Chapter 12

Solutions

From Chapter 1: Classification of matter

Matter

Homogeneous(visibly indistinguishable)

Heterogeneous (visibly distinguishable)

Elements

Compounds

Mixtures(multiple components)

Pure Substances(one component)

(Solutions)

Solution = Solute + Solvent

Vodka = ethanol + water Brass = copper + zinc

If solvent is water, the solution is called an aqueous solution.

LiquorBeer Wine

Ethanol Concentration

Four Concentrations

Unit: none

Unit: mol/L

%100solutionofmass

soluteofmasspercentMass (1)

solutionofliters

soluteofmoles(M)Molarity (2)

Four Concentrations

Unit: none

moles of solute Mole fraction of solute

moles of solution(3)

BA

AA nn

n

χ

moles of solvent Mole fraction of solvent

moles of solution

Unit: noneB

A B

n

n nBχ

A B 1χ χ

Four Concentrations

Unit: mol/kg

moles of soluteMolaliy (m)

kilograms of solvent(4)

A solution contains 5.0 g of toluene (C7H8) and 225 g of

benzene (C6H6) and has a density of 0.876 g/mL.

Calculate the mass percent and mole fraction of C7H8, and

the molarity and molality of the solution.

Practice on Example 12.4 on page 533 and compare your results with the answers.

Electrical Conductivity of Aqueous Solutions

solute

strong electrolyte

weak electrolyte

nonelectrolyte

strong acids

strong bases

salts

weak acids

weak bases

many organic compounds

van’t Hoff factor

dissolved solute of moles

solute from particles ofmolesi

nonelectrolyte: i = 1

strong electrolyte: depends on chemical formula

weak electrolyte: depends on degree of dissociation

Unit: none

MgBr2 MgSO4 FeCl3

Glucose

Mg3(PO4)2

dissolved solute of moles

solute from particles ofmolesi

NaOH

Hexane

Four properties of solutions

(1) Boiling point elevation

water = solvent

water + sugar = solution

Boiling point = 100 °C

Boiling point > 100 °C

Solution compared to pure solvent

Sugar Dissolved in Water to Make Candy Causes the Boiling Point to be Elevated

∆Tb = Tb,solution − Tb,solvent = i Kb m

i: van’t Hoff factor

m: molality

Kb: boiling-point elevation constant

Kb is characteristic of the solvent. Does notdepend on solute.

Units

Boiling point elevation can be used to find molar mass of solute.

∆Tb ― experiments

i ― electrolyte or nonelectrolyte

Kb ― table or reference book

b

b

Ki

ΔT

solventofkilogram

soluteofmolesm

b

bbb Ki

ΔTm m Ki ΔT

soluteofmoles

solute of masssolute of massmolar

A solution was prepared by dissolving 18.00 g glucose in 150.0 g

water. The resulting solution was found to have a boiling point of

100.34 °C. Calculate the molar mass of glucose. Glucose is

molecular solid that is present as individual molecules in solution.

b

b

Ki

ΔT

solventofkilogram

soluteofmolesm

soluteofmoles

solute of masssolute of massmolar

180 g/mol

Four properties of solutions

(1) Boiling point elevation

(2) Freezing point depression

water = solvent

water + salt = solution

freezing point = 0 °C

freezing point < 0 °C

Solution compared to pure solvent

∆Tf = Tf,solvent − Tf,solution = i Kf m

i: van’t Hoff factor

m: molality

Kf: freezing-point depression constant

Kf is characteristic of the solvent. Does notdepend on solute.

Units

Freezing point depression can be used to find molar mass of solute.

∆Tf ― experiments

i ― electrolyte or nonelectrolyte

Kf ― table or reference book

f

f

Ki

ΔT

solventofkilogram

soluteofmolesm

f

fff Ki

ΔTm m Ki ΔT

soluteofmoles

solute of masssolute of massmolar

A chemist is trying to identify a human hormone that controls

metabolism by determining its molar mass. A sample weighing

0.546 g was dissolved in 15.0 g benzene, and the freezing-point

depression was determined to be 0.240 °C. Calculate the molar

mass of the hormone.

f

f

Ki

ΔT

solventofkilogram

soluteofmolesm

soluteofmoles

solute of masssolute of massmolar

776 g/mol

The Addition of Antifreeze Lowers the Freezing Point of Water in a Car's Radiator

0 °C 100 °C

water

< 0 °C > 100 °C

antifreeze = water + ethylene glycol

Four properties of solutions

(1) Boiling point elevation

(2) Freezing point depression

(3) Osmotic pressure

Osmotic Pressure

Π = iMRT

Π ― osmotic pressuue

M ― molarity

R ― ideal gas constant

T ― temperature

i ― van’t Hoff factor

Π = iMRT

Units

Π ― atm

M ― mol/L

R ― atm·L·K−1·mol−1

T ― K

i ― none

Osmotic pressure can be used to find molar mass of solute.

iRT M iMRTΠ

Π ― experiments

i ― electrolyte or nonelectrolyte

R ― constant T ― experiments

RTisolutionofliters

soluteofmolesM

soluteofmoles

solute of masssolute of massmolar

To determine the molar mass of a certain protein, 1.00 x 10−3 g

of it was dissolved in enough water to make 1.00 mL of solution.

The osmotic pressure of this solution was found to be 1.12 torr

at 25.0 °C. Calculate the molar mass of the protein.

RTisolutionofliters

soluteofmolesM

soluteofmoles

solute of masssolute of massmolar

1.66 x 104 g/mol

Practice on Example 12.10 on page 546 and compare your results with the answers.

What concentration of NaCl in water is needed to produce an

aqueous solution isotonic with blood ( Π = 7.70 atm at 25 °C)?

0.158 mol/L

Four properties of solutions

(1) Boiling point elevation

(2) Freezing point depression

(3) Osmotic pressure

(4) Lowering the vapor pressure

Lowering Vapor Pressure

Nonvolatile solute to volatile solvent

The Presence of a Nonvolatile Solute Lowers the Vapor Pressure of the Solvent

pure solvent

Liquid SurfaceSurface Molecules

When you count the number of solute particles, use van’t Hoff factor i.

solvent + solute

Liquid Surface

Four Concentrations

Unit: none

moles of solute Mole fraction of solute

moles of solution(3)

BA

AA nn

n

χ

moles of solvent Mole fraction of solvent

moles of solution

Unit: noneB

A B

n

n nBχ

A B 1χ χ

Raoult’s Law: Case 1

solvent0solventsolution PP

solutionP ― vapor pressure of solution

P0solvent ― vapor pressure of pure solvent

solvent ― mole fraction of solvent

Nonvolatile solute in a Volatile solvent

For a Solution that Obeys Raoult's Law, a Plot of Psoln Versus Xsolvent, Give a Straight Line

Calculate the vapor pressure at 25 °C of a solution containing

99.5 g of sucrose (C12H22O11) and 300 mL of water. The vapor

pressure of pure water at 25 °C is 23.8 torr. Assume the

density of water to be 1.00 g/mL.

Example 12.6, page 537

23.4 torr

Predict the vapor pressure of a solution prepared by mixing

35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g

water at 25 °C. The vapor pressure of pure water at 25 °C is

23.76 torr.

When you count the number of solute particles, use van’t Hoff factor i.

solvent + solute

Liquid Surface

Predict the vapor pressure of a solution prepared by mixing

35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g

water at 25 °C. The vapor pressure of pure water at 25 °C is

23.76 torr.

22.1 torr

B0BA

0A

BAsolution

P P

P PP

Raoult’s Law: Case 2

Volatile solute in a Volatile solvent

Recall Dalton’s law of partial pressures

XA + XB = 1

Vapor Pressure for a Solution of Two Volatile Liquids

A mixture of benzene (C6H6) and toluene (C7H8) containing

1.0 mol of benzene and 2.0 mol of toluene. At 20 °C the vapor

pressures of pure benzene and toluene are 75 torr and 22 torr,

respectively. What is the vapor pressure of the mixture?

What is the mole fraction of benzene in the vapor?

0solvent

solution

solutesolvent

solventsolvent P

P χ

nn

n

Lowering vapor pressure can be used to find molar mass of solute.

0solvent

solutionsolventsolvent

0solventsolution P

Pχ χ PP

solutionP and P0solvent ― experiments

soluteofmoles

solute of masssolute of massmolar

At 25 °C of a solution is prepared by dissolving 99.5 g of

sucrose (nonelectrolyte, nonvolatile) into 300 mL of water. The

vapor pressure of the solution and pure water are 23.4 torr and

23.8 torr, respectively. Assume the density of water to be

1.00 g/mL. Calculate the molar mass of sucrose.

Modified Example 12.6, page 537

A solution that obeys Raoult’s Law is called an

ideal solution.

A solution is prepared by mixing 5.81 g acetone (molar mass =

58.1 g/mol) and 11.0 g chloroform (molar mass = 119.4 g/mol).

At 35 °C, this solution has a total vapor pressure of 260. torr.

Is this an ideal solution? The vapor pressure of pure acetone

and pure chloroform at 35 °C are 345 torr and 293 torr,

respectively.

What kind of solution is ideal?

10%

P0

# of molecules in vapor = 100 x 1 x 10% = 10

χ

pure solvent

10%

5%

15%

# of molecules in vapor = 100 x 0.8 x 5% = 4

# of molecules in vapor = 100 x 0.8 x 15% = 12

# of molecules in vapor = 100 x 0.8 x 10% = 8

χ

Raoult’s law:

Deviate fromRaoult’s law

P0

solvent + solute

Psln

What kind of solution is ideal?

Solute-solute, solvent-solvent, and solute-solvent

interactions are very similar.

Comparison to ideal gas.

A solution contains 3.95 g of carbon disulfide (CS2) and 2.43 g of acetone (CH3COCH3). The vapor pressures at 35 C of pure carbon disulfide and pure acetone are 515 torr and 332 torr, respectively. Assuming ideal behavior, calculate the vapor pressures of each of the components and the total vapor pressure above the solution. The experimentally measured total vapor pressure of the solution at 35 C is 645 torr. Is the solution ideal? If not, what can you say about the relative strength of carbon disulfide–acetone interactions compared to the acetone–acetone and carbon disulfide–carbon disulfide interactions?

Example 12.7, page 540

(1) Boiling point elevation: ∆Tb = i Kb m

(2) Freezing point depression: ∆Tf = i Kf m

(3) Osmotic pressure: Π = iMRT

(4) Lowering the vapor pressure: solvent0solventsolution PP

Four Colligative properties of solutions

Colligative: depend on the quantity (number of particles,

concentration) but not the kind or identity of the solute particles.