in 54.0 g of water.

Practice—Calculate the boiling point of a solution made by dissolving 1.00 g of glycerin, C3H8O3, in 54.0 g of water. Given: Find: 1.00 g C3H8O3, 54.0 g H2O Tb, sol’n, °C Conceptual Plan: Relationships: m DTb g C3H8O3, kg H2O m = mol/kg, DTb = m∙Kb, Kb for H2O = °C/m, BPH2O = °C, 1 mol C3H8O3 = g Solution: 113

Osmosis Osmosis is the flow of solvent from a solution of low concentration into a solution of high concentration. The solutions may be separated by a semipermeable membrane. A semipermeable membrane allows solvent to flow through it, but not solute.

Osmotic Pressure The amount of pressure needed to keep osmotic flow from taking place is called the osmotic pressure. The osmotic pressure, P, is directly proportional to the molarity of the solute particles. R = (atm∙L)/(mol∙K) P = MRT 115

Example 12. 10 What is the molar mass of a protein if 5
Example What is the molar mass of a protein if 5.87 mg per 10 mL gives an osmotic pressure of 2.45 torr at 25 °C? Given: Find: 5.87 mg/10 mL, P = 2.45 torr, T = 25 °C molar mass, g/mol Conceptual Plan: Relationships: P,T M P = MRT, T(K) = T(°C) , R = atm∙L/mol∙K M = mol/L, 1 mL = L, MM = g/mol, 1 atm = 760 torr mL L mol protein Solution:

Practice—Lysozyme is an enzyme used to cleave cell walls
Practice—Lysozyme is an enzyme used to cleave cell walls. A solution made by dissolving g of lysozyme in mL results in an osmotic pressure of 1.32 × 10−3 atm at 25 °C. Calculate the molar mass of lysozyme.

Practice—What is the molar mass of lysozyme if 0. 0750 g per 100
Practice—What is the molar mass of lysozyme if g per mL gives an osmotic pressure of 1.32 × 10−3 atm at 25 °C? Given: Find: g/100 mL, P = 1.32 × 10−3 atm, T = 25 °C molar mass, g/mol Conceptual Plan: Relationships: P,T M P = MRT, T(K) = T(°C) , R = atm∙L/mol∙K M = mol/L, 1 mL = L, MM = g/mol, 1 atm = 760 torr mL L mol protein Solution: 119

van’t Hoff Factors Ionic compounds produce multiple solute particles for each formula unit. The van’t Hoff factor, i, is the ratio of moles of solute particles to moles of formula units dissolved. Measured van’t Hoff factors are often lower than you might expect due to ion pairing in solution.

Example 12. 11 What is the van’t Hoff factor if 0
Example What is the van’t Hoff factor if m CaCl2(aq) has a freezing point of −0.27 ºC? Given: Find: 0.050 m CaCl2(aq), Tf = −0.27 °C i Conceptual Plan: Relationships: DTf = i∙m∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00 °C m, DTf i Solution: 122

Principles of Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro

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Principles of Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro

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