Chapter 12 Solutions. From Chapter 1: Classification of matter Matter Homogeneous (visibly...
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Transcript of Chapter 12 Solutions. From Chapter 1: Classification of matter Matter Homogeneous (visibly...
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.