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Chapter 11 – Solutions 11.1 (a) saturated solution; (b) aqueous solution; (c) molarity; (d) solvent; (e) entropy; (f) solubility; (g) parts per

million; (h) miscible; (i) osmosis; (j) colligative property; (k) percent by volume; (l) mass/volume percent 11.2 (a) supersaturated solution; (b) solution; (c) molality; (d) solute; (e) ion-dipole force; (f) concentration; (g)

parts per billion; (h) unsaturated solution; (i) semipermeable membrane; (j) titration; (k) percent by mass 11.3 The solute is the substance that is dissolved in the solution. The solvent is the substance that dissolves the

solute. Usually, the solute is the substance that is present in the lesser amount, and the solvent is the substance present in the greater amount in the solution.

Solute Solvent (a) sodium chloride (and other salts) water (b) carbon iron (c) oxygen (O2) water

11.4 The solute is the substance that is dissolved in the solution. The solvent is the substance that dissolves the

solute. Usually, the solute is the substance that is present in the lesser amount, and the solvent is the substance present in the greater amount in the solution.

Solute Solvent (a) the drug water (b) ethanol, water ethyl acetate (“mostly”) (c) CO2 water

11.5 Strong electrolytes are substances that fully ionize when dissolved in solution. For example, when sodium

chloride dissolves in water, it completely dissociates into sodium ions and chloride ions. Strong acids, strong bases, and soluble ionic compounds are all strong electrolytes. Most molecular compounds (with the notable exception of strong acids) do not dissociate in water and are not electrolytes (they are nonelectrolytes).

11.6 Generally, molecular compounds are nonelectrolytes (with the exception of strong acids). Most molecular

substances do not form ions when they dissolve in water; rather, they keep their molecular form. As a result, these compounds do not form conductive solutions.

11.7 The solute appears to be a molecular compound, and is likely a nonelectrolyte. If the solute were an

electrolyte, we would expect to see cations and anions formed from the ionization of the substance. However, this solution appears to contain only one type of particle.

11.8 The solute appears to be an electrolyte. When an electrolyte dissolves, it dissociates into its ions.

Nonelectrolyte molecular compounds remain in their molecular forms. 11.9 Acids, bases, and soluble salts are electrolytes because they dissociate (or ionize) in water. (a) HBr is an

acid that produces H+(aq) and Br−(aq) when it dissolves in water. (b) NH4Cl is a soluble salt that dissociates into NH4

+(aq) and Cl−(aq) when it dissolves in water. (c) Butanol is a molecular compound, so it does not form ions but retains its molecular form (CH3CH2CH2CH2OH(aq)) in water. Note: Even though butanol has an −OH group, it is not a base. Most bases are ionic compounds composed of metal ions and hydroxide (OH−) ions.

11.10 Acids, bases, and soluble salts are electrolytes because they dissociate (or ionize) in water. (a) LiOH is a

soluble ionic compound that produces Li+(aq) and OH−(aq) when it dissolves in water. (b) O2 is a

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molecular compound, so it retains its molecular form, O2(aq), when it dissolves in water. (c) Mg(NO3)2 is a soluble ionic compound that forms Mg2+(aq) and NO3

−(aq) when it dissolves in water. 11.11 When we count the spheres, we discover a two to one ratio of Cl− to Mg2+ in image A. Magnesium

chloride is a soluble salt, so it will produce one Mg2+ (the smaller spheres) and two Cl− (larger spheres) for each formula unit that dissolves. Images B and C are incorrect because they do not show the ions formed by MgCl2 in solution.

11.12 Sodium sulfate is a soluble salt. When it dissolves it forms two Na+(aq) ions and one SO4

2–(aq) ions. When we count the spheres, we find that there is a two-to-one ratio of sodium ions to sulfate ions in image B. Image A is incorrect because there is no ionization. Image C is incorrect because it shows sodium as diatomic ions.

11.13 Water-soluble ionic compounds separate into ions when they dissolve in water. Each ion is surrounded

with water molecules, a situation we indicate with the (aq) designation, which stands for aqueous. When water-soluble molecular substances (other than strong acids) dissolve, water molecules surround them in their entirety, something we indicate with the (aq) designation. (a) Ca(OH)2(s) 2H O→ Ca2+(aq) + 2 OH−(aq) (b) N2(g) 2H O→ N2(aq) (c) CH3OH(l) 2H O→ CH3OH(aq)

11.14 Water-soluble ionic compounds separate into ions when they dissolve in water. Each ion is surrounded

with water molecules, a situation we indicate with the (aq) designation, which stands for aqueous. When water-soluble molecular substances (other than strong acids) dissolve, water molecules surround them in their entirety, something we indicate with the (aq) designation. (a) HCl(g) → H+(aq) + Cl−(aq) (b) Cl2(g) → Cl2(aq) (c) Mg(NO3)2(s) → Mg2+(aq) + 2NO3

−(aq) 11.15 Attractions in both the solute and the solvent must be disrupted before a solute can dissolve in a solvent.

This means that because the solute is an ionic compound, the ionic bonds in the crystal lattice must break so the cations and anions can separate, and both the dipole-dipole and hydrogen bonding interactions in the water must be overcome. Finally, ion-dipole interactions form in the solution as the water molecules surround (solvate) the ions in the solution.

11.16 The dipole-dipole intermolecular forces in the polar molecular substance must break so the molecules can

separate from each other, and both the dipole-dipole and hydrogen bonding interactions in the water must also be overcome. As the polar solute dissolves, new dipole-dipole interactions between the solute and solvent molecules form.

11.17 The London dispersion forces in both the solute and the solvent must be overcome. The forces attracting

the solvent to the solute in the solution are also London dispersion forces. 11.18 Entropy is the natural tendency for the energy of matter to become more randomly distributed. Sometimes,

entropy drives the solution process, which is the case when nonpolar substances experience a slight solubility in water.

11.19 Because the solution cools, we know that energy is absorbed as the solution forms. This means that more

energy is required to overcome the attractions between the NH4+ and NO3

− ions in the solute and between the water molecules (the solvent) than is released when the water molecules surround (solvate) the ions to form the solution. Therefore, we can say that the relative strengths of the attractive forces between the solute and solvent are weaker than the attractive forces between particles of the pure substances.

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11.20 Because the solution warms, we know that energy is released as the solution forms. This means that less energy is required to overcome the attractions between the Na+ and OH− ions in the solute and between the water molecules (the solvent) than is released when the water molecules surround (solvate) the ions to form the solution. Therefore, we can say that the relative strengths of the attractive forces between the solute and solvent are stronger than the attractive forces between particles of the pure substances.

11.21 Energy and entropy changes drive the formation of solutions (Figure 11.11). When the intermolecular

forces which act between the solute and solvent are stronger than the intermolecular forces within the pure solvent and solute, the formation of the solution produces energy and solution formation is energetically favorable. Solution formation is also favored when the dissolving process creates more random distribution of energy.

11.22 Entropy is a measure of the tendency for the energy of matter to become more disordered or randomly

distributed. For the formation of most solutions, entropy increases as the solute dissolves into the solvent and both the solute and solvent particles become more random in their distribution. Pure substances have relatively high degrees of order, so their entropies are lower in the pure state than when they are mixed.

11.23 The intermolecular attractions between water molecules and between oil molecules are not the same

because water is polar and cooking oil is nonpolar. Water molecules are attracted to substances that are ionic or polar in nature and oil molecules are attracted to nonpolar substances. Because the intermolecular forces are so dissimilar (they are not “like” in the sense that “like dissolves like”) oil and water are immiscible.

11.24 Iodine is nonpolar. Because, like most hydrocarbons, hexane is also nonpolar, iodine is readily soluble in

it. Water molecules are very polar and form hydrogen bonds. Iodine does not dissolve in water because water is not strongly attracted to the iodine molecules.

11.25 The molecules of both substances (ethanol CH3CH2OH; water H2O) are polar and can form hydrogen

bonds. Substances that experience similar intermolecular forces tend to be soluble in each other because their molecules are mutually attracted to one another.

11.26 Substances that dissolve in water are usually ionic or polar. Because it does not dissolve in water, grease

must be nonpolar. 11.27 To apply the “like dissolves like” rule, we must first determine the types of intermolecular forces that exist

between molecules or formula units of each substance. Water molecules are polar and form hydrogen bonds. Substances that are polar, form hydrogen bonds, or are soluble ionic compounds are soluble in water. (a) Benzene, a hydrocarbon, is nonpolar so it does not dissolve in water. (b) From its formula we can tell that ethylene glycol has −OH groups, which allow it to form hydrogen bonds. Because ethylene glycol and water are both polar and form hydrogen bonds, ethylene glycol is soluble in water. (c) Potassium iodide is ionic and, based on the solubility rules, dissolves in water. Potassium iodide is soluble in water.

11.28 To apply the “like dissolves like” rule, we must first determine the types of intermolecular forces that exist

between molecules or formula units of each substance. Hexane, a hydrocarbon, experiences primarily London dispersion forces. Substances that experience London dispersion forces are soluble in hexane. Substances that are polar, form hydrogen bonds, or are ionic are not soluble in hexane. (a) Sodium nitrate is an ionic compound, so it is not soluble in hexane. (b) Methanol is a polar substance that forms hydrogen bonds (because of the –OH group) and will not dissolve in hexane. (c) Bromine is a nonpolar substance and is soluble in hexane.

11.29 Compared to other types of interactions, ion-ion interactions are very strong so it takes considerable energy

to separate ions from one another. Because nonpolar solvents are not strongly attracted to ions, not enough energy is released, nor entropy gained, through formation of ion-solvent interactions to cause the dissolution process to be favorable.

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11.30 It takes a relatively large amount of energy to disrupt the strong intermolecular forces experienced by polar solvents. Because polar solvents are not strongly attracted to nonpolar particles, not enough energy is released, nor entropy gained, through formation of the solution of a nonpolar solute in a polar solvent to make the process favorable.

11.31 The ions in NaCl are attracted by ionic bonding. Water molecules are attracted by dipole-dipole forces and

hydrogen bonding. Although disrupting the attractive forces in this solute and solvent requires considerable energy, formation of the ion-dipole interactions between solute and solvent particles releases enough energy to allow the ionic compound to dissolve.

11.32 Nonpolar substances (in this case, both the solute and solvent particles) are attracted by relatively weak

London dispersion forces. The solute-solvent attractions in a solution of these substances are also London dispersion forces. As a result, the amount of energy required to separate the solute and solvent molecules is about the same as the amount of energy released during the formation of the solution. The greater entropy of the solution compared with the entropies of the pure solvent and solute favors the process of solution formation.

11.33 The solubility of gases, such as O2, decreases as temperature increases. This happens because the increased

kinetic energy of the nonpolar gas molecules in the warmer solution [e.g. O2(aq)] disrupts their weak interactions with the solvent, allowing them to escape more easily into the gas phase.

11.34 Thermal pollution of water is the artificial heating of bodies of waters (rivers, lakes, oceans). One

consequence of this type of pollution is a decrease in the concentration of dissolved gases. Thermal water pollution occurs when river (or lake or ocean) water is used to cool power plants. When the water is returned to the lake or river, its temperature is higher than it originally was, so the concentration of dissolved oxygen is diminished. The reduced oxygen level, coupled with the increased temperature, can adversely affect aquatic life.

11.35 Gas solubility decreases as the partial pressure of the gas over the solution decreases. At higher elevations,

while the percent composition of the atmosphere is relatively unchanged, the atmospheric pressure is lower. This lower pressure results in lower partial pressures of oxygen and nitrogen. Because the solubility of these gases is directly proportional to their pressures above the solution, the solubilities of oxygen and nitrogen go down as altitude increases.

11.36 A diver experiences increasing pressures at increasing depths underwater. Because the solubility of blood

gases is directly proportional to the pressure of those gases above the solution (in this case the diver’s blood), the concentration of dissolved gases increases as the diver swims to lower depths of the ocean.

11.37 Divers get what is known as the bends when they ascend from a dive too rapidly. The rapid decrease in the

solubility of their blood gases results in the formation of air bubbles in the divers’ blood streams. Besides being very painful, the presence of these air bubbles leads to death. Putting divers in pressurized chambers causes the insoluble gases to re-dissolve into their blood. Then, as the chamber pressure slowly decreases (decompression), the divers can exhale the gases (primarily nitrogen) from their lungs.

11.38 Before a carbonated beverage is opened, there is a high partial pressure of CO2 in the space above the

carbonated beverage. This increases the solubility of CO2 in the solution. When the container is opened, the partial pressure of CO2 rapidly decreases. This causes a rapid decrease in the solubility of CO2 in the solution and bubbles appear in the liquid. As the solution goes “flat”, additional fizz is generated by the reaction:

H2CO3(aq) → H2O(l) + CO2(g)

When we open a bottle of carbonated beverage, the gas that escapes is primarily CO2 (there is also some water vapor). As the carbonic acid, H2CO3, and CO2 concentration in the beverage decreases, the beverage slowly goes flat.

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11.39 At higher temperatures, the solubility of gases decreases so fewer molecules remain in solution.

11.40 After the partial pressure of O2 above the solution increases, the concentration of dissolved oxygen will

increase.

11.41 We use the water solubility of KCl, 34.0 g per 100 g water, as a conversion factor for calculating the

amount of KCl that can dissolve into any mass of water at 20oC. We can express this conversion factor in two different ways:

2

34.0 g KCl100 g H O

and 2100 g H O34.0 g KCl

Because we are given the mass of water, we choose the expression that allows the units ‘g H2O’ to cancel and gives us the number of grams of KCl.

Mass KCl = 250.0 g H O2

34.0 g KCl100 g H O

× = 17.0 g KCl

11.42 We use the water solubility of NH4Cl, 41.1 g per 100 g water, as a conversion factor for calculating the

amount of NH4Cl that can dissolve into any mass of water at 30oC. We can express this conversion factor in two different ways:

4

2

41.1 g NH Cl100 g H O

and 2

4

100 g H O41.1 g NH Cl

Because we are given the mass of water, we choose the expression that allows the units ‘g H2O’ to cancel and gives us the number of grams of NH4Cl.

Mass NH4Cl = 2455.0 g H O 4

2

41.1 g NH Cl100 g H O

× = 187 g NH4Cl

11.43 On a macroscopic level, we can’t dissolve more solute in a saturated solution, but we can dissolve more

solute in an unsaturated solution. A saturated solution contains the maximum amount of solute that can be dissolved at a particular temperature, while an unsaturated solution contains less than this amount of dissolved solute. On a molecular level, the ions in a saturated solution are in equilibrium with the solid. This means that the solid is dissolving at a rate equal to the rate at which it is forming from the ions in the solution. In an unsaturated solution, none of the solid is present.

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11.44 A saturated solution contains a concentration of solute equal to the solubility of the solute at the solution temperature. A supersaturated solution contains a greater solute concentration than is specified by the solubility (Figure 11.23). Supersaturated solutions are generally unstable, and the excess solute will separate out if we disturb them. After this separation, the remaining solution is saturated.

11.45 We could add solute a little at a time to the solvent until no more solid dissolves and we see traces of

undissolved solute in the solution. At this point, the dissolved solute will be in equilibrium with the undissolved solute in the saturated solution.

11.46 If we dissolve the ammonium chloride at an elevated temperature, we can dissolve more solid than is

soluble at room temperature. Then, if we carefully cool the solution, none of the solid will precipitate and we will have a solution in which the solute is more concentrated than is indicated by its solubility at room temperature.

11.47 Because the solubility of Ca(OH)2 is 0.15 g/100g of water, only 0.15 g of the 1.00 g sample will dissolve

when we add it to the water (assuming the water temperature remains at 30oC). The resulting solution will be saturated, and 0.85 g of Ca(OH)2(s) will remain undissolved.

11.48 Because 0.15 g Ca(OH)2 is less Ca(OH)2 than can dissolve in 100.0 g H2O (assuming the temperature

remains at 10oC), all of the solid will dissolve, producing an unsaturated solution. 11.49 Percent by mass is:

Percent by mass = g solute 100%g solution

×

Note that the mass of the solution represents the mass of the solute plus the mass of the solvent.

Grams solute = 15.0 g NaCl

Grams solution = 15.0 g NaCl + 90.0 g H2O = 105.0 g

Percent by mass = 15.0 g

105.0 g100%× = 14.3%

11.50 Percent by mass is:

Percent by mass = g solute 100%g solution

×

Note that the mass of the solution represents the mass of the solute plus the mass of the solvent.

Grams solute = 25.0 g sucrose

Grams solution = 25.0 g sucrose + 100.0 g H2O = 125.0 g

Percent by mass = 25.0 g

125.0 g100%× = 20.0%

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11.51 We can write two different relationships that represent a 15.0% KI solution: 15.0 g KI

100 g solution and 100 g solution

15.0 g KI

Because we are given the mass of the solution, we choose the conversion that allows us to cancel ‘g of solution’ and gives us ‘g KI.’

Grams of KI = 700.0 g solution 15.0 g KI100 g solution

× = 105 g KI

11.52 We can write two different relationships that represent a 20.0% Na3PO4 solution:

3 420.0 g Na PO100 g solution

and 3 4

100 g solution20.0 g Na PO

Because we are given the mass of the solution, we choose the conversion that allows us to cancel ‘g solution’ and gives us ‘g Na3PO4.’

Grams of Na3PO4 = 255 g solution 3 420.0 g Na PO100 g solution

× = 51.0 g Na3PO4

11.53 To calculate mass percent of acetic acid in a solution, we need to know the mass of acetic acid (54.50 g)

and the mass of solution. We use the density of the solution and its volume to calculate the mass of solution, and then calculate the mass percent of acetic acid in the solution.

Volume in mL = 1.000 L 1000 mL×

1 L1.005 g1 mL

× = 1005 g

Percent by mass = 54.50 gg solute 100%

g solution× =

1005 g100%× = 5.423%

11.54 To calculate mass percent of H3PO4 in a solution, we need to know the mass of phosphoric acid (88.20 g)

and the mass of the solution. We use the density of the solution and its volume to calculate the mass of solution, and then calculate the mass percent of H3PO4 in the solution.

Volume in mL = 100.0 mL 1.40 g1 mL

× = 140. g

Percent by mass = 88.20 gg solute 100%

g solution× =

140. g100%× = 63.0%

11.55 We often express the concentrations of solutions composed of liquid solutes as percent by volume. We do

this primarily for convenience, because we usually measure liquids by volume. 11.56 Solution volumes are often not additive, so we can’t assume 1 L + 1 L = 2 L. As a result, we must measure

the final solution volume. For example, when we mix 500 mL of pure ethyl alcohol with 500 mL of pure water, the resulting solution has a volume of about 900 mL rather than 1000 mL that we would have predicted. This volume difference is the result of the relative strengths of intermolecular attractions within the two pure liquids and of the solvent-solute interactions in the solution.

11.57 Percent by volume is:

Percent by volume = volume solute 100%volume solution

×

Percent by volume = 35.0 mL115.0 mL

100%× = 30.4%

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11.58 Percent by volume is:

Percent by volume = volume solute 100%volume solution

×

Percent by volume = 2.0 mL40.0 mL

100%× = 5.0%

11.59 (a) Since the EPA has established the safe drinking level for TCE to be 5 ppb, we need to calculate the

concentration of TCE in the sample and see if it falls below this level. If we find that it is below, the water is considered safe to drink. The concentration, in ppb, is calculated using the formula:

Parts Per Billion (ppb) = 9mass solute 10mass solution

×

Technically, the mass of solution is the mass of the solvent plus the mass of the solute. However, since the mass of the solute is so small, the solute does not have a significant effect on the total mass:

msolution = 200.0 kg + 0.43 mg = 200.0 kg + 0.00000043 kg (not enough to make a difference!)

msolution = 200.0 kg

The mass of TCE needs to be expressed in kg so that the units will cancel properly:

mTCE = 310 g

0.43 mg−

×1 mg 3

1 kg10 g

× = 4.3 × 10–7 kg

Parts per billion (ppb) = 7

94.3 10 kg 10200.0 kg

−×× = 2.2 ppb

(b) The concentration is less than 5 ppb, so the water is safe to drink.

11.60 The formula for ppb is given by:

Parts per billion (ppb) = 9mass solute 10mass solution

×

Rearranging we find the mass of solute as:

mass solute (kg) = 9ppb mass solution

10× = 9

190 65 kg10× = 1.235 × 10–5 kg, rounded off to 1.2 × 10–5 kg

mass solute (g) = 51.2 10 kg−×310 g

kg× = 1.2 × 10–2 g

11.61 The molarity of a solution is the number of moles of solvent dissolved in 1 L of the solution. Molarity is a useful conversion between the number of moles of solute and the volume of solution in liters. For a 0.15 M HCl solution, we can write:

0.15 mol HCl1 L

and 1 L0.15 mol HCl

Because the solution volume in this case is given in milliliters (100.0 mL), we must express it in liters before we can use it in our calculation.

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Volume in liters = 100.0 mL 1 L1000 mL

× = 0.100 L

Moles of HCl = 0.100 L 0.15 mol HCl1 L

× = 0.015 mol HCl

11.62 The molarity of a solution is the number of moles of solute dissolved in 1 L of solution. For a 0.25 M

NaOH solution we can write:

0.25 mol NaOH1 L

or 1 L0.25 mol NaOH

Because we are looking for the volume of solution that will contain 0.050 mol NaOH, we select the expression that allows us to cancel ‘mol of NaOH’ and gives us ‘L’.

Moles of NaOH = 0.050 mol NaOH 1 L0.25 mol NaOH

× = 0.20 L

11.63 To calculate molality we need to know the number of moles of solute and the mass of the solvent (not the

solution) in kilograms. We determine the number of moles of solute from the mass (20.5 g NaCl) and molar mass of NaCl (58.44 g/mol).

Moles of NaCl = 20.5 g NaCl 1 mol NaCl58.44 g NaCl

× = 0.351 mol NaCl

Because we know the mass of the solution, we need to calculate the mass of solvent in kilograms:

Mass of solvent = 166.2 g solution − 20.5 g solute = 145.7 g solvent

Mass in kilograms = 145.7 g solvent 1 kg solvent1000 g solvent

× = 0.1457 kg solvent

Molality of solution = moles solute 0.351 mol NaClmass of solvent (kg) 0.1457 kg

= = 2.41 m

11.64 We know the molality of the solution (0.85 m) and its mass (50.0 g). Molality expresses the number of

moles of solute per kilogram of solvent. We are given the mass of the solution, and need to determine the moles of solute. From the solution concentration we know that there are 0.85 mol of C12H22O11 (342.97 g/mol) in one kilogram of solvent. We can express this number of moles, using the molar mass of sucrose:

Mass of solute = 12 22 110.85 mol C H O 12 22 11

12 22 11

342.97 g C H O1 mol C H O

× = 2.9 × 102 g C12H22O11

This means that there are 2.9 × 102 g C12H22O11 (0.85 mol) for every 1290 grams of solution (1000 g + 290 g).

Moles of solute in 50.0 g solution = 50.0 g solution 12 22 110.85 mol C H O1290 g solution

× = 0.033 mol C12H22O11

11.65 When potassium nitrate dissolves, it produces two ions:

KNO3(s) → K+(aq) + NO3–(aq)

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Because there are twice as many ions as there are KNO3 formula units, the ion concentration is two times the molarity, or 3.0 M. Going about this another way, we can also calculate the concentration of ions as shown below:

Moles of ions per liter = 31.5 mol KNO

3

2 mol ions1 L 1 mol KNO

× = 3.0 mol ions1 L

= 3.0 M ions

Similarly, the molal concentration of ions is 3.0 m.

Moles of ions per kilogram solvent = 31.5 mol KNO

3

2 mol ions1 kg 1 mol KNO

× = 3.0 mol ions1 kg

= 3.0 m ions

11.66 When magnesium chloride dissolves, it produces three ions:

MgCl2(s) → Mg2+(aq) + 2Cl–(aq)

Because there are three times as many ions as there are MgCl2 formula units, the ion concentration is three times the molarity, or 7.5 M. Going about this another way, we can also calculate the concentration of ions as shown below:

Moles of ions per liter = 22.5 mol MgCl

2

3 mol ions1 L 1 mol MgCl

× = 7.5 mol ions1 L

= 7.5 M ions

Similarly, the molal concentration of ions is 7.5 m.

Moles of ions per kilogram solvent = 32.5 mol KNO

3

3 mol ions1 kg 1 mol KNO

× = 7.5 mol ions1 kg

= 7.5 m ions

11.67 We determine the percent (or parts per hundred), parts per million (ppm), and parts per billion (ppb)

concentrations of a solution as shown:

Percent by mass = mass solute 100%mass solution

×

Parts per million = 6mass solute 10 ppmmass solution

× Parts per billion = 9mass solute 10 ppbmass solution

×

The mass units in the equation must be the same. Because the mass of solute (arsenic) is given as 2.4 mg, we convert milligrams to grams before we calculate the concentrations.

Mass in grams = 2.4 mg 1 g1000 mg

× = 2.4 × 10−3 g

Percent by mass = 32.4 10 g−×

250.0 g100 %× = 9.6 × 10−4%

Parts per million = 32.4 10 g−×

250.0 g610 ppm× = 9.6 ppm

Parts per billion b = 32.4 10 g−×

250.0 g910 ppb× = 9600 ppb

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11.68 We determine the percent (or parts per hundred), parts per million (ppm), and parts per billion (ppb) concentrations of a solution as shown:

Percent by mass = 2mass solute 10 %mass solution

×

Parts per million = 6mass solute 10 ppmmass solution

× Parts per billion = 9mass solute 10 ppbmass solution

×

The mass units in the equation must be the same. Because the mass of solute (chromium) is given as 37 mg, we convert milligrams to grams before we calculate the concentrations.

Mass in grams = 37 mg 1 g1000 mg

× = 3.7 × 10−2 g

Percent by mass = 23.7 10 g−×

375.0 g100 %× = 9.9 × 10−3%

Parts per million = 23.7 10 g−×

375.0 g610 ppm× = 99 ppm

Parts per billion = 23.7 10 g−×

375.0 g910 ppb× = 9.9 × 104 ppb

11.69 (a) A solution that is 0.0090 ppm in lead contains 0.0090 g of lead in every one million (1 × 106) grams of

solution. To express this concentration in ppb, we assume that we have 1×106 g of solution:

Parts per billion = 9mass solute 10 ppbmass solution

× = 0.0090 g

610 g910 ppb× = 9.0 ppb

The water is safe to drink because its lead concentration, 9.0 ppb, is lower than the EPA established limit of 15.0 ppb.

(b) From part (a) we know that 1× 106 g of water contains 0.0090 grams of lead. The density of water is 1.0 g/mL, so we can say that this mass of water represents 1 × 106 mL of water. To calculate the number of milligrams of lead per milliliter of water, we first convert the mass of lead to milligrams

mg lead = 20.0090 g Pb +2

2

1000 mg Pb

1 g Pb

+

+× = 9.0 mg

Concentration in mg/mL = 2

69.0 mg Pb

10 mL

+

= 9.0 × 10−6 mg/mL

(c) Mass of lead in 100.0 mL = 100.0 mL69.0 10 mg

mL

−×× = 9.0 × 10−4 mg

(d) To calculate the number of moles of lead ion in 100.0 mL of water, we convert the number of milligrams from part (c) to grams, and use the molar mass of lead (207.2 g/mol) to calculate the number of moles of lead in 100.0 mL of water.

Moles lead = 4 29.0 10 mg Pb− +×21 g Pb +

×21000 mg Pb +

2

2

1 mol Pb

207.2 g Pb

+

+× = 4.3 × 10−9 mol Pb2+

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11.70 (a) A solution that is 95 ppb in nitrate ions contains 95 grams of nitrate ions in every billion (1 × 109) grams of solution. To express this concentration in ppm, we assume that we have 1 × 109 g of solution:

Parts per million = 6mass solute 10 ppmmass solution

× = 95 g

910 g610 ppm× = 0.095 ppm

The water is safe to drink because its nitrate ion concentration, 0.095 ppm, is well below the EPA established limit of 10.0 ppm.

(b) From part (a) we know that 1 × 109 g of water contains 95 grams of nitrate. The density of water is 1.0 g/mL, so we can say that this mass of water represents 1 × 109 mL of water. To calculate the number of milligrams of nitrate ions per milliliter of water, we first convert the mass of nitrate to milligrams:

mg nitrate = 395 g NO − 3

3

1000 mg NO

1 g NO

−

−× = 9.5 × 104 mg

Concentration in mg/mL = 4

39

9.5 10 mg NO10 mL

−× = 9.5 × 10−5 mg/mL

(c) Mass of nitrate in 100.0 mL = 100.0 mL59.5 10 mg

mL

−×× = 9.5 × 10−3 mg

(d) To determine the number of moles of NO3− in 100.0 mL of water, we convert the milligrams of NO3

− from part (c) to grams, and use the molar mass of nitrate (62.01 g/mol) to calculate the number of moles of nitrate in 100.0 mL of water.

Moles nitrate = 339.5 10 mg NO− −× 31 g NO −

×31000 mg NO −

3

3

1 mol NO

62.01 g NO

−

−× = 1.5 × 10−7 mol nitrate

11.71 (a) For problems involving solution concentrations, we begin with the solution for which we have both a

volume and concentration (the Pb2+ solution), and use the following problem solving map: 2

2 2 21000 mL 1 L Pb mole ratio ImL Pb L Pb mol Pb mol I L I+ −

+ + + − −=→ → → →M M

Volume I− = 2100.0 mL Pb +21 L Pb +

×21000 mL Pb +

20.15 mol Pb +

×21 L Pb +

2 mol I−×

21 mol Pb +

1 L I

1.0 mol I

−

−× = 0.030 L I−

Because there is one mole of I− per mole of KI, we need 0.030 L of 1.0 M KI.

(b) We can use a problem solving map similar to the one we used in (a) to determine the mass of PbI2 that forms.

22 2 2 2

2 2PbI1000 mL 1 L Pb mole ratiomL Pb L Pb mol Pb mol PbI g PbI

++ + +=→ → → →

MMM

Mass PbI2 = 2100.0 mL Pb +21 L Pb +

×21000 mL Pb +

20.15 mol Pb +

21 L Pb +

21 mol PbI×

21 mol Pb +2

2

461.0 g PbI1 mol PbI

× = 6.9 g PbI2

11.72 (a) For problems involving solution concentrations, we begin with the solution for which we know the

volume and concentration (the HCl solution) and follow the problem solving map below:

33 3

AgNO1000 mL 1 L HCl mole ratiomL HCl L HCl mol HCl mol AgNO L AgNO=→ → → →MM

11−13

Volume AgNO3 = 250.0 mL HCl1 L HCl

×1000 mL HCl

0.100 mol HCl×

1 L HCl31 mol AgNO

×1 mol HCl

3

3

1 L AgNO1.20 mol AgNO

× = 0.0208 L

(b) We determine the mass of AgCl that forms by using a problem solving map similar to the one we used in part (a):

1000 mL 1 L HCl Mole ratio AgClmL HCl L HCl mol HCl mol AgCl g AgCl=→ → → →M MM

Mass AgCl = 250.0 mL HCl1 L HCl

×1000 mL HCl

0.100 mol HCl1 L HCl

1 mol AgCl×

1 mol HCl143.35 g AgCl1 mol AgCl

× = 3.58 g AgCl

11.73 The balanced chemical equation is:

Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl(aq)

We begin by determining the number of moles of CaCl2 present, and then determine the number of moles NaCO3 necessary to react with the CaCl2:

22 2 2 2 3

CaCl1000 mL 1 L mole ratiomL CaCl L CaCl mol CaCl mol Na CO=→ → →M

Moles Na2CO3 = 2850.0 mL CaCl 21 L CaCl×

21000 mL CaCl20.35 mol CaCl

×21 L CaCl

2 3

2

1 mol Na CO1 mol CaCl

× = 0.30 mol Na2CO3

11.74 The balanced chemical equation is:

Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl(aq)

We begin by determining the number of moles of Na2CO3 present, and then determine the number of moles of CaCl2 necessary to react with the Na2CO3:

2 32 3 2 3 2

Na CO mole ratioL Na CO mol Na CO mol CaCl→ →M

Moles CaCl2 = 2 32.5 L Na CO 2 30.75 mol Na CO×

2 31 L Na CO2

2 3

1 mol CaCl1 mol Na CO

× = 1.9 mol CaCl2

11.75 The balanced equation for the complete neutralization of H2SO4 is:

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

We use the following problem solving map to calculate the volume of NaOH required for the neutralization:

2 42 4 2 4

H SO mole ratio NaOHL H SO mol H SO mol NaOH L NaOH→ → →M M

First, we must express the volume of H2SO4 solution in liters:

Volume in Liters = 2 420.00 mL H SO 2 4

2 4

1 L H SO1000 mL H SO

× = 0.02000 L H2SO4

Volume NaOH = 2 40.02000 L H SO 2 40.1500 mol H SO×

2 41 L H SO2 mol NaOH

×2 41 mol H SO

1 L NaOH0.1050 mol NaOH

× = 0.05714 L

We need 0.05714 L or 57.14 mL of 0.1050 M NaOH for the complete neutralization of the acid.

11−14

11.76 The balanced reaction for the complete neutralization of H2SO4 is:

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

We use the following problem solving map to calculate the volume of H2SO4 required for the neutralization:

2 42 4 2 4

H SONaOH mole ratioL NaOH mol NaOH mol H SO L H SO→ → →MM

First, we must express the volume of NaOH solution in liters:

Volume in Liters = 20.00 mL NaOH 1 L NaOH1000 mL NaOH

× = 0.02000 L NaOH

Volume H2SO4 = 0.02000 L NaOH0.1033 mol NaOH

×1 L NaOH

2 41 mol H SO×

2 mol NaOH2 4

2 4

1 L H SO0.850 mol H SO

× = 0.00122 L

We need 0.00122 L or 1.22 mL of 0.850 M H2SO4 for the neutralization. 11.77 The balanced equation for the complete neutralization of H2SO4 is:

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

To calculate the molarity of the H2SO4 solution, we first need to determine the number of moles of H2SO4 (the volume is given as 20.00 mL) present. Then we can calculate the molarity of the H2SO4 solution.

2 4NaOH mole ratioL NaOH mol NaOH mol H SO→ →M

The volume of NaOH is 35.77 mL or 0.03577 L.

Moles H2SO4 = 0.03577 L NaOH0.9854 mol NaOH

×1 L NaOH

2 41 mol H SO2 mol NaOH

× = 0.01762 mol H2SO4

To calculate the molarity of H2SO4 we divide the number of moles by the volume of the solution, in liters (20.00 mL or 0.02000 L).

Molarity = 2 40.01762 mol H SO0.02000 L

= 0.8812 M H2SO4

11.78 The balanced equation for the complete neutralization of H2SO4 is:

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

To calculate the molarity of the NaOH solution, we first need to determine the number of moles of NaOH (the volume is given as 44.68 mL). Then we can calculate the molarity of the NaOH solution.

2 42 4 2 4

H SO mole ratioL H SO mol H SO mol NaOH→ →M

The volume of H2SO4 is 20.00 mL or 0.02000 L.

Moles NaOH = 2 40.02000 L H SO 2 41.005 mol H SO×

2 41 L H SO 2 4

2 mol NaOH1 mol H SO

× = 0.04020 mol NaOH

To calculate the molarity of NaOH solution, we divide the number of moles of NaOH by the volume of solution, in liters (44.68 mL or 0.04468 L).

Molarity = 0.04020 mol NaOH0.04468 L

= 0.8997 M NaOH

11−15

11.79 We base the calculations for each part of this problem on the problem solving map below:

acid mole ratio with hydroxide NaOHL acid mol acid mol NaOH L NaOH→ → →M M

We determine the mole ratio from the chemical equation. We can take a shortcut if we realize that each mole of hydrogen ion we neutralize consumes one mole of NaOH. This means that 1 mol of HCl or of HNO3 reacts with 1 mol of NaOH; and 1 mol of H3PO4 reacts with 3 mol of NaOH. We also need to remember to convert volumes to liters at the beginning of the each calculation.

(a) Volume NaOH = 0.01000 L HCl0.1000 mol HCl

×1 L HCl

1 mol NaOH×

1 mol HCl1 L NaOH

0.1000 mol NaOH×

= 0.01000 L NaOH (or 10.00 mL)

(b) Volume NaOH = 30.01500 L HNO 30.3500 mol HNO×

31 L HNO1 mol NaOH

×31 mol HNO

1 L NaOH0.1000 mol NaOH

×

= 0.05250 L NaOH (or 52.50 mL)

(c) Volume NaOH = 3 40.02500 L H PO 3 40.0500 mol H PO×

3 41 L H PO3 mol NaOH

×3 41 mol H PO

1 L NaOH0.1000 mol NaOH

×

= 0.0375 L NaOH (or 37.5 mL) 11.80 We base the calculations for each part of this problem on the problem solving map below:

base mole ratio with HCl HClL base mol base moles HCl L HCl→ → →M M

We determine the mole ratio from the chemical equation. We can take a shortcut if we realize that each mole of hydroxide ion we neutralize consumes 1 mol of hydrogen ion. This means that 1 mol of NaOH or of NH3 reacts with 1 mol of HCl, and 1 mol of Ba(OH)2 reacts with 2 mol of HCl. We also need to remember to convert volumes to liters before we begin each calculation.

(a) Volume HCl = 0.03500 L NaOH0.0500 mol NaOH

×1 L NaOH

1 mol HCl×

1 mol NaOH1 L HCl

0.5000 mol HCl×

= 0.00350 L HCl (or 3.50 mL)

(b) Volume HCl = 20.01000 L Ba(OH) 20.200 mol Ba(OH)×

21 L Ba(OH)2 mol HCl

×21 mol Ba(OH)

1 L HCl0.5000 mol HCl

×

= 0.00800 L HCl (or 8.00 mL)

(c) The reaction for HCl and ammonia is:

HCl(aq) + NH3(aq) → NH4Cl(aq)

Volume HCl = 30.01500 L NH 30.2500 mol NH×

31 L NH1 mol HCl

×31 mol NH

1 L HCl0.5000 mol HCl

×

= 0.007500 L HCl (or 7.5 mL)

11−16

11.81 A colligative property is a physical property of a solution that varies with the number of solute particles dissolved in the solvent, and does not depend on the identity of those solute particles.

11.82 Boiling point elevation, freezing point depression, and osmotic pressure are examples of colligative

properties. 11.83 Osmosis occurs when two solutions with different solute particle concentrations are separated by a

membrane that allows solvent, but not solute, particles to pass through (a semipermeable membrane). The volume of the more concentrated solution increases (often seen as increasing height of solution). Solvent particles pass through the membrane to equalize the concentrations of the solutions on both sides.

11.84 No. If the solute concentration in the pure water is zero, then water would flow into the solution in an

effort to dilute the solution to the point that its solute concentration is zero. Reaching a solute concentration of zero would be impossible.

11.85 Reverse osmosis occurs when solvent is forced to travel in a direction opposite the direction it would

naturally travel during osmosis. For example, if we separate a salt solution and pure water with a semipermeable membrane, osmosis would cause water to flow through the membrane into the salt solution. By applying pressure (osmotic pressure) to the salt solution, we can cause water to pass from the salt solution into the pure water, increasing the concentration of the salt solution.

11.86 Desalinization is the process of making drinking water from salt water. One method uses reverse osmosis

to separate water from salt water. 11.87 The red blood cell will shrink as water moves from the inside of the cell into the aqueous solution

(Figure 11.30). 11.88 Water will begin to flow into the cell by osmosis, causing the cell to swell and eventually burst. This

happens because the blood cell is hypertonic to the pure water. 11.89 (a) vapor pressure decreases; (b) boiling point increases; (c) freezing point decreases 11.90 (a) osmotic pressure decreases; (b) boiling point decreases; (c) freezing point increases 11.91 Addition of sugar, a nonvolatile solute, decreases the vapor pressure of the solution. This vapor pressure

decrease results in an increase in the boiling point of the solution (the sweet tea). 11.92 Addition of a solute to water causes the freezing point of the solution to be lower than that of pure water.

Because the arctic fish have “antifreeze” in their blood, it is not as likely to freeze in the cold arctic weather.

11.93 We calculate boiling point elevation using the equation: ∆Tb = Kb m where the boiling point elevation

constant, Kb, depends on the solvent. The boiling point elevation constant for water is 0.52°C/m. (a) We calculate the difference in boiling points using the boiling point elevation constant for water and

the molality of the solute particles. Because C6H12O6 is a molecular compound, it produces only one solute particle per molecule.

∆Tb = 0.52 C 2.5°× m

m = 1.3°C

The difference in boiling points is 1.3°C. (b) We calculate the boiling point of the solution by adding the boiling point elevation to the boiling point

of the pure solvent (100.0°C for water).

Tb = 100.0°C + 1.3°C = 101.3°C

11−17

11.94 Freezing point depression is given by the equation: ∆Tf = Kf m where the freezing point depression constant, Kf, depends on the solvent. The freezing point depression constant for water is −1.86°C/m. (a) We calculate the freezing point depression of the solution using the freezing point depression constant

for water and the molality of the solute particles. Because ethylene glycol is a molecular compound, it produces only one solute particle per molecule.

∆Tf = 1.86 C 5.0− °× m

m = −9.3°C

The difference in freezing points is −9.3°C. (b) We determine the freezing point of the solution by adding the freezing point depression to the freezing

point of the pure solvent (0.0°C for water).

Tf = 0.0°C − 9.3°C = −9.3°C

11.95 The sodium chloride solution should have the higher boiling point because one formula unit of NaCl

produces two solute particles when it dissolves in water (Na+(aq) and Cl−(aq)). Glucose is a molecular compound, so it produces only one solute particle per molecule.

11.96 When it dissociates, MgCl2 produces more ions per formula unit than NaNO3 produces when it dissociates:

MgCl2(s) → Mg2+(aq) + 2Cl−(aq) (three ions produced)

NaNO3(s) → Na+(aq) + NO3–(aq) (two ions produced)

Because boiling point elevation depends on the number of solute particles in solution, a 1.0 m MgCl2 solution should have a higher boiling point than a 1 m NaNO3 solution.

11.97 Solution B has the highest concentration of dissolved particles, and therefore it has the lowest freezing

point. Glucose is a molecular compound, and images B and C show molecular compounds. Sodium chloride produces two ions of different size, as shown in image A.

11.98 Because the solution in image B shows the highest concentration of solute particles, it has the lowest

freezing point. Images A and B show solutions of NaNO3 because the same ions are present in both images. Image A represents the less concentrated NaNO3 solution because there are fewer particles in the image. Image C represents the MgCl2 solution because the ions are different from those in images A and B, and there is a 2 to 1 ratio of ions (two chloride ions for each magnesium) in image C.

11.99 Non-acidic molecular compounds produce 1 solute particle per molecule. Ionic compounds produce 1

particle per ion in the formula unit. Sodium chloride, for example, has two ions (Na+ and Cl−) in its formula unit so it produces two solute particles (ions) per formula unit. (a) CH3OH is a molecular compound. The solution contains 2.0 mol of particles per kilogram of solvent. (b) NaCl is an ionic compound that produces 2 ions per formula unit. The solution contains 5.0 mol of ions (2.5 mol of Na+ ions and 2.5 mol of Cl− ions). (c) Al(NO3)3 is an ionic compound that produces 4 mol of ions per formula unit. The solution contains 4.0 mol of ions (1 mol of Al3+ ions and 3 mol of NO3

− ions). 11.100 Non-acidic molecular compounds produce 1 solute particle per molecule. Ionic compounds produce 1

particle per ion in the formula unit. Sodium chloride, for example, has two ions (Na+ and Cl−) in its formula unit so it produces two solute particles per formula unit. (a) HCl is an acid. The solution contains 4.0 mol of ions (2 mol of H+ ions and 2 mol of Cl− ions). (b) Na2SO4 is an ionic compound that produces 3 ions per formula unit. The solution contains 0.45 mol of ions (0.30 mol of Na+ ions and 0.15 mol of SO4

2- ions). (c) C12H22O11 is molecular compound. The solution contains 1.0 mol of solute particles.

11.101 One formula unit of magnesium chloride produces 3 particles when dissolved (one Mg2+ ion and two Cl−

ions), whereas one formula unit of NaNO3 produces 2 ions (one Na+ ion and one NO3− ion). The

magnitude of colligative properties depends on the number of solute particles in the solution. Therefore, the osmotic pressure of the MgCl2 solution will be greater than that of the NaNO3 solution.

11−18

11.102 One formula unit of sodium chloride produces two ions (Na+ and Cl−) when it dissolves. Methanol is a molecular compound, so it produces one solute particle per molecule as it dissolves. Because osmotic pressure depends on the number of solute particles dissolved in a solvent, a 1.0 m NaCl solution will have a higher osmotic pressure than a 1.0 m CH3OH solution.

11.103 Vitamin D is nonpolar. We make that assumption based on the “like dissolves like” rule. Substances that

are soluble in each other have similar intermolecular forces. 11.104 Water is polar and forms hydrogen bonds, so we can assume that Vitamin B is polar and/or forms hydrogen

bonds. We make this assumption based on the “like dissolves like” rule. Substances that are soluble in each other have similar intermolecular forces.

11.105 Energy is required to separate molecules or oppositely charged ions. This is because disruption of a crystal

lattice and disruption of intermolecular attractions between water molecules both require energy. The formation of ion-dipole interactions between water and the ions releases energy. In general, energy is released any time two objects that are attracted to one another come together.

11.106 The solubility of a gas in a solution increases when the temperature decreases and when the partial pressure

of the gas above the solution increases. 11.107 We can express the solubility of NaCl in water (36.3 g per 100 g water) in two ways:

2

36.3 g NaCl100 g H O

and 2100 g H O36.3 g NaCl

We know the mass of water, and can use the solubility information to cancel ‘g H2O’ and give us the desired units, ‘g NaCl.’

Mass NaCl = 2500.0 g H O2

36.3 g NaCl 100 g H O

× = 182 g NaCl

11.108 No. A solution is saturated only when the solute concentration is the same as its solubility at the specified

temperature. Because the KNO3 concentration in the warmed solution is lower than the solubility of KNO3 at the higher temperature, the warmed solution is unsaturated.

11.109 Molarity expresses the number of moles of solute dissolved in 1 L of solution. To calculate the molarity of

a solution, we need to find the number of moles of solute and the solution volume. Assume we have exactly 100 g of the 48.0% HBr solution. This means that we have 48.0 g of HBr. Using the molar mass of HBr (80.91 g/mol), we can calculate the equivalent number of moles of HBr represented by 48.0 g. Also, because we know the solution density, we can determine the volume occupied by 100 g of the solution.

Moles HBr = 48.0 g HBr 1 mol HBr80.91 g HBr

× = 0.593 mol HBr

Volume in mL = 100 g HBr 1 mL1.50 g HBr

× = 66.7 mL

Volume in L = 1 L66.7 mL1000 mL

× = 6.67 × 10−2 L

Molarity = 20.593 mol

6.67 10 L−× = 8.90 M

11.110 Molarity expresses the number of moles of solute dissolved in 1 L of solution. To calculate the molarity

we need to find the number of moles of solute and the solution volume. Assume that we have exactly 100 g of solution. This means that our solution contains 24.0 g of H2SO4. Using the molar mass of H2SO4

11−19

(98.08 g/mol), we can calculate the number of moles of H2SO4 present in the solution. In addition, because we know its density, we can determine the volume occupied by 100 g of the solution.

Moles H2SO4= 2 424.0 g H SO 2 4

2 4

1 mol H SO98.08 g H SO

× = 0.245 mol H2SO4

Volume in mL = 2 4100 g H SO2 4

1 mL1.17 g H SO

× = 85.5 mL

Volume in L = 1 L85.5 mL1000 mL

× = 8.55 × 10−2 L

Molarity = 20.245 mol

8.55 10 L−× = 2.86 M

11.111 The volume of a solution changes with temperature. Concentration units based only on mass or number of

moles will not change with temperature changes. Percent by mass (a) and molality (d) do not change with temperature. This is one of the reasons we use molality in colligative property calculations.

11.112 The molality of a solution describes the number of moles of solute per kilogram of solvent. Assume that

we have exactly one liter of 1.0 M acetic acid. This solution contains 1.0 mol or 60. g of acetic acid, HC2H3O2 (60.05 g/mol). From the density of the solution 1.005 g/mL, we can determine the mass of the solution:

kg solution = 2 3 21 L HC H O 1000 mL×

2 3 21 L HC H O1.005 g

×1 mL

1 kg1000 g

× = 1.005 kg

To calculate the mass of the solvent, we subtract the mass of the solute from the mass of the solution.

kg solute = 1 kg60. g1000 g

× = 0.060 kg

kg solvent = 1.005 kg − 0.060 kg = 0.945 kg

To calculate the molality of the solution, we divide the number of moles of solute by the mass of the solvent:

m = 1.0 mol0.945 kg

= 1.1 m

11.113 The densities of dilute aqueous solutions are approximately equal to the density of water. The mass of one

liter of water is one kilogram. Therefore, the molarity (number of moles of solute/liter of solution) and molality (number of moles of solute/kilogram of solvent) are very similar.

11.114 The balanced chemical equation for each of these situations is the same:

2AgNO3(aq) + Na2CO3(aq) → Ag2CO3(s) + 2NaNO3(aq)

We are given the volume and concentration of both reactants, so we must determine which, if either, of the two is the limiting reactant. To determine this, we calculate the number of moles of Ag2CO3 that each reactant would produce at the point at which it is consumed. The reactant that produces the smallest number of moles of product is the limiting reactant, and the mass of Ag2CO3 produced by that reactant is the theoretical yield for this reaction. The general problem solving map for these calculations is:

2 3reactant mole ratioL reactant mol reactant mol Ag CO→ →M

We must remember to express solution volumes in liters. (a) The number of moles of Ag2CO3 produced by 100.0 mL (0.1000 L) of 0.200 M AgNO3:

11−20

Moles Ag2CO3 = 30.1000 L AgNO 30.200 mol AgNO×

31 L AgNO2 3

3

1 mol Ag CO2 mol AgNO

× = 0.0100 mol Ag2CO3

The number of moles of Ag2CO3 produced by 100.0 mL (0.1000L) of 0.200 M Na2CO3:

Moles Ag2CO3 = 2 30.1000 L Na CO 2 30.200 mol Na CO×

2 31 L Na CO2 3

2 3

1 mol Ag CO1 mol Na CO

× = 0.0200 mol Ag2CO3

Because the AgNO3 is consumed when only 0.0100 mol Ag2CO3 forms, AgNO3 is the limiting reactant, and the theoretical reaction yield is 0.0100 mol of Ag2CO3. We use the molar mass of Ag2CO3 (275.81 g/mol) to calculate the mass of product produced.

Mass Ag2CO3 = 2 30.0100 mol Ag CO 2 3

2 3

275.81 g Ag CO1 mol Ag CO

× = 2.76 g Ag2CO3

(b) Number of moles of Ag2CO3 produced by 200.0 mL (0.2000 L) of 0.200 M AgNO3:

Moles Ag2CO3 = 30.2000 L AgNO 30.200 mol AgNO×

31 L AgNO2 3

3

1 mol Ag CO2 mol AgNO

× = 0.0200 mol Ag2CO3

Number of moles of Ag2CO3 produced by 100.0 mL (0.1000L) of 0.200 M Na2CO3:

Moles Ag2CO3 = 2 30.1000 L Na CO 2 30.200 mol Na CO×

2 31 L Na CO2 3

2 3

1 mol Ag CO1 mol Na CO

× = 0.0200 mol Ag2CO3

The amounts of product produced by both reagents are the same, so both reactants are limiting and 0.0200 mol of Ag2CO3 is produced. We use the molar mass of Ag2CO3 to calculate the mass of product produced.

Mass Ag2CO3 = 2 30.0200 mol Ag CO 2 3

2 3

275.81 g Ag CO1 mol Ag CO

× = 5.52 g Ag2CO3

(c) Number of moles of Ag2CO3 produced by 100.0 mL (0.1000 L) of 0.200 M AgNO3:

Moles Ag2CO3 = 30.1000 L AgNO 30.200 mol AgNO×

31 L AgNO2 3

3

1 mol Ag CO2 mol AgNO

× = 0.0100 mol Ag2CO3

Number of moles of Ag2CO3 produced by 200.0 mL (0.2000L) of 0.200 M Na2CO3:

Moles Ag2CO3 = 2 30.2000 L Na CO 2 30.200 mol Na CO×

2 31 L Na CO2 3

2 3

1 mol Ag CO1 mol Na CO

× = 0.0400 mol Ag2CO3

Because the AgNO3 is consumed when only 0.0100 mol Ag2CO3 is formed, this is the theoretical yield of the reaction. We use the molar mass of Ag2CO3 to calculate the mass of product produced.

Mass Ag2CO3 = 2 30.0100 mol Ag CO 2 3

2 3

275.81 g Ag CO1 mol Ag CO

× = 2.76 g Ag2CO3

11.115 The balanced equation for each of these situations is the same:

BaCl2(aq) + K2SO4(aq) → 2KCl(aq) + BaSO4(s)

We are given the volumes and concentrations of two solutions, so we must determine which, if either, of the two solutions is the limiting reactant. Because BaCl2 and K2SO4 react in a 1:1 molar ratio, the solution that provides the smaller number of moles of reactant is limiting.

11−21

(a) 0.5000 L × 0.100 M BaCl2 = 0.0500 mol; 0.0900 L × 0.500 M K2SO4 = 0.0450 mol; K2SO4 is limiting.

Mass of BaSO4 produced = 2 40.0450 mol K SO 41 mol BaSO×

2 41 mol K SO4

4

233.36 g BaSO1 mol BaSO

× = 10.5 g BaSO4

(b) 0.1000 L × 0.100 M BaCl2 = 0.0100 mol; 0.1000 L × 0.500 M K2SO4 = 0.0500 mol; BaCl2 is limiting.

Mass of BaSO4 produced = 20.0100 mol BaCl 41 mol BaSO×

21 mol BaCl4

4

233.36 g BaSO1 mol BaSO

× = 2.33 g BaSO4

(c) 0.1000 L × 0.100 M BaCl2 = 0.0100 mol; 0.5000 L × 0.500 M K2SO4 = 0.250 mol; BaCl2is limiting

Mass of BaSO4 produced = 20.0100 mol BaCl 41 mol BaSO×

21 mol BaCl4

4

233.36 g BaSO1 mol BaSO

× = 2.33 g BaSO4

11.116 The balanced chemical equation is:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

To answer this question, we use the following problem solving map:

HCl mole ratio NaOHL HCl mol HCl mol NaOH g NaOH→ → →M MM

The molar mass of NaOH is 40.00 g/mol.

Mass NaOH = 1.8 L HCl2.0 mol HCl

×1 L HCl

1 mol NaOH×

1 mol HCl40.00 g NaOH1 mol NaOH

× = 1.4 × 102 g NaOH

11.117 The balanced chemical equation is:

H2SO4(aq) + 2NaHCO3(s) → Na2SO4(aq) + 2H2O(l) + 2CO2(g)

To answer this question, we use the following problems solving map:

32 42 4 2 4 3 3

NaHCOH SO mole ratioL H SO mol H SO mol NaHCO g NaHCO→ → →MMM

The molar mass of NaHCO3 is 84.01 g/mol.

Mass NaHCO3 = 2 42.0 L H SO 2 46.0 mol H SO×

2 41 L H SO32 mol NaHCO

×2 41 mol H SO

3

3

84.01 g NaHCO1 mol NaHCO

× = 2.0 × 103 g

11.118 Because one formula unit of Ca(NO3)2 is composed of 3 ions (one Ca2+ ion and 2 NO3

− ions), the molal concentration of the solution is three times the molarity of the Ca(NO3)2, or 9.0 m. The boiling point elevation is given by the equation, b b∆ =T K m. For water, Kb is 0.52°C/m.

∆Tb = 0.52 C 9.0°× m

m = 4.7°C

To calculate the boiling point of solution, we add the boiling point elevation to the normal boiling point of water (100.0°C). The boiling point of the solution is 104.7°C.

11.119 Freezing point depression depends on the number of moles of solute particles in the solution. Solution B

has the lowest freezing point (the largest freezing point depression) so it is the solution with the highest NaCl concentration.

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11.120 Using the equation for boiling point elevation, b b∆ =T K m, we can calculate the molality of the solution. The normal boiling point of water is 100.0°C, so we know that the boiling point elevation is 1.5°C. The boiling point elevation constant for water is 0.52°C/m. Rearranging the equation for molality, m, we have:

ob

ob

1.5 C0.52 C/

∆= =

Tm

K m = 2.9 m

The units of molality are moles of solute per kilogram of solvent. To calculate the number of moles of C12H22O11 in 100.0 grams of solution, we need to calculate the number of moles of C12H22O11 per gram of solution. If we assume that we have a solution with 1 kg (1000 g) of solvent, it will contain 2.9 moles C12H22O11 (342.97 g/mol).

Mass C12H22O11 = 12 22 112.9 mol C H O12 22 11

342.97 g1 mol C H O

× = 990 g C12H22O11

The total mass of the solution is: Solution mass = 1000 g + 990 g = 1990 g Knowing that there are 2.9 moles of C12H22O11 in 1990 g solution, we can calculate the number of moles and the mass of C12H22O11 in 100 g of the solution as shown below:

Moles C12H22O11 in 100.0 g solution = 100.0 g solution 12 22 112.9 mol C H O1990 g solution

× = 0.15 mol C12H22O11

Mass C12H22O11 in 100.0 g solution = 100.0 g solution 12 22 11990 g C H O1990 g solution

× = 50. g C12H22O11

11.121 Olive oil is nonpolar and vinegar is a solution of polar acetic acid in polar water. Olive oil molecules are attracted more strongly to one another than to acetic acid or to water, so these molecules will separate out of the mixture. Water and acetic acid molecules are attracted more strongly to one another than to olive oil molecules, so they will also separate out of the mixture.

11.122 Power plants produce heat. The plant would be located next to a river so the river water can be used for

cooling. This results in water downstream from the power plant being at a higher temperature than water upstream from the power plant. Oxygen, like all gases, is less soluble in hotter water than in colder water, so less oxygen will be dissolved in the water downstream from the power plant.

11.123 Freezing point decreases as the concentration of particles in solution increases. All the solutions listed are

strong electrolytes, so the concentration of particles will be the concentration of ions. NaCl has two ions per formula unit, so the concentration of ions is 0.40 m. Na2SO4 has three ions per formula unit, so the concentration of ions is 0.30 m. KBr has two ions per formula unit, so the concentration of ions is 0.2 m. VCl3 has four ions per formula unit, so the concentration of ions is 0.60 m. The order of decreasing freezing point is : 0.10 m KBr, 0.10 m Na2SO4, 0.20 m NaCl, 0.15 m VCl3.

11.124 Vapor pressure decreases as the concentration of particles in solution increases. Since the concentration of

the two solutes is the same, the one with the higher vapor pressure must produce fewer particles. That is, it is a weaker electrolyte than the one with the lower vapor pressure. Thus, HCO2H is a weak electrolyte, and HBr is a strong electrolyte.

11.125 Lettuce becomes limp when its cells lose water. Placing the lettuce in water causes water molecules to

cross the cell membrane into the cells by osmosis since there is a higher concentration of ions inside the cells than outside.

11.126 The sodium acetate solution must be supersaturated since addition of a crystal causes precipitation of

sodium acetate. Sodium acetate will precipitate from the supersaturated solution until the solution becomes saturated.

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11.127 Barium sulfate is a very insoluble ionic salt (Table 11.1). Since very little of the barium sulfate dissolves, the digestive system contains very few barium ions. In addition, barium sulfate has a high density, so it passes through the digestive system rather quickly.

11.128 Calcium chloride is stockpiled during the winter by northern cities for use in lowering the freezing point of

water to melt ice on city streets. 11.129 Blood cells are found in polar water in the bloodstream. Fatty tissues are largely nonpolar. Following the

rule “like-dissolves-like”, we can predict that nonpolar vitamin A is more soluble in fatty tissues and therefore more likely to be found there.

11.130 Glucose has a molar mass of 180.2 g/mol. We can use this quantity to convert g/L to mol/L:

54.30 gMolarity = 1 mol

1 L 180.2 g× = 0.3013 mol/L or 0.3013 M

CONCEPT REVIEW

11.131 Answer: C; For each mole of magnesium nitrate that dissolves, 1 mole of Mg2+ and 2 moles of NO3− ions

are present.

A. The solution is 0.20 M in formula units of Mg(NO3)2 . B. The solution is 0.20 M in Mg2+ and 0.40 M NO3

− ions. D. The solution is 0.60 M in total ions present. E. Magnesium nitrate is soluble in water so it is expected to exist as ions in the solution.

11.132 Answer: B; Hexane is a nonpolar solvent so it is expected to dissolve nonpolar solutes like CCl4. All the

other compounds are polar so they would be most soluble in polar solvents. 11.133 Answer: A; The ionic bonds in KBr and some hydrogen bonds between water molecules have to be broken.

These are both processes that require energy. When new bonds or attractive forces form, energy is released.

B. Energy is required to break bonds and intermolecular forces (hydrogen bonds). When the new ion-dipole forces form, energy is released.

C. Energy is absorbed to break the ionic bonds in KBr, but to separate some of the water molecules, hydrogen bonds, not ionic bonds, must be broken. When the new ion-dipole forces form, energy is released.

D. Energy is absorbed to break the ionic bonds in KBr, but some of the hydrogen bonds must be broken between water molecules. When the new ion-dipole forces form, energy is released.

E. Energy is released when the ion-dipole forces form between the ions and water. 11.134 Answer: C; When 0.3 g of PbI2 is placed in water, only 0.25 grams of it will dissolve:

2 2grams PbI dissolved = 250 mL H O 2

2

0.1 g PbI

100 mL H O× 2 0.25 g H O=

0.3 g PbI2 – 0.25 g PbI2 = 0.05 g PbI2 undissolved

A. Only some (0.25 g) of the PbI2 will dissolve. B. Only 0.1 PbI2 will dissolve in 100 mL of water. In 250 mL of water, 0.25 g PbI2 will dissolve. D. The dissolved solid would exist in solution as Pb2+ and I− ions. E. According to the solubility rules, PbI2 is considered insoluble. However, a very small amount can

dissolve in water as described by its solubility, 0.1 g PbI 2 per 100 mL of water. 11.135 Answer: E; This solution is 15.0 percent-by-mass as determined by the calculation:

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( )2

30.0 g KClPercent-by-mass = 100 15.0% KCl

30.0 g KCl + 170.0 g H O× =

A. ( )2

30.0 g KClPercent-by-mass = 100 13.0% KCl

30.0 g KCl + 200.0 g H O× =

B. mol KClMass KCl = 30.0 74.55 g KCl

1 mol KCl

× 2240 g KCl=

( )2

2240 g KClPercent-by-mass = 100 91.8% KCl

2240 g KCl + 200.0 g H O× =

This solution couldn’t possibly form. It would be impossible for 2240 g KCl to dissolve in such as small quantity of water.

D. ( )2

170.0 g KClPercent-by-mass = 100 85.0% KCl

170.0 g KCl + 30.0 g H O× =

This solution couldn’t possibly form. It would be impossible for 170.0 g KCl to dissolve in such a small quantity of water.

11.136 Answer: A; Molality is defined as moles of solute

kg of solvent

B. molarity C. density of solution D. mass-volume percent E. volume percent

11.137 Answer: D; Molarity is defined as moles solute per liter of solution. This calculation converts grams solute

to moles, then divides by the volume of solution in units of liters.

A. This calculation takes the ratio of grams solute per milliliter of solution. B. The molar mass of solute is incorrect. It should be shown in the mathematical operation as

3 2

3 2

1 mol Mg(NO )

148.33 g Mg(NO ). Also, the volume should be expressed in units of liters.

C. This calculation takes the ratio of moles solute per milliliter of solution. The volume of solution should be expressed in units of liters.

E. This mathematical operation would calculate the inverse of molarity.

11.138 Answer: B; Quantities of both reactants are mixed so we’ll have to determine which one is the limiting reactant.

3 2 43 3 2 4

3 3

1.00 mol AgNO 1 mol K SO0.0250 L AgNO solution × = 0.0250 mol AgNO × = 0.0125 mol K SO

1 L AgNO solution 2 mol AgNOThis is the moles K2SO4 required to react with all the AgNO3 present. We can then calculate the moles of K2SO4 actually present:

2 42 4 2 4

2 4

2.00 mol K SO0.0250 L K SO solution × = 0.0500 mol K SO present

1 L K SO solution

The limiting reactant is AgNO3 so its quantity determines the amount of Ag2SO4 produced:

2 43 2 4

3

1 mol Ag SO0.0250 mol AgNO × = 0.0125 mol Ag SO produced

2 mol AgNO

According to the solubility rules, Ag2SO4 is insoluble, so this product is expected to precipitate.

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A. 0.0125 mole K2SO4 reacts with 0.0250 mol AgNO3. C. 0.0250 mol of aqueous KNO3 should be produced:

30.0250 mol AgNO 3

3

2 mol KNO ×

2 mol AgNO 3 = 0.0250 mol KNO produced

D. The final reaction mixture will contain products and unreacted K2SO4. E. 0.0375 mol of K2SO4 remains unreacted.

11.139 Answer: B; The solution that produces the greatest number of particles will have the lowest freezing point.

For each formula unit of Al(NO3)3 that dissolves, four ions are produced (one Al3+ and three NO3−):

Al(NO3)3(s) → Al3+(aq) + 3NO3−(aq)

A. Two ions per formula unit dissolved: KBr(s) → K+(aq) + Br−(aq) C. Three ions per formula unit dissolved: Na2CO3(s) → 2Na+(aq) + CO3

2−(aq) D. Does not dissociate: CH3OH(s) → CH3OH(aq) E. Three ions per formula unit dissolved: MgCl2(s) → Mg2+(aq) + 2Cl−(aq)

Each solution will cause the osmotic pressure and boiling point to increase, and the vapor pressure and freezing point to decrease compared to pure water. The order of increasing effect on these properties is: D < A < C = E < B