Silberberg7e solution manual ch 13

The intermolecular forces for the more soluble solute will be more similar to the intermolecular forces in the solvent than the forces in the less soluble solute.. 13.6A Plan: To find th

CHAPTER 13 THE PROPERTIES OF

MIXTURES: SOLUTIONS AND COLLOIDS

FOLLOW–UP PROBLEMS

13.1A Plan: Compare the intermolecular forces in the solutes with the intermolecular forces in the solvent The

intermolecular forces for the more soluble solute will be more similar to the intermolecular forces in the solvent than the forces in the less soluble solute

Solution:

a) 1,4–Butanediol is more soluble in water than butanol Intermolecular forces in water are primarily hydrogen

bonding The intermolecular forces in both solutes also involve hydrogen bonding with the hydroxyl groups Compared to butanol, each 1,4–butanediol molecule will form more hydrogen bonds with water because 1,4–butanediol contains two hydroxyl groups in each molecule, whereas butanol contains only one –OH group Since 1,4–butanediol has more hydrogen bonds to water than butanol, it will be more soluble than butanol

b) Chloroform is more soluble in water than carbon tetrachloride because chloroform is a polar molecule and

carbon tetrachloride is nonpolar Polar molecules are more soluble in water, a polar solvent

13.1B Plan: Compare the intermolecular forces in the solutes with the intermolecular forces in the solvent The

intermolecular forces for the solvent that dissolves more solute will be more similar to the intermolecular forces in the solute than the forces in the solvent that dissolves less solute

Solution:

a) Both chloroform and chloromethane are polar molecules that experience dipole-dipole and dispersion

intermolecular forces Methanol, on the other hand, has an O–H bond and, thus, can participate in hydrogen

bonding in addition to dipole-dipole and dispersion forces Chloroform dissolves more chloromethane than

methanol because of similar dipole-dipole forces

b) Pentanol has a polar O–H group that can participate in hydrogen bonding, dipole-dipole and dispersion forces However, it has a much larger non-polar section (CH3CH2CH2CH2-) that experiences only dispersion forces Because the nonpolar portion of the pentanol is much larger than the polar portion, the nonpolar portion

determines the overall solubility of the molecule Thus, pentanol will be more soluble in nonpolar solvents like

hexane than polar solvents like water Hexane dissolves more pentanol due to dispersion forces

13.2A Plan: Use the relationship Hsolution = Hlattice +Hhydration of the ions Given Hsolution and Hlattice, Hhydration of the ions

can be calculated

Solution:

The two ions in potassium nitrate are K+ and NO3 Hhydration of the ions = Hhydration(K+) + Hhydration(NO3)

Hsolution = Hlattice +Hhydration of the ions

Hhydration of the ions = Hsolution – Hlattice = 34.89 kJ/mol – 685 kJ/mol = –650.11 = –650 kJ/mol

13.2B Plan: Use the relationship Hsolution = Hlattice +Hhydration of the ions Given Hsolution and Hlattice, Hhydration of the ions

can be calculated Hhydration of the ions and Hhydration (Na+) can then be used to solve for Hhydration(CN–)

Solution:

The two ions in sodium cyanide are Na+ and CN–

Hsolution = Hlattice +Hhydration of the ions

Hhydration of the ions = Hsolution – Hlattice = 1.21 kJ/mol – 766 kJ/mol = –764.79 = –765 kJ/mol

Hhydration of the ions = Hhydration(Na+) + Hhydration(CN–)

Hhydration(CN–) = Hhydration of the ions – Hhydration(Na+)= –765 kJ/mol – (–410 kJ/mol) = –355 kJ/mol

13.3A Plan: Solubility of a gas can be found from Henry’s law: Sgas = kH  Pgas The problem gives kH for N2 but not its

partial pressure To calculate the partial pressure, use the relationship from Chapter 5: Pgas = Xgas x Ptotal where X

represents the mole fraction of the gas

Solution:

To find partial pressure use the 78% N2 given for the mole fraction:

Pgas = Xgas x Ptotal

13.3B Plan: Solubility of a gas can be found from Henry’s law: Sgas = kH  Pgas The problem gives kH for N2O but not its

partial pressure To calculate the partial pressure, use the relationship from Chapter 5: Pgas = Xgas x Ptotal where X

represents the mole fraction of the gas

Solution:

To find partial pressure use the 40.% N2O given for the mole fraction:

Pgas = Xgas x Ptotal

S (N2O) = (2.5x10–2 mol/L•atm)(0.48 atm) = 0.012 mol/L

13.4A Plan: Molality (m) is defined as amount (mol) of solute per kg of solvent Use the molality and the mass of solvent

given to calculate the amount of glucose in moles Then convert amount (mol) of glucose to mass (g) of glucose

by multiplying by its molar mass

13.4B Plan: Molality (m) is defined as amount (mol) of solute per kg of solvent Calculate the moles of solute, I2, from

the mass Find the mass (kg) of the solvent, diethyl ether, from the given number of moles Then divide the moles

of solute by the kg of solvent to find the molality

Solution:

Amount (mol) of solute = (15.20 g I2) 1 mol I2

253.8 g I2 = 0.05989 mol I2Mass of solvent (kg) = (1.33 mol (CH3CH2)2O 74.12 g (CH3CH2)2O

13.5A Plan: Mass percent is the mass (g) of solute per 100 g of solution For each alcohol, divide the mass of the alcohol

by the total mass Multiply this number by 100 to obtain mass percent To find mole fraction, first find the amount (mol) of each alcohol, then divide by the total moles

Solution:

Mass % propanol = mass of propanol x 100%

mass of propanol + mass of ethanol = 35.0 150 g35.0 g  x 100%

= 18.9189 = 18.9% propanol

Mass % of ethanol = mass of ethanol x 100%

mass of propanol + mass of ethanol = 35.0 150 g150 g  x 100%

Xpropanol = 0.5824596 mol propanol

0.5824596 mol propanol + 3.2559149 mol ethanol = 0.151746 = 0.152

Xethanol = 3.2559149 mol ethanol

0.5824596 mol propanol + 3.2559149 mol ethanol = 0.84825 = 0.848

13.5B Plan: Mass percent is the mass (g) of solute per 100 g of solution For each component of the mixture, divide the

mass of the component by the total mass Multiply this number by 100 to obtain mass percent To find mole percent, first find the amount (mol) of each component, then divide by the total moles and multiply by 100% Solution:

Mass % ethanol = mass of ethanol

mass of ethanol + mass of iso-octane + mass of heptane x 100% =

1.87 g 1.87 g + 27.4 g + 4.10 g x 100% = 5.6038 = 5.60% ethanol

Mass % iso-octane = mass of iso-octane

mass of ethanol + mass of iso-octane + mass of heptane x 100% =

27.4 g 1.87 g + 27.4 g + 4.10 g x 100% = 82.1097 = 82.1% iso-octane

Mass % heptane = mass of heptane

mass of ethanol + mass of iso-octane + mass of heptane x 100% =

4.10 g 1.87 g + 27.4 g + 4.10 g x 100% = 12.2865 = 12.3% heptane

Moles of ethanol = (1.87 g ethanol) 1 mole ethanol

46.07 g ethanol = 0.0406 mol ethanol Moles of iso-octane = (27.4 g iso-octane) 1 mole iso-octane

114.22 g iso-octane = 0.240 mol iso-octane Moles of heptane = (4.10 g heptane) 1 mole heptane

100.20 g heptane = 0.0409 mol heptane Mole percent ethanol = moles of ethanol

moles of ethanol + moles of iso-octane + moles of heptane x 100% =

0.0406 mol 0.0406 mol + 0.240 mol + 0.0409 mol x 100% = 12.6283 = 12.6% ethanol

Mole percent iso-octane = moles of iso-octane

moles of ethanol + moles of iso-octane + moles of heptane x 100% =

0.240 mol 0.0406 mol + 0.240 mol + 0.0409 mol x 100% = 74.6501 = 74.6% iso-octane

Mole percent heptane = moles of heptane

moles of ethanol + moles of iso-octane + moles of heptane x 100% =

0.0409 mol 0.0406 mol + 0.240 mol + 0.0409 mol x 100% = 12.7216 = 12.7% heptane

(Slight differences from 100% are due to rounding.)

13.6A Plan: To find the mass percent, molality and mole fraction of HCl, the following is needed:

1) Moles of HCl in 1 L solution (from molarity) 2) Mass of HCl in 1L solution (from molarity times molar mass of HCl) 3) Mass of 1L solution (from volume times density)

4) Mass of solvent in 1L solution (by subtracting mass of solute from mass of solution) 5) Moles of solvent (by dividing the mass of solvent by molar mass of water)

Mass percent is calculated by dividing mass of HCl by mass of solution and multiplying by 100

Molality is calculated by dividing moles of HCl by mass of solvent in kg

Mole fraction is calculated by dividing mol of HCl by the sum of mol of HCl plus mol of solvent

Solution:

Assume the volume is exactly 1 L

1) Mole of HCl in 1 L solution = 1.0 L 11.8 mol HCl

Mass (kg) of solvent in 1 L solution = 759.772 g 1 kg3

2

2

1 mol H O759.772 g H O

18.02 g H O

  = 42.1627 mol solvent Mass percent of HCl = mass HCl  100

2

mol HClmol HCl + mol H O =

11.8 mol11.8 mol + 42.1627 mol = 0.21866956 = 0.219

13.6B Plan: To find the molality, molarity and mole percent of CaBr2, the following is needed:

1) Mass of CaBr2 and mass of H2O in 100 g of CaBr2

2) Moles of CaBr2 and mass of H2O in 100 g of CaBr2

3) Volume of 100 g of solution Molality is calculated by dividing moles of CaBr2 by mass of solvent in kg

Molarity is calculated by dividing the moles of CaBr2 by the volume of the solution (L)

Mole percent is calculated by dividing moles of CaBr2 by the total moles in the solution and multiplying by 100 Solution:

Assume you have 100 g of solution

1) Mass of CaBr2 in 100 g solution = (100 g solution) 52.1 g CaBr2

100 g solution = 52.1 g CaBr2

Mass of H2O in 100 g solution = 100 g – 52.1 g = 47.9 g H2O

2) Moles of CaBr2 in 100 g solution = (52.1 g CaBr2) 1 mol CaBr2

199.88 g CaBr2 = 0.261 mol CaBr2 Moles of H2O in 100 g solution = (47.9 g H2O) 1 mol H2O

18.02 g H 2 O = 2.66 mol H2O3) Volume (L) of 100 g solution = (100 g solution) 1 mL

1.70 g

1 L

1000 mL = 0.0588 L solution Molality = moles solute

kg solvent = 0.261 mol CaBr2

47.9 g H 2 O

1000 g

1 kg = 5.45 m

Molarity = moles solute

L solution = 0.261 mol CaBr2

0.0588 L = 4.44 M

Mole percent = moles solute

moles solute moles solvent x 100% = 0.261 mol + 2.66 mol0.261 mol x 100% = 8.94%

13.7A Plan: Raoult’s law states that the vapor pressure of a solvent is proportional to the mole fraction of the solvent:

Psolvent = Xsolvent  P°solvent To calculate the drop in vapor pressure, a similar relationship is used with the mole fraction of the solute substituted for that of the solvent

P = Xaspirin P°methanol = (7.06381x10–3) (101 torr) = 0.71344 = 0.713 torr

13.7B Plan: The drop in vapor pressure of a solvent is calculated by the following equation derived from Raoult’s law:

ΔP = Xsolute  P°solvent

Solution:

Amount (mol) menthol = (6.49 g menthol) 1 mol menthol

156.26 g menthol = 0.0415 mol menthol Amount (mol) ethanol = (25.0 g ethanol) 1 mol ethanol

46.07 g ethanol = 0.543 mol ethanol

13.8A Plan: Find the molality of the solution by dividing the moles of P4 by the mass of the CS2 in kg The change in

freezing point is calculated from Tf = iKfm, where Kf is 3.83°C/m when CS2 is the solvent, i is the van’t Hoff factor, and m is the molality of particles in solution Since P4 is a covalent compound and does not ionize in water,

i = 1 Once Tf is calculated, the freezing point is determined by subtracting it from the freezing point of pure CS2(–111.5°C) The boiling point of a solution is increased relative to the pure solvent by the relationship Tb = iKbm

where Kb is 2.34°C/m when CS2 is the solvent, i is the van’t Hoff factor, and m is the molality of particles in

solution P4 is a nonelectrolyte (it is a molecular compound) so i = 1 Once Tb is calculated, the boiling point is determined by adding it to the boiling point of pure CS2 (46.2°C)

Solution:

Molality of P4 solution:

Moles of P4 = (8.44 g P4) 1 mol P4

123.88 g P4 = 0.0681 mol Molality of P4 = moles solute

kg solvent = 0.0681 mol P4

60.0 g CS2

1000 g

1 kg = 1.14 m P4

The boiling point is 46.2°C + 2.67°C = 48.9°C

13.8B Plan: The question asks for the concentration of ethylene glycol that would prevent freezing at 0.00°F First,

convert 0.00°F to °C The change in freezing point of water will then be this temperature subtracted from the freezing point of pure water, 0.00°C Use this value for T in Tf = 1.86°C/m  molality of solution and solve for molality

 = 9.557956 = 9.56 m

The minimum concentration of ethylene glycol would have to be 9.56 m in order to prevent the water from

freezing at 0.00°F

13.9A Plan: Use the osmotic pressure, the temperature, and the gas constant to calculate the molarity of the solution with

the equation  = MRT Multiply the molarity of the solution by the volume of the sample to calculate the number

of moles in the sample The molar mass is calculated by dividing the mass of the sample by the number of moles

13.10A Plan: Write the formula of potassium phosphate, determine if the compound is soluble and, if soluble, how many

cations and anions result when one unit of the compound is placed in water Calculate the molality of the solution The boiling point of a solution is increased relative to the pure solvent by the relationship Tb = iKbm where Kb is

0.512°C/m for water, i is the van’t Hoff factor (equal to the number of particles or ions produced when one unit of the compound is dissolved in the solvent), and m is the molality of particles in solution Once Tb is calculated, the boiling point is determined by adding it to the boiling point of pure H2O (100.00°C)

Solution:

a) The formula for potassium phosphate is K3PO4 When it is placed in water, 3 K+ ions and 1 PO43– ion are formed Scene B is the only scene that shows separate ions in a 3K+/1PO43– ratio

b)

Molality of K3PO4 solution:

Moles of K3PO4 = (31.2 g K3PO4) 1 mol K3PO4

212.27 g K3PO4 = 0.147 mol K3PO4Molality of K3PO4 = moles solute

kg solvent = 0.147 mol K3 PO4

85.0 g H2O

1000 g

1 kg = 1.73 m K3PO4Because 4 ions total are formed when each unit of K3PO4 is placed in water, i = 4

Tb = iKbm = (4)(0.512 °C/m)(1.73 m) = 3.54°C

The boiling point is 100.00°C + 3.54°C = 103.54°C

13.10B Plan: Calculate the molarity of the solution, then calculate the osmotic pressure of the magnesium chloride

solution Do not forget that MgCl2 is a strong electrolyte, and ionizes to yield three ions per formula unit This will result in a pressure three times as great as a nonelectrolyte (glucose) solution of equal concentration

 = iMRT = 3 0.099640 mol 0.0821L • atm  273.2 20.0 K 

rise Scene C represents this situation

CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B13.1 a) The colloidal particles in water generally have negatively charged surfaces and so repel each other, slowing the

settling process Cake alum, Al2(SO4)3, is added to coagulate the colloids The Al3+ ions neutralize the negative surface charges and allow the particles to aggregate and settle

b) Water that contains large amounts of divalent cations (such as Ca2+ and Mg2+) is called hard water During cleaning, these ions combine with the fatty-acid anions in soaps to produce insoluble deposits

c) In reverse osmosis, a pressure greater than the osmotic pressure is applied to the solution, forcing the water back through the membrane and leaving the ions behind

d) Chlorine may give the water an unpleasant odor, and can form carcinogenic chlorinated compounds

e) The high concentration of NaCl displaces the divalent and polyvalent ions from the ion-exchange resin B13.2 Plan: Osmotic pressure is calculated from the molarity of particles, the gas constant, and temperature Convert the

mass of sucrose to moles using the molar mass, and then to molarity Sucrose is a nonelectrolyte so i = 1

volume of solution =

0.01037102 mol1.0 L = 1.037102x10

–2 M sucrose

 = iMRT = (1)(1.037102x10–2 mol/L)(0.0821 L•atm/mol•K)(293 K) = 0.2494780 = 0.249 atm

A pressure greater than 0.249 atm must be applied to obtain pure water from a 3.55 g/L solution

END–OF–CHAPTER PROBLEMS

13.1 The composition of seawater, like all mixtures, is variable The components of seawater (water and various

ions) have not been changed and thus retain some of their properties For example, seawater has a salty taste due

to the presence of salts such as NaCl

13.2 When a salt such as NaCl dissolves, ion-dipole forces cause the ions to separate, and many water molecules cluster

around each of them in hydration shells Ion-dipole forces hold the first shell Additional shells are held by hydrogen bonding to inner shells

13.3 In CH3(CH2)n COOH, as n increases, the hydrophobic (CH) portion of the carboxylic acid increases and the

hydrophilic part of the molecule stays the same, with a resulting decrease in water solubility

13.4 Sodium stearate would be a more effective soap because the hydrocarbon chain in the stearate ion is longer than

the chain in the acetate ion A soap forms suspended particles called micelles with the polar end of the soap interacting with the water solvent molecules and the nonpolar ends forming a nonpolar environment inside the micelle Oils dissolve in the nonpolar portion of the micelle Thus, a better solvent for the oils in dirt is a more nonpolar substance The long hydrocarbon chain in the stearate ion is better at dissolving oils in the micelle than the shorter hydrocarbon chain in the acetate ion

13.5 Hexane and methanol, as gases, are free from any intermolecular forces and can simply intermix with each other

As liquids, hexane is a nonpolar molecule, whereas methanol is a polar molecule “Like dissolves like.”

13.6 Hydrogen chloride (HCl) gas is actually reacting with the solvent (water) and thus shows a higher solubility than propane (C3H8) gas, which does not react

13.7 Plan: A more concentrated solution will have more solute dissolved in the solvent Determine the types of

intermolecular forces in the solute and solvents A solute tends to be more soluble in a solvent whose

intermolecular forces are similar to its own

Solution:

Potassium nitrate, KNO3, is an ionic compound and can form ion-dipole forces with a polar solvent like water, thus dissolving in the water Potassium nitrate is not soluble in the nonpolar solvent CCl4 Because potassium

nitrate dissolves to a greater extent in water, a) KNO 3 in H 2 O will result in the more concentrated solution

13.8 b) Stearic acid in CCl 4 Stearic acid will not dissolve in water It is nonpolar while water is very polar Stearic

acid will dissolve in carbon tetrachloride, as both are nonpolar

13.9 Plan: To identify the strongest type of intermolecular force, check the formula of the solute and identify the forces

that could occur Then look at the formula for the solvent and determine if the forces identified for the solute would occur with the solvent Ionic forces are present in ionic compounds; dipole-dipole forces are present in polar substances, while nonpolar substances exhibit only dispersion forces The strongest force is ion-dipole followed by dipole-dipole (including H bonds) Next in strength is ion–induced dipole force and then dipole–induced dipole force The weakest intermolecular interactions are dispersion forces

Solution:

a) Ion-dipole forces are the strongest intermolecular forces in the solution of the ionic substance cesium chloride

in polar water

b) Hydrogen bonding (type of dipole-dipole force) is the strongest intermolecular force in the solution of polar

propanone (or acetone) in polar water

c) Dipole–induced dipole forces are the strongest forces between the polar methanol and nonpolar carbon

tetrachloride

13.10 a) metallic bonding

13.11 Plan: To identify the strongest type of intermolecular force, check the formula of the solute and identify the forces

that could occur Then look at the formula for the solvent and determine if the forces identified for the solute would occur with the solvent Ionic forces are present in ionic compounds; dipole-dipole forces are present in polar substances, while nonpolar substances exhibit only dispersion forces The strongest force is ion-dipole followed by dipole-dipole (including H bonds) Next in strength is ion–induced dipole force and then dipole–induced dipole force The weakest intermolecular interactions are dispersion forces

Solution:

a) Hydrogen bonding occurs between the H atom on water and the lone electron pair on the O atom in dimethyl

ether (CH3OCH3) However, none of the hydrogen atoms on dimethyl ether participates in hydrogen bonding

because the CH bond does not have sufficient polarity

b) The dipole in water induces a dipole on the Ne(g) atom, so dipole–induced dipole interactions are the

strongest intermolecular forces in this solution

c) Nitrogen gas and butane are both nonpolar substances, so dispersion forces are the principal attractive forces

13.12 a) dispersion forces

13.13 Plan: CH3CH2OCH2CH3 is polar with dipole-dipole interactions as the dominant intermolecular forces Examine

the solutes to determine which has intermolecular forces more similar to those in diethyl ether This solute is the one that would be more soluble

Solution:

a) HCl would be more soluble since it is a covalent compound with dipole-dipole forces, whereas NaCl is an ionic

solid Dipole-dipole forces between HCl and diethyl ether are more similar to the dipole forces in diethyl ether than the ion-dipole forces between NaCl and diethyl ether

b) CH 3 CHO (acetaldehyde) would be more soluble The dominant interactions in H2O are hydrogen bonding, a

stronger type of dipole-dipole force The dominant interactions in CH3CHO are dipole-dipole The solute-solvent interactions between CH3CHO and diethyl ether are more similar to the solvent intermolecular forces than the forces between H2O and diethyl ether

c) CH 3 CH 2 MgBr would be more soluble CH3CH2MgBr has a polar end (–MgBr) and a nonpolar end

(CH3CH2–), whereas MgBr2 is an ionic compound The nonpolar end of CH3CH2MgBr and diethyl ether would interact with dispersion forces, while the polar end of CH3CH2MgBr and the dipole in diethyl ether would interact with dipole-dipole forces

13.14 a) CH 3 CH 2 -O-CH 3(g), due to its smaller size (smaller molar mass)

b) CH 2 Cl 2, because it is more polar than CCl4

molecules

13.15 No, river water is a heterogeneous mixture, with its composition changing from one segment to another

13.16 Plan: Determine the types of intermolecular forces present in the two compounds and in water and hexane

Substances with similar types of forces tend to be soluble while substances with different type of forces tend to be insoluble

Solution:

Gluconic acid is a very polar molecule because it has –OH groups attached to every carbon The abundance of –OH bonds allows gluconic acid to participate in extensive H bonding with water, hence its great solubility in water On the other hand, caproic acid has a five carbon, nonpolar, hydrophobic (“water hating”) tail that does not easily dissolve in water The dispersion forces in the nonpolar tail are more similar to the dispersion forces in hexane, hence its greater solubility in hexane

13.17 There may be a disulfide linkage (a covalent disulfide bridge) between the ends of two cysteine side chains

that bring together parts of the chain There may be salt links between ions –COO– and –NH3 groups There may

be hydrogen bonding between the C=O of one peptide bond and the N–H of another

13.18 The nitrogen bases hydrogen bond to their complimentary bases The flat, N-containing bases stack above each

other, which allow extensive interaction through dispersion forces The exterior negatively charged

sugar-phosphate chains form ion-dipole and hydrogen bonds to the aqueous surroundings, but this is of minor

importance to the structure

13.19 While an individual hydrogen bond is not too strong, there are very large numbers of hydrogen bonds present in

DNA The energy of so many hydrogen bonds keeps the chains together But the hydrogen bonds are weak enough that a few are easily broken when the two chains in DNA must separate

13.20 The more carbon and hydrogen atoms present, the more soluble the substance is in nonpolar oil droplets

Therefore, sodium propanoate is not as effective a soap as sodium stearate with the longer hydrocarbon chain 13.21 Dispersion forces are present between the nonpolar portions of the molecules within the bilayer Polar groups are

present to hydrogen bond or to form ion-dipole interactions with the aqueous surroundings

13.22 In soluble proteins, polar groups are found on the exterior and nonpolar groups on the interior In proteins

embedded in a membrane, the exterior of the protein that lies within the bilayer consists of nonpolar amino acid side chains, whereas the portion lying outside the bilayer has polar side chains

13.23 Amino acids with side chains that may be ionic are necessary Two examples are lysine and glutamic acid

13.24 The Hsolvent and Hmix components of the heat of solution combined together represent the enthalpy change

during solvation, the process of surrounding a solute particle with solvent particles When the solvent is water, solvation is called hydration

13.25 For a general solvent, the energy changes needed to separate solvent into particles (Hsolvent), and that needed to

mix the solvent and solute particles (Hmix) would be combined to obtain Hsolution

13.26 a) Charge density is the ratio of the ion’s charge to its volume An ion’s charge and size affect its charge density

b) – < + < 2– < 3+

c) The higher the charge density, the more negative the Hhydration Hhydration increases with increasing charge and decreases with increasing size

13.27 The solution cycle for ionic compounds in water consists of two enthalpy terms: the lattice energy, and the

combined heats of hydration of the cation and anion

Hsolution = Hlattice +Hhydration of ionsFor a heat of solution to be zero (or very small)

Hlattice  Hhydration of ions, and they would have to have opposite signs

13.28 a) Endothermic

b) The lattice energy term is much larger than the combined ionic heats of hydration

c) The increase in entropy outweighs the increase in enthalpy, so ammonium chloride dissolves

13.29 This compound would be very soluble in water A large exothermic value in Hsolution (enthalpy of solution)

means that the solution has a much lower energy state than the isolated solute and solvent particles, so the system tends to the formation of the solution Entropy that accompanies dissolution always favors solution formation Entropy becomes important when explaining why solids with endothermic Hsolution values (and higher energy states) are still soluble in water

13.30 Plan: Hsolution = Hlattice +Hhydration Lattice energy values are always positive as energy is required to

separate the ions from each other Hydration energy values are always negative as energy is released when intermolecular forces between ions and water form Since the heat of solution for KCl is endothermic, the lattice energy must be greater than the hydration energy for an overall input of energy

13.31 Lattice energy values are always positive as energy is required to separate the ions from each other

Hydration energy values are always negative as energy is released when intermolecular forces between

ions and water form Since the heat of solution for NaI is exothermic, the negative hydration energy must be greater than the positive lattice energy

13.32 Plan: Charge density is the ratio of an ion’s charge (regardless of sign) to its volume An ion’s volume is

related to its radius For ions whose charges have the same sign (+ or –), ion size decreases as a group in

the periodic table is ascended and as you proceed from left to right in the periodic table Charge density increases with increasing charge and increases with decreasing size

Solution:

a) Both ions have a +1 charge, but the volume of Na + is smaller, so it has the greater charge density

b) Sr 2+ has a greater ionic charge and a smaller size (because it has a greater Zeff), so it has the greater charge density

c) Na + has a smaller ion volume than Cl–, so it has the greater charge density

d) O 2– has a greater ionic charge and similar ion volume, so it has the greater charge density

e) OH – has a smaller ion volume than SH– (O is smaller than S), so it has the greater charge density

f) Mg 2+ has the higher charge density because it has a smaller ion volume

g) Mg 2+

has the higher charge density because it has both a smaller ion volume and greater charge

h) CO 3 2– has the higher charge density because it has both a smaller ion volume and greater charge

13.33 a) I – has a smaller charge density (larger ion volume) than Br–

b) Ca 2+ has a lower ratio than Sc3+, due to its smaller ion charge

c) Br – has a lower ratio than K+, due to its larger ion volume

d) Cl – has a lower ratio than S2–, due to its smaller ion charge

e) Sc 3+ has a lower ratio than Al3+, due to its larger ion volume

f) ClO 4 has a lower ratio due to its smaller ion charge

g) Fe 2+ has a lower ratio due to its smaller ion charge

h) K + has a lower ratio due to its smaller ion charge

13.34 Plan: The ion with the greater charge density will have the larger Hhydration

Solution:

a) Na + would have a larger Hhydration than Cs+ since its charge density is greater than that of Cs+

b) Sr 2+ would have a larger Hhydration than Rb+

c) Na + would have a larger Hhydration than Cl–

d) O 2– would have a larger Hhydration than F–

e) OH – would have a larger Hhydration than SH–

f) Mg 2+ would have a larger Hhydration than Ba2+

g) Mg 2+ would have a larger Hhydration than Na+

h) CO 3 2– would have a larger Hhydration than NO3

13.35 a) I – b) Ca 2+ c) Br – d) Cl – e) Sc 3+ f) ClO 4 g) Fe 2+ h) K +

13.36 Plan: Use the relationship Hsolution = Hlattice +Hhydration Given Hsolution and Hlattice, Hhydration can be

calculated Hhydration increases with increasing charge density, and charge density increases with increasing charge and decreasing size

Solution:

a) The two ions in potassium bromate are K+ and BrO3

Hsolution = Hlattice +Hhydration

Hhydration = Hsolution – Hlattice = 41.1 kJ/mol – 745 kJ/mol = –703.9 = –704 kJ/mol

b) K + ion contributes more to the heat of hydration because it has a smaller size and, therefore, a greater charge density

13.37 a) Hhydration = Hsolution – Hlattice

Hhydration = 17.3 kJ/mol –763 kJ/mol

Hhydration = –745.7 = – 746 kJ/mol

b) Na + ion contributes more to the heat of hydration due to its smaller size (larger charge density)

13.38 Plan: Entropy increases as the possible states for a system increase, which is related to the freedom of motion of

its particles and the number of ways they can be arranged

Solution:

a) Entropy increases as the gasoline is burned Gaseous products at a higher temperature form

b) Entropy decreases as the gold is separated from the ore Pure gold has only the arrangement of gold atoms next

to gold atoms, while the ore mixture has a greater number of possible arrangements among the components of the mixture

c) Entropy increases as a solute dissolves in the solvent

13.39 a) Entropy increases

b) Entropy decreases

c) Entropy increases

13.40 Hsolution = Hlattice + Hhydration

Hsolution = 822 kJ/mol + (– 799 kJ/mol)

Hsolution = 23 kJ/mol

13.41 Add a pinch of the solid solute to each solution A saturated solution contains the maximum amount of dissolved

solute at a particular temperature When additional solute is added to this solution, it will remain undissolved An unsaturated solution contains less than the maximum amount of dissolved solute and so will dissolve added solute

A supersaturated solution is unstable and addition of a “seed” crystal of solute causes the excess solute to

crystallize immediately, leaving behind a saturated solution

13.42 KMnO4(s) + H2O(l) + heat → KMnO4(aq)

Prepare a mixture of more than 6.4 g KMnO4/100 g H2O and heat it until the solid completely dissolves Then carefully cool it, without disturbing it or shaking it, back to 20°C If no crystals form, you would then have a supersaturated solution

13.43 An increase in temperature produces an increase in kinetic energy; the gaseous solute molecules overcome the

weak intermolecular forces, which results in a decrease in solubility of any gas in water In nearly all cases, gases dissolve exothermically (Hsoln < 0)

13.44 Plan: The solubility of a gas in water decreases with increasing temperature and increases with increasing

pressure

Solution:

a) Increasing pressure for a gas increases the solubility of the gas according to Henry’s law

b) Increasing the volume of a gas causes a decrease in its pressure (Boyle’s law), which decreases the solubility

of the gas

13.45 a) increase b) stay the same

13.46 Plan: Solubility for a gas is calculated from Henry’s law: Sgas = kH  Pgas We know kH and Pgas, so Sgas can be calculated with units of mol/L To calculate the mass of oxygen gas, convert moles of O2 to mass of O2 using the

13.48 The solution is saturated

13.49 Plan: Solubility for a gas is calculated from Henry’s law: Sgas = kH  Pgas We know kH and Pgas, so Sgas can be calculated with units of mol/L

Solution:

Sgas = kH  Pgas = (3.7x10–2 mol/L•atm)(5.5 atm) = 0.2035 = 0.20 mol/L

13.50 Solubility of gases increases with increasing partial pressure of the gas, and the goal of these devices is to increase

the amount of oxygen dissolving in the bloodstream

13.51 Molarity is defined as the number of moles of solute dissolved in one liter of solution Molality is defined as the

number of moles of solute dissolved in 1000 g (1 kg) of solvent Molal solutions are prepared by measuring masses of solute and solvent, which are additive and not changed by temperature, so the concentration in molality does not change with temperature and is the preferred unit when the temperature of the solution may change 13.52 Plan: Refer to the table of concentration definitions for the different methods of expressing concentration

Solution:

b) Parts-by-mass (% w/w) include the mass of solution directly (Others may involve the mass indirectly.)

13.53 No, 21 g solute/kg of solvent would be 21 g solute/1.021 kg solution

13.54 Converting between molarity and molality involves conversion between volume of solution and mass of solution

Both of these quantities are given so interconversion is possible To convert to mole fraction requires that the mass of solvent be converted to moles of solvent Since the identity of the solvent is not given, conversion to mole fraction is not possible if the molar mass is not known

13.55 % w/w, mole fraction, and molality are weight-to-weight relationships that are not affected by changes in

temperature % w/v and molarity are affected by changes in temperature, because the volume is temperature dependant

13.56 Plan: The molarity is the number of moles of solute in each liter of solution: M = mol of solute

V(L) of solution Convert the masses to moles and the volumes to liters and divide moles by volume

a) Mconc = 6.25 M HCl Vconc = 25.5 mL Mdil = ? Vdil = 0.500 L

Mdil =   

 

conc conc dil

M



= 0.01375 = 0.0138 M

13.60 Plan: For part a), find the number of moles of KH2PO4 needed to make 365 mL of a solution of this molarity

Convert moles to mass using the molar mass of KH2PO4 For part b), use the relationship MconcVconc = MdilVdil to

find the volume of 1.25 M NaOH needed

1 mol KH PO

  = 4.24703 = 4.25 g KH2PO4 Add 4.25 g KH 2 PO 4 to enough water to make 365 mL of aqueous solution

b) Mconc = 1.25 M NaOH Vconc = ? Mdil = 0.335 M NaOH Vdil = 465 mL

M

M = 124.62 = 125 mL

Add 125 mL of 1.25 M NaOH to enough water to make 465 mL of solution

13.61 a) Find the number of moles NaCl needed to make 2.5 L of this solution Convert moles to mass using the molar

Add 95 g NaCl to enough water to make 2.5 L of aqueous solution

b) Use the relationship MconcVconc = MdilVdil to find the volume of 2.1 M urea needed

Mconc = 2.1 M urea Vconc = ? Mdil = 0.3 M urea Vdil = 15.5 L

M

M = 2.21429 = 2 L Add 2 L of 2.1 M urea to enough water to make 15.5 L of solution

Note because of the uncertainty in the concentration of the dilute urea (0.3 M), only one significant figure

is justified in the answer

13.62 Plan: To find the mass of KBr needed in part a), find the moles of KBr in 1.40 L of a 0.288 M solution and

convert to grams using the molar mass of KBr To find the volume of the concentrated solution that will be diluted

to 255 mL in part b), use MconcVconc = MdilVdil and solve for Vconc

To make the solution, weigh 48.0 g KBr and then dilute to 1.40 L with distilled water

b) Mconc = 0.264 M LiNO3 Vconc = ? Mdil = 0.0856 M LiNO3 Vdil = 255 mL

M

M = 82.68182 = 82.7 mL

To make the 0.0856 M solution, measure 82.7 mL of the 0.264 M solution and add distilled water to make a total

of 255 mL

13.63 a) To find the mass of Cr(NO3)3 needed, find the moles of Cr(NO3)3 in 57.5 mL of a 1.53x10–3 M solution and

convert to grams using molar mass of Cr(NO3)3

3 3

1.53x10 mol Cr(NO ) 238.03 g Cr(NO )

10 L57.5 mL

To make the solution, weigh 0.0209 g Cr(NO 3 ) 3 and then dilute to 57.5 mL with distilled water

b) To find the volume of the concentrated solution that will be diluted to 5.8x103 m3 use MconcVconc = MdilVdil and solve for Vconc

Mconc = 2.50 M NH4NO3 Vconc = ? Mdil = 1.45 M NH4NO3 Vdil = 5.8x103 m3

M

M = 3.364x103 = 3.4x103 m3

To make the 1.45 M solution, measure 3.4x103

m 3 of the 2.50 M solution and add distilled water to make

5.8x103 m3

13.64 Plan: Molality, m, = moles of solute

kg of solvent Convert the mass of solute to moles and divide by the mass of solvent in

13.66 Plan: Molality, m, = moles of solute

kg of solvent Use the density of benzene to find the mass and then the moles of benzene; use the density of hexane to find the mass of hexane and convert to units of kg Divide the moles of benzene by the mass of hexane

2.66 mL CCl

mL 153.81 g CCl 10 g

1.33 g 1 kg76.5 mL CH Cl

13.68 Plan: In part a), the total mass of the solution is 3.10x102 g, so masssolute + masssolvent = 3.10x102 g

Assume that you have 1000 g of the solvent water and find the mass of C2H6O2 needed to make a 0.125 m

solution Then a ratio can be used to find the mass of C2H6O2 needed to make 3.10x102 g of a 0.125 m solution

Part b) is a dilution problem First, determine the amount of solute in your target solution and then determine the amount of the concentrated acid solution needed to get that amount of solute

Mass (g) of C2H6O2 for 3.10x102 g of solution = 7.75875 g C H O2 6 2  2 

3.10x10 g solution1007.75875 g solution

= 2.386695 g C2H6O2

Masssolvent = 3.10x102 g – masssolute = 3.10x102 g – 2.386695 g C2H6O2 = 307.613305 = 308 g H2O

Therefore, add 2.39 g C 2 H 6 O 2 to 308 g of H 2O to make a 0.125 m solution

b) Mass (kg) of HNO3 in the 2.20% solution = 1.20 kg 2.20%

100%

  = 0.0264 kg HNO3 (solute) Mass % = mass of solute  100

mass of solutionMass of 52.0% solution containing 0.0264 kg HNO3 = mass of solute 100  0.0264 kg 100 

= mass % 52.0%

= 0.050769 = 0.0508 kg

Mass of water added = mass of 2.2% solution – mass of 52.0% solution

= 1.20 kg – 0.050769 kg = 1.149231 = 1.15 kg

Add 0.0508 kg of the 52.0% (w/w) HNO 3 to 1.15 kg H 2 O to make 1.20 kg of 2.20% (w/w) HNO3

13.69 a) The total weight of the solution is 1.50 kg, so

masssolute + masssolvent = 1.50 kg

Mass (g) of C2H5OH for 1.50 kg of solution = 1.635485 g C H OH2 5  

1500 g solution 1001.635485 g solution

= 2.449222 = 2.45 g C2H5OH

Masssolvent = 1500 g – masssolute = 1500 g – 2.449222 g C2H5OH = 1497.551 = 1498 g H2O

Therefore, add 2.45 g C 2 H 5 OH to 1498 g of H 2O to make a 0.0355 m solution

b) This is a disguised dilution problem First, determine the amount of solute in your target solution:

Mass (kg) of HCl in the 13.0% solution = 445 g 13.0%

100%

  = 57.85 g HCl (solute) Then determine the amount of the concentrated acid solution needed to get 57.85 g solute:

Mass % = mass of solute  100

mass of solutionMass of 34.1% solution containing 57.85 g HCl = mass of solute 100  57.85 g 100 

= mass % 34.1% = 169.6481 = 170 g Mass of water added = mass of 13.0% solution – mass of 34.1% solution

= 445 g – 169.6481 g = 275.35191 = 275 g

Add 170 g of the 34.1% (w/w) HCl to 275 g H 2 O

13.70 Plan: You know the moles of solute (C3H7OH) and the moles of solvent (H2O) Divide moles of C3H7OH by the

total moles of C3H7OH and H2O to obtain mole fraction To calculate mass percent, convert moles of solute and solvent to mass and divide the mass of solute by the total mass of solution (solute + solvent) For molality, divide the moles of C3H7OH by the mass of water expressed in units of kg

Solution:

a) Mole fraction is moles of isopropanol per total moles

Xisopropanol = moles of isopropanol

moles of isopropanol + moles of water = 0.35 mol isopropanol0.350.85 mol = 0.2916667 = 0.29

(Notice that mole fractions have no units.)

b) Mass percent = mass of solute  100

mass of solution From the mole amounts, find the masses

of isopropanol and water:

3 7

3 7

60.09 g C H OH0.35 mol C H OH

1 mol H O

  = 15.317 g water Mass percent = mass of solute  100

mass of solution = 21.0315 g isopropanol21.0315 15.317 g   100 = 57.860710 = 58%

c) Molality of isopropanol is moles of isopropanol per kg of water

m = moles of solute

kg of solvent =

3

0.35 mol isopropanol 10 g15.317 g water 1 kg

  = 22.85043 = 23 m isopropanol 13.71 a) Mole fraction is moles of NaCl per total moles

XNaCl = 0.100 8.60 mol0.100 mol NaCl  = 0.01149425 = 0.0115 (Notice that mole fractions have no units.)

b) Mass percent is the mass of NaCl per 100 g of solution

Mass (g) of NaCl = (0.100 mol NaCl)(58.44 g/mol) = 5.844 g NaCl

Mass (g) of water = (8.60 mol water)(18.02 g/mol) = 154.972 g water

5.844 g NaCl

x100%

5.844 154.972 g = 3.63396677 = 3.63% NaCl c) Molality of NaCl is moles of NaCl per kg of solvent

Molality NaCl =

3

0.100 mol NaCl 10 g154.972 g water 1 kg

  = 0.645277856 = 0.645 m NaCl

13.72 Plan: Molality = moles of solute

kg of solvent Use the density of water to convert the volume of water to mass Multiply the mass of water in kg by the molality to find moles of cesium bromide; convert moles to mass To find the mole fraction, convert the masses of water and cesium bromide to moles and divide moles of cesium bromide by the total moles of cesium bromide and water To calculate mass percent, divide the mass of cesium bromide by the total mass of solution and multiply by 100

Moles of H2O = (400 g H2O)(1 mol H2O/18.02 g H2O) = 22.197558 mol H2O

Moles of KI = (0.30 g KI)(1 mol KI /166.0 g KI) = 1.80723x10–3 mol KI

XKI =

3 3

1.80723x10 mol KI1.80723x10 22.197558 mol



–5 = 8.1x10 –5

13.74 Plan: You are given the mass percent of the solution Assuming 100 g of solution allows us to

express the mass % as the mass of solute, NH3 To find the mass of solvent, subtract the mass of NH3 from the mass of solution and convert to units of kg To find molality, convert mass of NH3 to moles and divide by the mass of solvent in kg To find molarity, you will need the volume of solution Use the density of the solution to convert the 100 g of solution to volume in liters; divide moles of NH3 by volume of solution To find the mole fraction, convert mass of solvent to moles and divide moles of NH3 by the total moles

Mass (kg) of H2O =  2  3

1 kg92.00 g H O

18.02 g H O

  = 5.1054 mol H2O Volume (L) of solution = 100.00 g solution 1 mL solution 10 L3

Using the above fundamental quantities and the definitions of the various units:

Molality = m = moles of solute

kg of solvent =

3 2

0.469759 mol NH0.09200 kg H O

 = 5.106076 = 5.11 m NH3 Molarity = M = moles of solute

L of solution =

3

0.469759 mol NH0.103616 L

  = 4.53365 = 4.53 M NH3 Mole fraction = X = moles of NH3

total moles = 0.469759 5.1054 mol0.469759 mol NH 3 = 0.084259 = 0.0843

13.75 The information given is 28.8 mass % FeCl3 solution with a density of 1.280 g/mL

For convenience, choose exactly 100.00 g of solution

Determine some fundamental quantities:

Mass (g) of FeCl3 = (100.00 g solution)(28.8% FeCl3/100%) = 28.8 g FeCl3

Mass (g) of H2O = mass of solution – mass FeCl3 = (100.00 – 28.8) g = 71.20 g H2O

Moles of FeCl3 = (28.80 g FeCl3)(1 mol FeCl3/162.20 g FeCl3) = 0.1775586 mol FeCl3

Moles of H2O = (71.20 g H2O)(1 mol H2O/18.02 g H2O) = 3.951165 mol H2O

Volume of solution = (100.00 g solution)(1 mL/1.280 g)(10–3 L/1 mL) = 0.078125 L

Using the above fundamental quantities and the definitions of the various units:

Molality = M = moles solute/kg solvent =

3 3 2

0.1775586 mol FeCl 10 g71.20 g H O 1 kg

0.1775586 3.951165 mol0.1775586 mol FeCl 3 = 0.043005688 = 0.0430

13.76 Plan: Use the equation for parts per million, ppm Use the given density of solution to find the mass of solution;

divide the mass of each ion by the mass of solution and multiply by 1x106

13.77 The information given is that ethylene glycol has a density of 1.114 g/mL and a molar mass of 62.07 g/mol Water

has a density of 1.00 g/mL The solution has a density of 1.070 g/mL

For convenience, choose exactly 1.0000 L as the equal volumes mixed Ethylene glycol will be

designated EG

Determine some fundamental quantities:

Mass (g) of EG = (1.0000 L EG)(1mL/10–3 L)(1.114 g EG/mL) = 1114 g EG

Mass (g) of H2O = (1.0000 L H2O)(1mL/10–3 L)(1.00 g H2O/mL) = 1.00x103 g H2O

Moles of EG = (1114 g EG)(1 mol EG/62.07 g EG) = 17.94747865 mol EG

Moles of H2O = (1.00x103 g H2O)(1 mol H2O/18.02 g H2O) = 55.49389567 mol H2O

Volume (L) of solution = (1114 g EG + 1.00x103 g H2O)(1 mL/1.070 g)(10–3 L/1 mL)

= 1.97570 L Using the above fundamental quantities and the definitions of the various units:

a) Volume percent = (1.0000 L EG/1.97570 L)100% = 50.61497 = 50.61% v/v

b) Mass percent = [(1114 g EG)/(1114 + 1.00x103) g]100% = 52.6963 = 52.7% w/w

c) Molarity = moles solute/L solution = 17.94747865 mol EC

1.97570 L = 9.08411 = 9.08 M ethylene glycol d) Molality = moles solute/kg solvent =

3 3

2

17.94747865 mol EG 10 g

1 kg1.00 x10 g H O

13.78 Colligative properties of a solution are affected by the number of particles of solute in solution The density of a

solution would be affected by the chemical formula of the solute

13.79 A nonvolatile nonelectrolyte is a covalently bonded molecule that does not dissociate into ions or evaporate when

dissolved in a solvent In this case, the colligative concentration is equal to the molar concentration, simplifying calculations

13.80 The “strong” in “strong electrolyte” refers to the ability of an electrolyte solution to conduct a large current This

conductivity occurs because solutes that are strong electrolytes dissociate completely into ions when dissolved in water

13.81 Raoult’s law states that the vapor pressure of solvent above the solution equals the mole fraction of the solvent

times the vapor pressure of the pure solvent Raoult’s law is not valid for a solution of a volatile solute in solution Both solute and solvent would evaporate based upon their respective vapor pressures