© 2014 Pearson Education, Inc.

**Chapter 12 ** **Solutions **

### Sherril Soman,

**Grand Valley State University **

### Lecture Presentation

### 12.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater 544

### 12.2 Types of Solutions and Solubility 546 12.3 Energetics of Solution Formation 551 12.4 Solution Equilibrium and Factors

### Affecting Solubility 555

### 12.5 Expressing Solution Concentration 559 12.6 Colligative Properties: Vapor Pressure Lowering, Freezing Point

### Depres sion, Boiling Point Elevation, and Osmotic Pressure 567

### 12.7 Colligative Properties of Strong Electrolyte Solutions 579

### 12.8 Colloids 582

### Key Learning Outcomes 587

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### 12.1 Thirsty Solutions: Why You Shouldn’t **Drink Seawater **

**Thirsty Seawater **

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**12.2 Types of Solutions and Solubility **

### When solutions with different solute concentrations

### come in contact, they spontaneously mix to result in a

### uniform distribution of solute throughout the solution.

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**Common Types of Solutions **

### • A solution may be compost of a solid and a

### liquid, a gas an a liquid, or other combinations.

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### • The majority component of a solution is called the solvent.

### • The minority component is called the solute.

**Solutions **

### • In aqueous

### solutions, water

### is the solvent.

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### • Many physical systems tend toward lower potential energy.

### • But 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.

**Nature’s Tendency Toward Mixing: Entropy **

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### • Energy changes in the formation of most solutions also involve differences in attractive forces

### between the particles.

**Solutions: Effect of Intermolecular Forces **

### • Chemist’s rule of thumb – like dissolves like

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*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.1 ** **Solubility **

Vitamins are often categorized as either fat soluble or water soluble. Water-soluble vitamins dissolve in body fluids and are easily eliminated in the urine, so there is little danger of overconsumption. Fat-soluble vitamins, on the other hand, can accumulate in the body’s fatty deposits. Overconsumption of a fat-soluble vitamin can be dangerous to your health. Examine the structure of each vitamin shown here and classify it as either fat soluble or water soluble.

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.1 ** **Solubility **

**a. The four —OH bonds in vitamin C make it highly polar and allow it to hydrogen bond with water. Vitamin C **
is water soluble.

**b. The C — C bonds in vitamin K**_{3} are nonpolar and the C — H bonds are nearly so. The C O bonds are
polar, but the bond dipoles oppose and largely cancel each other, so the molecule is dominated by the
nonpolar bonds. Vitamin K_{3} is fat soluble.

**Solution **

Continued
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*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.1 ** **Solubility **

Continued

**c. The C — C bonds in vitamin A are nonpolar and the C — H bonds are nearly so. The one polar —OH bond **
may increase its water solubility slightly, but overall vitamin A is nonpolar and therefore fat soluble.

**d. The three —OH bonds and one —NH bond in vitamin B**_{5} make it highly polar and allow it to hydrogen bond
with water. Vitamin B_{5} is water soluble.

**For Practice 12.1 **

Determine whether each compound is soluble in hexane.

**a. water (H**_{2}O) **b. propane (CH**_{3}CH_{2}CH_{3})
**c. ammonia (NH**_{3}) **d. hydrogen chloride (HCl) **

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**Solution Interactions **

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### • 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 increase in entropy from mixing.

**Relative Interactions and Solution **

**Formation **

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### • To make a solution you must

### 1. overcome all attractions between the solute particles;

### therefore, ΔH

_{solute}

### is endothermic.

### 2. overcome some attractions between solvent molecules;

### therefore, ΔH

_{solvent}

### is endothermic.

### 3. form new attractions between solute particles and solvent molecules; therefore, ΔH

_{mix}

### is exothermic.

### • The overall ΔH for making a solution depends on the relative sizes of the ΔH for these three processes.

### ΔH

_{sol’n}

### = ΔH

_{solute}

### + ΔH

_{solvent}

### + ΔH

_{mix}

**12.3 Energetics of Solution Formation: **

**The Enthalpy of Solution **

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### Step 1: Separating the solute into its constituent particles

**Solution Process **

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### Step 2: Separating the solvent particles from each other to make room for the solute particles

**Solution Process **

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### Step 3: Mixing the solute particles with the solvent particles

**Solution Process **

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### 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.

**Energetics of Solution Formation **

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### 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.

**Energetics of Solution Formation **

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**Ion–Dipole Interactions **

### • When ions dissolve in water they become **hydrated. **

### – Each ion is surrounded by water molecules.

### • The formation of these ion–dipole attractions

### causes the heat of hydration to be very exothermic.

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**Heats of Solution for Ionic Compounds **

### • For an aqueous solution of an ionic compound, the ΔH _{solution} is the difference between the heat of

### hydration and the lattice energy.

### ΔH _{solution} = −ΔH lattice energy + ΔH _{hydration} ΔH _{solution} = ΔH _{solute} + ΔH _{solvent} + ΔH _{mix}

### ΔH _{solution} = −ΔH lattice energy + ΔH _{solvent} + ΔH _{mix}

### ΔH _{solution} = ΔH _{hydration} − ΔH lattice energy

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### ΔH _{solution} = ΔH _{hydration} − ΔH lattice energy

**Heat of Hydration **

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**12.4 Solution Equilibrium and Factors ** **Affecting Solubility **

**Solution Equilibrium **

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**Adding a Crystal of NaC** _{2} **H** _{3} **O** _{2} ** to a ** **Supersaturated Solution **

_{2}

_{3}

_{2}

**saturated. ** **supersaturated. **

**unsaturated. **

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**Solubility Curves **

### • Solubility is

### generally given in grams of solute that will

### dissolve in

**100 g of **

**water. **

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**Purification by Recrystallization **

### • One of the common operations performed by a chemist is

### removing impurities from a solid compound.

### • 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.

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**Temperature Dependence of Solubility of **

**Gases in Water **

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**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.

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**Henry’s Law **

### • The solubility of a gas *(S* _{gas} ) is directly

### proportional to its partial *pressure, (P* _{gas} ).

*S* _{gas} * = k* _{H} *P* _{gas}

*• k* _{H} is called the Henry’s

**law constant. **

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*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.2 ** **Henry’s Law **

What pressure of carbon dioxide is required to keep the carbon dioxide concentration in a bottle of club soda at 0.12 M at 25 °C ?

**Sort **

You are given the desired solubility of carbon dioxide and asked to find the pressure required to achieve this solubility.

**Given: **

**Find: **

**Strategize **

Use Henry’s law to find the required pressure from the solubility. You will need the Henry’s law constant for carbon dioxide, which is listed in Table 12.4.

**Conceptual Plan **

**Relationships Used **

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*Chemistry: A Molecular Approach, 3rd Edition *
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**Example 12.2 ** **Henry’s Law **

Continued

**Solve **

Solve the Henry’s law equation for and substitute the other quantities to calculate it.

**Solution **

**Check **

The answer is in the correct units and seems reasonable. A small answer (for example, less than 1 atm) would be suspect because you know that the soda is under a pressure greater than atmospheric pressure when you open it. A very large answer (for example, over 100 atm) would be suspect because an ordinary can or bottle probably could not sustain such high pressures without bursting.

**For Practice 12.2 **

Determine the solubility of oxygen in water at 25 °C exposed to air at 1.0 atm. Assume a partial pressure for oxygen of 0.21 atm.

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### • Solutions have variable composition.

### • To describe a solution, you need to describe the **components and their relative amounts. **

**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

** 12.5 Expressing Solution Concentration **

**Concentrations **

### Molarity and Molality

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Slide 32 of 46 General Chemistry: Chapter 13

*Molarity (M) = * *Amount of solute (in moles) * Volume of solution (in liters)

*Molality (m) = * *Amount of solute (in moles) *

### Mass of solvent (in kilograms)

### Parts per Million, Parts per Billion, and Parts per Trillion

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Slide 33 of 46 General Chemistry: Chapter 13

### Very low solute concentrations are expressed as:

### ppm: parts per million (g/g, mg/L) ppb: parts per billion (ng/g, g/L) ppt: parts per trillion (pg/g, ng/L)

### note that 1.0 L 1.0 g/mL = 1000 g

### ppm, ppb, and ppt are properly m/m or v/v.

### Mole Fraction and Mole Percent

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Slide 34 of 46 General Chemistry: Chapter 13

###

_{i }

### = *Amount of component i (in moles) *

### Total amount of all components (in moles)

###

_{1 }

### +

_{2}

### +

_{3}

### + …

_{n}

### = 1

*Mole % i = *

_{i}

### 100%

### Mass Percent (m/m) Volume Percent (v/v) Mass/Volume percent (m/v)

### Isotonic saline is prepared by dissolving 0.9 g of NaCl in 100 mL of water and is said to be:

### 0.9% NaCl (mass/volume)

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General Chemistry: Chapter 13 Slide 35 of 46

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**Concentrations **

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### • Moles of solute per 1 liter of solution

### • 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

**Solution Concentration: Molarity **

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### • 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

**Solution Concentration: Molality, m **

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### • 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 ⇒ 5 g solute ≡ 100 g solution

### – “Dissolve the grams of solute in enough solvent to total the total amount of solution.”

**Preparing a Solution **

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**Preparing a Solution **

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### • Parts can be measured by mass or volume.

### • Parts are generally measured in the same units.

### – By mass in grams, kilogram, lbs, etc.

### – By volume in mL, L, gallons, etc.

### – Mass and volume combined in grams and mL

**Parts Solute in Parts Solution **

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### • 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.

**Parts Solute in Parts Solution **

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### • 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.

**PPM **

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**Parts Per Billion Concentration **

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*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.3 ** **Using Parts by Mass in Calculations **

What volume (in mL) of a soft drink that is 10.5% sucrose (C_{12}H_{22}O_{11}) by mass contains 78.5 g of sucrose? (The
density of the solution is 1.04 g/mL.)

**Sort **

You are given a mass of sucrose and the concentration and density of a sucrose solution, and you are asked to find the volume of solution containing the given mass.

**Given: 78.5 g C**_{12}H_{22}O_{11 }
10.5% C_{12}H_{22}O_{11} by mass
density = 1.04 g/mL

**Find: mL **

**Strategize **

Begin with the mass of sucrose in grams. Use the mass percent concentration of the solution (written as a ratio, as shown under relationships used) to find the number of grams of solution containing this quantity of sucrose. Then use the density of the solution to convert grams to milliliters of solution.

**Conceptual Plan **

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*Chemistry: A Molecular Approach, 3rd Edition *
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Continued

**Relationships Used **

**Solve **

Begin with 78.5 g C_{12}H_{22}O_{11} and multiply by the conversion factors to arrive at the volume of solution.

**Solution **

**Check **

The units of the answer are correct. The magnitude seems correct because the solution is approximately 10% sucrose by mass. Since the density of the solution is approximately 1 g/mL, the volume containing 78.5 g sucrose should be roughly 10 times larger, as calculated (719 ≈ 10 × 78.5).

**Example 12.3 ** **Using Parts by Mass in Calculations **

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Continued

**For Practice 12.3 **

What mass of sucrose (C_{12}H_{22}O_{11}), in g, is contained in 355 mL (12 ounces) of a soft drink that is 11.5% sucrose by
mass? (Assume a density of 1.04 g/mL.)

**For More Practice 12.3 **

A water sample is found to contain the pollutant chlorobenzene with a concentration of 15 ppb (by mass). What
volume of this water contains 5.00 × 10^{2} mg of chlorobenzene? (Assume a density of 1.00 g/mL.)

**Example 12.3 ** **Using Parts by Mass in Calculations **

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### • 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%

**Solution Concentrations: Mole Fraction, X** _{A}

**Solution Concentrations: Mole Fraction, X**

_{A}© 2014 Pearson Education, Inc.

### 1. Write the given concentration as a ratio.

### 2. Separate the numerator and denominator.

### – Separate into the solute part and solution part

### 3. Convert the solute part into the required unit.

### 4. Convert the solution part into the required unit.

### 5. Use the definitions to calculate the new concentration units.

**Converting Concentration Units **

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**Example 12.4 ** **Calculating Concentrations **

A solution is prepared by dissolving 17.2 g of ethylene glycol (C_{2}H_{6}O_{2}) in 0.500 kg of water. The final volume of
the solution is 515 mL. For this solution, calculate the concentration in each unit.

**a. molarity ** **b. molality ** **c. percent by mass **

**d. mole fraction ** **e. mole percent **

**Solution **

**a. To calculate molarity, first find the amount of ethylene glycol in moles from the mass and molar mass. Then **
divide the amount in moles by the volume of the solution in liters

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Continued

**Example 12.4 ** **Calculating Concentrations **

**b. To calculate molality, use the amount of ethylene glycol in moles from part (a), and divide by the mass of the **
water in kilograms.

**c. To calculate percent by mass, divide the mass of the solute by the sum of the masses of the solute and solvent **
and multiply the ratio by 100%.

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Continued

**Example 12.4 ** **Calculating Concentrations **

**d. To calculate mole fraction, first determine the amount of water in moles from the mass of water and its molar **
mass. Then divide the amount of ethylene glycol in moles (from part (a)) by the total number of moles.

**e. To calculate mole percent, multiply the mole fraction by 100%. **

**For Practice 12.4 **

A solution is prepared by dissolving 50.4 g sucrose (C_{12}H_{22}O_{11}) in 0.332 kg of water. The final volume of the
solution is 355 mL. Calculate the concentration of the solution in each unit.

**a. molarity ** **b. molality ** **c. percent by mass **

**d. mole fraction ** **e. mole percent **

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*Chemistry: A Molecular Approach, 3rd Edition *
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**Example 12.5 ** **Converting between Concentration Units **

What is the molarity of a 6.56% by mass glucose (C_{6}H_{12}O_{6}) solution? (The density of the solution is 1.03 g/mL.)

**Sort **

You are given the concentration of a glucose solution in percent by mass and the density of the solution. Find the concentration of the solution in molarity.

**Given: 6.56% C**_{6}H_{12}O_{6 }
density = 1.03 g/mL
**Find: M **

**Strategize **

Begin with the mass percent concentration of the solution written as a ratio, and separate the numerator from the
denominator. Convert the numerator from g C_{6}H_{12}O_{6} to mol C_{6}H_{12}O_{6}. Convert the denominator from g soln to
mL of solution and then to L solution. Then divide the numerator (now in mol) by the denominator (now in L) to
obtain molarity.

**Conceptual Plan **

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*Chemistry: A Molecular Approach, 3rd Edition *
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**Example 12.5 ** **Converting between Concentration Units **

Continued

**Relationships Used **

**Solve **

Begin with the numerator (6.56 g C_{6}H_{12}O_{6}) and use the molar mass to convert to mol C_{6}H_{12}O_{6}.

Convert the denominator (100 g solution) into mL of solution (using the density) and then to L of solution.

Finally, divide mol C_{6}H_{12}O_{6} by L solution to arrive at molarity.

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**• Colligative properties are properties whose value ** **depends only on the number of solute **

**particles, and not on what they are. **

### – Value of the property depends on the concentration of the solution.

**12.6 Colligative Properties: Vapor Pressure **

**Lowering, Freezing Point Depression, Boiling Point **

**Elevation, and Osmotic Pressure **

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**Vapor Pressure of Solutions **

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**Thirsty Solutions ** • The vapor pressure of a volatile solvent above a solution is equal to its normal vapor pressure, *P°, multiplied by its *

### mole fraction in the *solution. *

*P* solvent in solution = χ _{solvent} *∙P* °

### – Because 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.

**Raoult’s Law **

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*Chemistry: A Molecular Approach, 3rd Edition *
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**Example 12.6 ** **Calculating the Vapor Pressure of a Solution **

**Containing a Nonelectrolyte and Nonvolatile Solute **

Calculate the vapor pressure at 25 °C of a solution containing 99.5 g sucrose (C_{12}H_{22}O_{11}) and 300.0 mL water. The
vapor pressure of pure water at 25 °C is 23.8 torr. Assume the density of water to be 1.00 g/mL.

**Sort **

You are given the mass of sucrose and volume of water in a solution. You are also given the vapor pressure and density of pure water and asked to find the vapor pressure of the solution.

**Given: **

**Find: P**_{solution }

**Strategize **

Raoult’s law relates the vapor pressure of a solution to the mole fraction of the solvent and the vapor pressure of the pure solvent. Begin by calculating the amount in moles of sucrose and water.

Calculate the mole fraction of the solvent from the calculated amounts of solute and solvent.

Then use Raoult’s law to calculate the vapor pressure of the solution.

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.6 ** **Calculating the Vapor Pressure of a Solution **

**Containing a Nonelectrolyte and Nonvolatile Solute **

Continued

**Conceptual Plan **

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*Chemistry: A Molecular Approach, 3rd Edition *
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**Example 12.6 ** **Calculating the Vapor Pressure of a Solution **

**Containing a Nonelectrolyte and Nonvolatile Solute **

Continued

**Solve **

Calculate the number of moles of each solution component.

Use the number of moles of each component to calculate the mole fraction of the solvent (H_{2}O).

Use the mole fraction of water and the vapor pressure of pure water to calculate the vapor pressure of the solution.

**Solution **

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.6 ** **Calculating the Vapor Pressure of a Solution **

**Containing a Nonelectrolyte and Nonvolatile Solute **

Continued

**Check **

The units of the answer are correct. The magnitude of the answer seems right because the calculated vapor pressure of the solution is just below that of the pure liquid, as you would expect for a solution with a large mole fraction of solvent.

**For Practice 12.6 **

Calculate the vapor pressure at 25 °C of a solution containing 55.3 g ethylene glycol (HOCH_{2}CH_{2}CH_{2}OH) and
285.2 g water. The vapor pressure of pure water at 25 °C is 23.8 torr.

**For Practice 12.6 **

A solution containing ethylene glycol and water has a vapor pressure of 7.88 torr at 10 °C. Pure water has a vapor pressure of 9.21 torr at 10 °C. What is the mole fraction of ethylene glycol in the solution?

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### • The vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent.

### • The vapor pressure of the solution is directly proportional to the amount of the solvent in the solution.

### • The difference between the vapor pressure of the pure solvent and the vapor pressure of the solvent in solution is called the vapor pressure lowering.

*ΔP = P°* _{solvent} * − P* _{solution} * = * χ _{solute} * ∙ * *P°* _{solvent }

**Vapor Pressure Lowering **

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**Deviations from Raoult’s Law **

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*Chemistry: A Molecular Approach, 3rd Edition *
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**Example 12.7 ** **Calculating the Vapor Pressure of a Two-Component ** **Solution **

Continued

Use the mole fraction of each component along with Raoult’s law to calculate the partial pressure of each component. The total pressure is the sum of the partial pressures.

**Conceptual Plan **

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*Chemistry: A Molecular Approach, 3rd Edition *
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**Example 12.7 ** **Calculating the Vapor Pressure of a Two-Component ** **Solution **

Continued

**Relationships Used **

**Solve **

Begin by converting the mass of each component to the amounts in moles.

Then calculate the mole fraction of carbon disulfide.

Calculate the mole fraction of acetone by subtracting the mole fraction of carbon disulfide from 1.

Calculate the partial pressures of carbon disulfide and acetone by using Raoult’s law and the given values of the vapor pressures of the pure substances.

Calculate the total pressure by summing the partial pressures.

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*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.7 ** **Calculating the Vapor Pressure of a Two-Component ** **Solution **

Continued

Lastly, compare the calculated total pressure for the ideal case to the experimentally measured total pressure. Since the experimentally measured pressure is greater than the calculated pressure, we can conclude that the interactions between the two components are weaker than the interactions between the components themselves.

**Solution **

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*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.7 ** **Calculating the Vapor Pressure of a Two-Component ** **Solution **

Continued

P_{tot}(ideal) = 285 torr + 148 torr

= 433 torr

* P*_{tot}(exp) = 645 torr
* P*_{tot}*(exp) > P*_{tot} (ideal)

The solution is not ideal and shows positive deviations from Raoult’s law. Therefore, carbon disulfide–acetone interactions must be weaker than acetone–acetone and carbon disulfide–carbon disulfide interactions.

**Check **

The units of the answer (torr) are correct. The magnitude seems reasonable given the partial pressures of the pure substances.

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.7 ** **Calculating the Vapor Pressure of a Two-Component ** **Solution **

Continued

**For Practice 12.7 **

A solution of benzene (C_{6}H_{6}) and toluene (C_{7}H_{8}) is 25.0% benzene by mass. The vapor pressures of pure

benzene and pure toluene at 25 °C are 94.2 torr and 28.4 torr, respectively. Assuming ideal behavior, calculate the following:

**a. The vapor pressure of each of the solution components in the mixture. **

**b. The total pressure above the solution. **

**c. The composition of the vapor in mass percent. **

Why is the composition of the vapor different from the composition of the solution?

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**Freezing Point Depression and Boiling Point ** **Elevation **

### (FP _{solvent} – FP _{solution} ) = ΔT _{f} = *m∙K* _{f}

### (BP _{solution} – BP _{solvent} ) = ΔT _{b}

### = m∙K _{b}

### with glucose,

### which acts as an

### antifreeze.

Slide 70 of 33

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### Which of the following aqueous solutions will have the highest boiling point?

### 1. 3 molal glucose 2. 4 molal ethanol 3. 2.5 molal NaCl 4. 2.5 molal CaCl

_{2 }

### 5. Cannot tell without K

_{b}

### for water

Slide 71 of 33

Copyright © 2011 Pearson Canada Inc.

### Which of the following aqueous solutions will have the highest boiling point?

### 1. 3 molal glucose 2. 4 molal ethanol 3. 2.5 molal NaCl 4. 2.5 molal CaCl

_{2 }

### 5. Cannot tell without K

_{b}

### for water

Slide 72 of 33

Copyright © 2011 Pearson Canada Inc.

### A 2.0 molal aqueous solution of glucose (C

_{6}

### O

_{6}

### H

_{12}

### ) is found to boil at 101

^{o}

### C. What would the boiling point of a 2.0 molal solution of

### sucrose(C

_{12}

### O

_{11}

### H

_{22}

### ) be?

### glucose (C

_{6}

### O

_{6}

### H

_{12}

### )

### sucrose (C

_{12}

### O

_{11}

### H

_{22}

### ) 1. 102

^{o}

### C

### 2. 100.5

^{o}

### C 3. 101

^{o}

### C

### 4. Slightly higher than 100.5

^{o}

### C

### 5. Cannot determine without K

_{b }

Slide 73 of 33

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### A 2.0 molal aqueous solution of glucose (C

_{6}

### O

_{6}

### H

_{12}

### ) is found to boil at 101

^{o}

### C. What would the boiling point of a 2.0 molal solution of

### sucrose(C

_{12}

### O

_{11}

### H

_{22}

### ) be?

### glucose (C

_{6}

### O

_{6}

### H

_{12}

### )

### sucrose (C

_{12}

### O

_{11}

### H

_{22}

### ) 1. 102

^{o}

### C

### 2. 100.5

^{o}

### C 3. 101

^{o}

### C

### 4. Slightly higher than 100.5

^{o}

### C

### 5. Cannot determine without K

_{b }

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.8 ** **Freezing Point Depression **

*Calculate the freezing point of a 1.7 m aqueous ethylene glycol solution *

**Sort **

You are given the molality of a solution and asked to find its freezing point.

**Given: 1.7 m solution **

**Find: freezing point (from ΔT**_{f})

**Strategize **

To solve this problem, use the freezing point depression equation.

**Conceptual Plan **

**Solve **

*Substitute into the equation to calculate ΔT*_{f}.

*The actual freezing point is the freezing point of pure water (0.00 °C) − ΔT*_{f}.

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.8 ** **Freezing Point Depression **

**Solution **

**Check **

The units of the answer are correct. The magnitude seems about right. The expected range for freezing points of an aqueous solution is anywhere from −10 °C to just below 0 °C. Any answers out of this range would be suspect.

**For Practice 12.8 **

*Calculate the freezing point of a 2.6 m aqueous sucrose solution. *

Continued

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.9 ** **Boiling Point Elevation **

What mass of ethylene glycol (C_{2}H_{6}O_{2}), in grams, must be added to 1.0 kg of water to produce a solution that
boils at 105.0 °C?

**Sort **

You are given the desired boiling point of an ethylene glycol solution containing 1.0 kg of water and asked to find the mass of ethylene glycol you need to add to achieve the boiling point.

**Given: ΔT**_{b} = 5.0 °C, 1.0 kg H_{2}O
**Find: g C**_{2}H_{6}O_{2 }

**Strategize **

To solve this problem, use the boiling-point elevation equation to calculate the desired molality of the solution
**from ΔT**_{b}.

Then use that molality to determine how many moles of ethylene glycol are needed per kilogram of water.

Finally, calculate the molar mass of ethylene glycol and use it to convert from moles of ethylene glycol to mass of ethylene glycol.

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.9 ** **Boiling Point Elevation **

**Conceptual Plan **

**Relationships Used **

C_{2}H_{6}O_{2} molar mass = 62.07 g/mol
*ΔT*_{b}* = m × K*_{b} (boiling point elevation)

**Solve **

Begin by solving the boiling point elevation equation for molality and substituting the required quantities to
*calculate m. *

Continued

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.9 ** **Boiling Point Elevation **

**Solution **

**Check **

The units of the answer are correct. The magnitude might seem a little high initially, but the boiling point elevation constant is so small that a lot of solute is required to raise the boiling point by a small amount.

**For Practice 12.9 **

*Calculate the boiling point of a 3.60 m aqueous sucrose solution. *

Continued

© 2014 Pearson Education, Inc.

**K** _{f}

**K**

_{f}

© 2014 Pearson Education, Inc.

### • 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 = 0.08206 (atm∙L)/(mol∙K) *

**Π = MRT **

**Π = MRT**

**Osmotic Pressure **

### • A semipermeable

**membrane allows **

**solvent to flow **

### through it, but not

### solute.

© 2014 Pearson Education, Inc.

**• 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.

**Osmosis **

Slide 82 of 33

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### A B A B

### A B

### A B

### To the right is a diagram of a closed system containing two salt water solutions. The solution labeled A is more concentrated than the one labeled B. Which of the diagrams below best represents the system at an infinite time after preparation?

### 1. 2. 3.

Slide 83 of 33

Copyright © 2011 Pearson Canada Inc.

### A B A B

### A B

### A B

### To the right is a diagram of a closed system containing two salt water solutions. The solution labeled A is more concentrated than the one labeled B. Which of the diagrams below best represents the system at an infinite time after preparation?

### 1. 2. 3.

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.10 Osmotic Pressure **

The osmotic pressure of a solution containing 5.87 mg of an unknown protein per 10.0 mL of solution is 2.45 torr at 25 °C. Find the molar mass of the unknown protein.

**Sort **

You are given that a solution of an unknown protein contains 5.87 mg of the protein per 10.0 mL of solution. You are also given the osmotic pressure of the solution at a particular temperature and asked to find the molar mass of the unknown protein.

**Given: 5.87 mg protein **
10.0 mL solution

Π = 2.45 torr
*T = 25 °C *

**Find: molar mass of protein (g/mol) **

**Strategize **

Step 1: Use the given osmotic pressure and temperature to find the molarity of the protein solution.

Step 2: Use the molarity calculated in step 1 to find the number of moles of protein in 10 mL of solution.

Step 3: Finally, use the number of moles of the protein calculated in step 2 and the given mass of the protein in 10.0 mL of solution to find the molar mass

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.10 Osmotic Pressure **

**Conceptual Plan **

**Relationships Used **

*Π = MRT (osmotic pressure equation) *
Continued

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.10 Osmotic Pressure **

Continued

**Solve **

Step 1: Begin by solving the osmotic pressure equation for molarity and substituting in the required quantities in the correct units to calculate M.

Step 2: Begin with the given volume, convert to liters, then use the molarity to find the number of moles of protein.

Step 3: Use the given mass and the number of moles from step 2 to calculate the molar mass of the protein.

**Solution **

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.10 Osmotic Pressure **

Continued

**Check **

The units of the answer are correct. The magnitude might seem a little high initially, but proteins are large molecules and therefore have high molar masses.

**For Practice 12.10 **

Calculate the osmotic pressure (in atm) of a solution containing 1.50 g ethylene glycol (C_{2}H_{6}O_{2}) in 50.0 mL of
solution at 25 °C.

© 2014 Pearson Education, Inc.

**12.7 Colligative Properties of Strong ** **Electrolyte Solutions **

**van’t Hoff factor, i, **

**van’t Hoff factor, i,**

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.11 Van’t Hoff Factor and Freezing Point Depression **

*The freezing point of an aqueous 0.050 m CaCl*_{2}* solution is −0.27 °C. What is the van’t Hoff factor (i) for CaCl*_{2}
*at this concentration? How does it compare to the expected value of i? *

**Sort **

*You are given the molality of a solution and its freezing point. You are asked to find the value of i, the van’t Hoff *
factor, and compare it to the expected value.

**Given: 0.050 m CaCl**_{2} solution,
*ΔT*_{f} = 0.27 °C

**Find: i **

**Strategize **

To solve this problem, use the freezing point depression equation including the van’t Hoff factor.

**Conceptual Plan **

*ΔT*

_{f}

*= im × K*

_{f }

**Solve **

*Solve the freezing point depression equation for i, substituting in the given quantities to calculate its value. *

*The expected value of i for CaCl*_{2} is 3 because calcium chloride forms 3 mol of ions for each mole of calcium
chloride that dissolves. The experimental value is slightly less than 3, probably because of ion pairing.

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.11 Van’t Hoff Factor and Freezing Point Depression **

**Solution **

**Check **

*The answer has no units, as expected since i is a ratio. The magnitude is about right since it is close to the value *
you would expect upon complete dissociation of CaCl_{2}.

**For Practice 12.11 **

*Calculate the freezing point of an aqueous 0.10 m FeCl*_{3} solution using a van’t Hoff factor of 3.2.

Continued

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.12 Calculating the Vapor Pressure of a Solution ** **Containing an Ionic Solute **

A solution contains 0.102 mol Ca(NO_{3})_{2} and 0.927 mol H_{2}O. Calculate the vapor pressure of the solution at 55 °C.

The vapor pressure of pure water at 55 °C is 118.1 torr. (Assume that the solute completely dissociates.)

**Sort **

You are given the number of moles of each component of a solution and asked to find the vapor pressure of the solution. You are also given the vapor pressure of pure water at the appropriate temperature.

**Given: 0.102 mol Ca(NO**_{3})_{2}
0.927 mol H_{2}O

= 118.1 torr (at 55 C)
**Find: P**_{solution}

**Strategize **

*To solve this problem, use Raoult’s law as you did in Example 12.6. Calculate χ*_{solvent }from the given amounts of
solute and solvent.

**Conceptual Plan **

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.12 Calculating the Vapor Pressure of a Solution ** **Containing an Ionic Solute **

Continued

**Solve **

The key to this problem is to understand the dissociation of calcium nitrate. Write an equation showing the dissociation.

Since 1 mol of calcium nitrate dissociates into 3 mol of dissolved particles, the number of moles of calcium nitrate must be multiplied by 3 when computing the mole fraction.

Use the mole fraction of water and the vapor pressure of pure water to calculate the vapor pressure of the solution.

**Solution **

© 2014 Pearson Education, Inc.

*Chemistry: A Molecular Approach, 3rd Edition *
Nivaldo J. Tro

**Example 12.12 Calculating the Vapor Pressure of a Solution ** **Containing an Ionic Solute **

Continued

**Check **

The units of the answer are correct. The magnitude also seems right because the calculated vapor pressure of the solution is significantly less than that of the pure solvent, as you would expect for a solution with a significant amount of solute.

**For Practice 12.12 **

A solution contains 0.115 mol H_{2}O and an unknown number of moles of sodium chloride. The vapor pressure of
the solution at 30 °C is 25.7 torr. The vapor pressure of pure water at 30 C is 31.8 torr. Calculate the number of
moles of sodium chloride in the solution.

© 2014 Pearson Education, Inc.

### isosmotic hyperosmotic solution Conc.

### hyposmotic solution Dilu.

### 0.9 g NaCl per

### 100 mL of solution.

### 1 g NaCl per

### 100 mL of solution.

### 0.5 g NaCl per

### 100 mL of solution.

Slide 95 of 33

Copyright © 2011 Pearson Canada Inc.

20 g C_{3}H_{6}O_{3 } 10 g C_{6}H_{12}O_{6 }

### Two aqueous solutions of equal volume, one containing 20 g of lactic acid (C

_{3}

### H

_{6}

### O

_{3}

### ) and one containing 10 g of glucose (C

_{6}

### H

_{12}

### O

_{6}

### ) are separated by a semi permeable membrane (allowing only H

_{2}

### O to pass). The net flow of water through the membrane is?

### 1. Left 2. Right

### 3. There is no net flow.

### 4. Cannot tell without knowing the

### exact volume of the

### solutions.

Slide 96 of 33

Copyright © 2011 Pearson Canada Inc.

20 g C_{3}H_{6}O_{3 } 10 g C_{6}H_{12}O_{6 }

### Two aqueous solutions of equal volume, one containing 20 g of lactic acid (C

_{3}

### H

_{6}

### O

_{3}

### ) and one containing 10 g of glucose (C

_{6}

### H

_{12}

### O

_{6}

### ) are separated by a semi permeable membrane (allowing only H

_{2}

### O to pass). The net flow of water through the membrane is?

### 1. Left 2. Right

### 3. There is no net flow.

### 4. Cannot tell without knowing the

### exact volume of the

### solutions.

© 2014 Pearson Education, Inc.

### • Solutions = homogeneous

### • Suspensions = heterogeneous, separate on standing

### • Colloids = heterogeneous, do not separate on standing

### – Particles can coagulate

### – Cannot pass through semipermeable membrane – Hydrophilic

### • Stabilized by attraction for solvent (water)

### – Hydrophobic

### • Stabilized by charged surface repulsions

### • Show the Tyndall effect and Brownian motion.

**12.8 Colloids **

**Mixtures **

© 2014 Pearson Education, Inc.

**The Tyndall Effect **

© 2014 Pearson Education, Inc.

**Brownian Motion **

© 2014 Pearson Education, Inc.