Solution Composition in Chemistry
Chapter 11
Properties of Solutions
DE Chemistry
Dr. Walker
Solutions
•
In a solution, the excess material is the
solvent
, the smaller amount is the
solute
Measuring Solution Composition
•
Molarity is most common, covered previously
in Chapter 4
Measuring Solution Composition
•
Mass Percent
Mass Percent Example
•
What is the weight percent of ethanol in a
solution made by dissolving 5.3 g of ethanol
(C
2
H
5
OH) in 85.0 mL of water?
Mass Percent Example
•
What is the weight percent of ethanol in a solution
made by dissolving 5.3 g of ethanol (C
2
H
5
OH) in 85.0
mL of water?
•
Density of water = 1.0 g/mL, therefore 85 mL of
water = 85 g
•
Mass of solute/mass of solution x 100 = mass %
•
5.3 g ethanol/(5.3 g ethanol + 85.0 g water) =
5.9%
Measuring Solution Composition
•
Mole Fraction
Mole Fraction Example
•
What is the weight percent of ethanol in a
solution made by dissolving 5.30 g of ethanol
(C
2
H
5
OH) in 85.0 mL of water?
•
Note: If the density of water is 1.0 g/mL, 85
mL = 85 g
Mole Fraction Example
•
What is the weight percent of ethanol in a
solution made by dissolving 5.30 g of ethanol
(C
2
H
5
OH) in 85.0 mL of water?
•
Note: If the density of water is 1.0 g/mL, 85
mL = 85 g
•
Calculate moles of ethanol:
5.30 g C
2
H
5
OH
1 mole C
2
H
5
OH
46.08 g C
2
H
5
OH
= 0.115 moles C
2
H
5
OH
Mole Fraction Example
•
What is the weight percent of ethanol in a
solution made by dissolving 5.30 g of ethanol
(C
2
H
5
OH) in 85.0 mL of water?
•
Note: If the density of water is 1.0 g/mL, 85
mL = 85 g
•
Calculate moles of water:
85.0 g H
2
O
1 mole H
2
O
= 4.72 moles H
2
O
18.02 g H
2
O
Mole Fraction Example
•
What is the weight percent of ethanol in a solution
made by dissolving 5.30 g of ethanol (C
2
H
5
OH) in
85.0 mL of water?
•
Note: If the density of water is 1.0 g/mL, 85 mL = 85
g
•
Mole Fraction of ethanol
•
Mole Fraction of Water
0.115 moles C
2
H
5
OH + 4.72 moles H
2
O
0.115 moles C
2
H
5
OH
= 0.024
0.115 moles C
2
H
5
OH + 4.72 moles H
2
O
4.72 moles H
2
O
= 0.976
Measuring Solution Composition
•
Molality
Example
•
What is the weight percent of ethanol in a
solution made by dissolving 5.30 g of ethanol
(C
2
H
5
OH) in 85.0 mL of water?
•
Note: If the density of water is 1.0 g/mL, 85
mL = 85 g
Example
•
What is the weight percent of ethanol in a solution
made by dissolving 5.30 g of ethanol (C
2
H
5
OH) in
85.0 mL of water?
•
Note: If the density of water is 1.0 g/mL, 85 mL = 85
g
•
Molality = moles solute/kg solvent
5.30 g C
2
H
5
OH
1 mole C
2
H
5
OH
46.08 g C
2
H
5
OH
= 0.115 moles C
2
H
5
OH
Measuring Solution Composition
•
Normality
•
Equivalents of acids and bases
–
Mass that donates or accepts a mole of protons
(usually a subscript of H
+
or OH
-
)
•
Equivalents of oxidizing and reducing agents
–
Mass that provides or accepts a mole of electrons
(usually a subscript)
Normality
•
Find molarity first
•
Normality = Molarity x number of acidic
protons
•
1.5 M HCl = 1.5 N HCl (1 acidic proton)
•
1.5 M H
2
SO
4
= 3.0 N H
2
SO
4
(2 acidic protons)
•
1.5 M H
3
PO
4
= 4.5 N H
3
PO
4
(3 acidic protons)
Energy of Solution Formation
•
“Like Dissolves Like”
–
Polar molecules and ionic compounds tend to
dissolve in polar solvents
–
Nonpolar molecules dissolve in nonpolar
compounds
Water - polar
Butane - nonpolar
Heat of Solution
The
Heat of Solution
is the amount of heat energy
absorbed (endothermic) or released (exothermic)
when a specific amount of solute dissolves in a
solvent.
Steps In Solution Formation
Steps in Solution Formation
H
H
1
1
Step 1 -
Step 1 -
Expanding the solute
Separating the solute into individual
components – usually endothermic
Steps in Solution Formation
H
H
2
2
Step 2 -
Step 2 -
Expanding the solvent
Overcoming intermolecular forces of the solvent
molecules – usually endothermic
Steps in Solution Formation
H
H
3
3
Step 3 -
Step 3 -
Interaction of solute and
solvent to form the solution – usually
exothermic
Enthalpy of Solution
•
May be positive or negative
–
Negative is favorable (exothermic)
–
Positive values are endothermic, does not dissolve
spontaneously
–
Solution formation increases entropy (favorable)
Factors Affecting Solubility
•
Structure Effects
–
Polar (hydrophilic) dissolves in polar
•
Water soluble vitamins
–
Nonpolar (hydrophobic) in nonpolar
•
Fat soluble vitamins
Pressure Effects – Henry’s Law
The concentration of a dissolved gas in a solution
is directly proportional to the pressure of the gas
above the solution
Applies most accurately for dilute solutions of
gases that do not dissociate or react with the
solvent
Yes
Yes
CO
CO
2
2
, N
, N
2
2
, O
, O
2
2
No
No
HCl, HI
HCl, HI
remember,
remember,
these dissociate in solution!
these dissociate in solution!
Henry’s Law Example
Henry’s Law Example
Unopened bottle:
C =
k
P C = (3.1 x 10
-2
mol/L
.
Atm)(5.0 atm) =
0.16 mol/L
Opened bottle:
C =
k
P C = (3.1 x 10
-2
mol/L
.
Atm)(4.0 x 10
-4
atm) =
1.2 x 10
-5
mol/L
Solubility – Temperature Effects
•
Solids
–
Increases in temperature always cause dissolving
to occur more rapidly
–
Increases in temperature usually increases
solubility (the amount that can be dissolved)
•
Gases
–
Solubility of gases always decreases with
increasing temperatures
Vapor Pressure – Nonvolatile solutes
•
Nonvolatile electrolytes lower the vapor
pressure of a solute
–
Nonvolatile molecules do not enter the vapor
phase
–
Fewer molecules are available to enter the vapor
phase
–
Dissociation of ionic compounds has nearly two,
three or more times the vapor pressure lowering
of nonionic (nonelectrolyte) solutes.
Raoult’s Law
The presence of a nonvolatile solute lowers
the vapor pressure of the solvent.
P
P
solution
solution
= Observed Vapor pressure of
the solution
P
P
0
0
solvent
solvent
= Vapor pressure of the pure solvent
solvent
solvent
= Mole fraction of the solvent
This should make sense, since you’ve learned previously that dissolved
Solutes raise boiling points. Lower vapor pressure = higher boiling point
Raoult’s Law Example
•
1.5 moles of cherry Kool-Aid are added to a
pitcher containing 2 liters of water on a nice
day at 25
o
C. The vapor pressure of water
alone is 23.8 mm Hg at 25
o
C. What is the new
vapor pressure of Kool-Aid?
Raoult’s Law Example
•
1.5 moles of cherry Kool-Aid are added to a pitcher containing
2 liters of water on a nice day at 25
o
C. The vapor pressure of
water alone is 23.8 mm Hg at 25
o
C. What is the new vapor
pressure of Kool-Aid?
•
Usually, you must solve for the mole fraction of the solvent
first:
2000 g H
2
O
1 mole H
2
O
18.02 g H
2
O
= 110.9 moles H
2
O
Mole Fraction of water (remember 1 L = 1000 g)
1.5 moles kool aid + 110.99 moles H
2
O
110.99 moles H
2
O
= 0.987
Raoult’s Law Example
•
1.5 moles of cherry Kool-Aid are added to a
pitcher containing 2 liters of water on a nice day
at 25
o
C. The vapor pressure of water alone is
23.8 mm Hg at 25
o
C. What is the new vapor
pressure of Kool-Aid?
Using the mole fraction from the previous page
P
Kool-Aid
=
H2O
P
pure H2O
= (.987)(23.8 mm Hg) = 23.5 mm Hg
Liquid-liquid solutions in which both
components are volatile (non-ideal solutions)
Modified Raoult's Law:
Modified Raoult's Law:
P
P
0
0
is the vapor pressure of the pure solvent
P
P
A
A
and
P
P
B
B
are the partial pressures
This is a variation on Dalton’s Law of Partial
Pressures that accounts for Raoult’s Law
Modified Raoult’s Law Example
What is the vapor pressure
of the solution?
Modified Raoult’s Law Example
Modified Raoult’s Law Example
What is the vapor pressure
of the solution?
Ideal Solutions
•
Liquid-liquid solution that obeys Raoult’s Law
•
Like gases, none are ideal, but some are close
•
Negative Deviations
–
Lower than predicted vapor pressure
–
Solute and solvent are similar, strong forces of
attraction
In a solution of acetone and
ethanol, there is a negative
deviation due to the hydrogen
bonding interactions shown
Ideal Solutions
•
Positive Deviations
–
Higher than predicted vapor pressure
–
Particles easily escape attractions in solution to
enter the vapor phase
Ethanol and hexane are not attracted to each other due to differences
In polarity. As a result, its solution is a positive deviation (higher
H)
From Raoult’s Law.
Colligative Properties
•
Properties dependent on the number of
solute particles but not on their identity
–
Boiling-Point elevation
–
Freezing-Point depression
–
Osmotic Pressure
Boiling Point Elevation
Boiling Point Elevation
Each mole of solute particles raises the boiling
point of 1 kilogram of water by 0.51 degrees
Celsius.
K
K
b
b
= 0.51
C
kilogram/mol
m
m
= molality of the solution
i
i
=
van’t Hoff
van’t Hoff
factor
Freezing Point Depression
Freezing Point Depression
Each mole of solute particles lowers the freezing
point of 1 kilogram of water by 1.86 degrees
Celsius.
K
K
f
f
= 1.86
C
kilogram/mol
m
m
= molality of the solution
i
i
=
van’t Hoff
van’t Hoff
factor
Boiling Point Elevation
•
A radiator in a car has a capacity of 6 L. If a
50/50 (v/v) solution of ethylene glycol
(C
2
H
6
O
2
, d=1.11 g/mL) and water is used, what
will be the new boiling point?
Boiling Point Elevation
•
A radiator in a car has a capacity of 6 L. If a
50/50 (v/v) solution of ethylene glycol
(C
2
H
6
O
2
, d=1.11 g/mL) and water is used, what
will be the new boiling point?
Van’t Hoff factor = 1 (not an ionic salt)
K
b
= 0.512 kg/mol
Molality = moles/kg
3000 mL C
2
H
6
O
2
1.11 g C
2
H
6
O
2
1 mL C
2
H
6
O
2
1 mole C
2
H
6
O
2
62.08 g C
2
H
6
O
2
= 53.15
moles
Boiling Point Elevation
•
A radiator in a car has a capacity of 6 L. If a
50/50 (v/v) solution of ethylene glycol
(C
2
H
6
O
2
, d=1.11 g/mL) and water is used, what
will be the new boiling point?
Van’t Hoff factor = 1 (not an ionic salt)
K
b
= 0.512 kg/mol
Molality = moles/kg
3 L water = 3 kg water
m = moles solute/kg solvent
m = 53.16 moles C
2
H
6
O
2
/3 kg water = 17.72 m
T = i
.
K
b
.
m= 1 x 0.512 x 17.72 = 9.15 C + 100 = 109.07
degrees C
Freezing Point Depression
•
Give the freezing point depression from
adding 10 g ethylene glycol to 100 g water.
DT = (1) (1.86) (m)
10 g C2H6O2 / 620.8g g/mol = 0.16 moles
m = 0.16 moles/0.1 kg = 1.6 m
DT = (1)(1.86)(1.6) = 2.88 degrees
Determination of Molar Mass by
Freezing Point Depression
Determination of Molar Mass by
Freezing Point Depression
•
Rearrange equation
–
M
solute
=
T/K
f
–
M
solute
= 0.240 C / 5.12 C
.
kg/mol = 4.69 x 10
-2
mol/kg
Determination of Molar Mass by
Freezing Point Depression
•
Rearrange equation
–
M
solute
=
T/K
f
–
M
solute
= 0.240 C / 5.12 C
.
kg/mol = 4.69 x 10
-2
mol/kg
0.015 kg benzene
Moles hormone
=
4.69 x 10
-2
mol/kg
Determination of Molar Mass by
Freezing Point Depression
0.015 kg benzene
Moles hormone
=
4.69 x 10
-2
mol/kg
Molality
Moles hormone = 7.04 x 10
-4
moles
Determine molar mass
Van’t Hoff Factor,
i
•
For ionic compounds, the expected value of
i
is an integer greater than 1
–
NaCl,
i
= 2
–
BaCl
2
,
i
= 3
–
Al
2
(SO
4
)
3
,
i
= 5
–
Total number of ions in solution
Dissociation Equations and the
Dissociation Equations and the
Determination of
Determination of
i
i
NaCl(s)
AgNO
3
(s)
MgCl
2
(s)
Na
2
SO
4
(s)
AlCl
3
(s)
Na
+
(aq) + Cl
-
(aq)
Ag
+
(aq) + NO
3
-
(aq)
Mg
2+
(aq) + 2 Cl
-
(aq)
2 Na
+
(aq) + SO
4
2-
(aq)
Al
3+
(aq) + 3 Cl
-
(aq)
i
i
= 2
= 2
i
i
= 2
= 2
i
i
= 3
= 3
i
i
= 3
= 3
i
i
= 4
= 4
Freezing Point Depression and Boiling
Point Elevation Constants,
C/
m
Osmotic Pressure
•
Semipermeable Membrane
–
Membrane which allows
solvent but not solute
molecules to pass through.
–
As time passes…
•
One side of the membrane is mostly solvent
•
The other side (which didn’t pass through) is more
concentrated since the solute can’t go through the
membrane
–
Osmosis
–
The flow of solvent into the solution through the
semipermeable membrane
Osmotic Pressure
Osmotic Pressure
The minimum pressure that stops the osmosis is
equal to the osmotic pressure of the solution
http://chemwiki.ucdavis.edu/Textbook_Maps/General_Chemistry_Textbook_Maps/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Phys
ical_Properties/13.07%3A_Osmotic_Pressure
Osmotic Pressure Calculations
Osmotic Pressure Calculations
= Osmotic pressure
M
M
= Molarity of the solution
R
R
= Gas Constant = 0.08206 L
atm/mol
K
i
i
= van’t Hoff Factor
See page 509 in text for example
Example
Dialysis
•
Transfer of solvent molecules as well as small
solute molecules and ions
–
Remember in osmosis, only solute molecules are
transferred
–
This description fits the “dialysis” used to filter the
blood when the kidneys do not work properly.
Kidney Dialysis
•
Dialyser
contains ions
and small
molecules in
blood, but
none of the
waste
products.
Osmotic Pressure and Living Cells
•
Crenation
–
Cells placed in a hypertonic (higher osmotic
pressure) solution lose water to the solution, and
shrink
•
Hemolysis
–
Cells placed in a hypotonic (lower osmotic
pressure) solution gain water from the solution
and swell, possibly bursting
Reverse Osmosis
•
External pressure applied to a solution can
cause water to leave the solution
–
Concentrates impurities (such as salt) in the
remaining solution
–
Pure solvent (such as water) is recovered on the
other side of the semipermeable membrane
–
Applicable to desalination plants which can make
drinking water from ocean water.
Colloids
•
Tiny particles suspended in some medium
•
Particles range in size from 1 to 1000 nm
•
Noticeable by shining light through the
medium
–
Particles are large enough that they scatter light
Examples of Colloids
Tyndall Effect
•
Scattering of light by
particles
–
Light passes through a
solution
–
Light is scattered in a
colloid
http://www.dorthonion.com/drcmcm/CHEMISTRY/Lessons/Lectures/images/14TyndallEffect.jpg
Exploring the properties of solutions, this content delves into measuring solution composition using techniques like molarity, mass percent, and mole fraction. Through examples, it explains how to calculate the weight percent of solutes in solutions, offering a comprehensive guide on understanding essential concepts in chemistry.
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Presentation Transcript
Chapter 11 Properties of Solutions DE Chemistry Dr. Walker
Solutions In a solution, the excess material is the solvent, the smaller amount is the solute
Measuring Solution Composition Molarity is most common, covered previously in Chapter 4
Measuring Solution Composition Mass Percent
Mass Percent Example What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol (C2H5OH) in 85.0 mL of water?
Mass Percent Example What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol (C2H5OH) in 85.0 mL of water? Density of water = 1.0 g/mL, therefore 85 mL of water = 85 g Mass of solute/mass of solution x 100 = mass % 5.3 g ethanol/(5.3 g ethanol + 85.0 g water) = 5.9%
Measuring Solution Composition Mole Fraction
Mole Fraction Example What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water? Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Mole Fraction Example What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water? Note: If the density of water is 1.0 g/mL, 85 mL = 85 g Calculate moles of ethanol: 5.30 g C2H5OH 1 mole C2H5OH = 0.115 moles C2H5OH 46.08 g C2H5OH
Mole Fraction Example What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water? Note: If the density of water is 1.0 g/mL, 85 mL = 85 g Calculate moles of water: 85.0 g H2O 1 mole H2O 18.02 g H2O = 4.72 moles H2O
Mole Fraction Example What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water? Note: If the density of water is 1.0 g/mL, 85 mL = 85 g Mole Fraction of ethanol 0.115 moles C2H5OH = 0.024 0.115 moles C2H5OH + 4.72 moles H2O Mole Fraction of Water 4.72 moles H2O = 0.976 0.115 moles C2H5OH + 4.72 moles H2O
Measuring Solution Composition Molality
Example What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water? Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Example What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water? Note: If the density of water is 1.0 g/mL, 85 mL = 85 g Molality = moles solute/kg solvent 5.30 g C2H5OH 1 mole C2H5OH = 0.115 moles C2H5OH 46.08 g C2H5OH
Measuring Solution Composition Normality Equivalents of acids and bases Mass that donates or accepts a mole of protons (usually a subscript of H+ or OH-) Equivalents of oxidizing and reducing agents Mass that provides or accepts a mole of electrons (usually a subscript)
Normality Find molarity first Normality = Molarity x number of acidic protons 1.5 M HCl = 1.5 N HCl (1 acidic proton) 1.5 M H2SO4 = 3.0 N H2SO4 (2 acidic protons) 1.5 M H3PO4 = 4.5 N H3PO4 (3 acidic protons)
Energy of Solution Formation Like Dissolves Like Polar molecules and ionic compounds tend to dissolve in polar solvents Nonpolar molecules dissolve in nonpolar compounds Water - polar Butane - nonpolar
Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent. Substance Heat of Solution (kJ/mol) -44.51 +25.69 +34.89 -74.84 NaOH NH4NO3 KNO3 HCl
Steps in Solution Formation H1 Step 1 - Expanding the solute Separating the solute into individual components usually endothermic
Steps in Solution Formation H2 Step 2 - Expanding the solvent Overcoming intermolecular forces of the solvent molecules usually endothermic
Steps in Solution Formation H3 Step 3 - Interaction of solute and solvent to form the solution usually exothermic
Enthalpy of Solution May be positive or negative Negative is favorable (exothermic) Positive values are endothermic, does not dissolve spontaneously Solution formation increases entropy (favorable)
Factors Affecting Solubility Structure Effects Polar (hydrophilic) dissolves in polar Water soluble vitamins Nonpolar (hydrophobic) in nonpolar Fat soluble vitamins
Pressure Effects Henrys Law The concentration of a dissolved gas in a solution is directly proportional to the pressure of the gas above the solution kP C= Applies most accurately for dilute solutions of gases that do not dissociate or react with the solvent Yes CO2, N2, O2 No HCl, HI remember, these dissociate in solution!
Henrys Law Example Unopened bottle: C = kP C = (3.1 x 10-2 mol/L . Atm)(5.0 atm) = 0.16 mol/L Opened bottle: C = kP C = (3.1 x 10-2 mol/L . Atm)(4.0 x 10-4 atm) = 1.2 x 10-5 mol/L
Solubility Temperature Effects Solids Increases in temperature always cause dissolving to occur more rapidly Increases in temperature usually increases solubility (the amount that can be dissolved) Gases Solubility of gases always decreases with increasing temperatures
Vapor Pressure Nonvolatile solutes Nonvolatile electrolytes lower the vapor pressure of a solute Nonvolatile molecules do not enter the vapor phase Fewer molecules are available to enter the vapor phase Dissociation of ionic compounds has nearly two, three or more times the vapor pressure lowering of nonionic (nonelectrolyte) solutes.
Raoults Law The presence of a nonvolatile solute lowers the vapor pressure of the solvent. P = 0 solvent P solution solvent Psolution = Observed Vapor pressure of the solution solvent= Mole fraction of the solvent P0solvent = Vapor pressure of the pure solvent This should make sense, since you ve learned previously that dissolved Solutes raise boiling points. Lower vapor pressure = higher boiling point
Raoults Law Example 1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25oC. The vapor pressure of water alone is 23.8 mm Hg at 25oC. What is the new vapor pressure of Kool-Aid?
Raoults Law Example 1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25oC. The vapor pressure of water alone is 23.8 mm Hg at 25oC. What is the new vapor pressure of Kool-Aid? Usually, you must solve for the mole fraction of the solvent first: Mole Fraction of water (remember 1 L = 1000 g) 2000 g H2O 1 mole H2O = 110.9 moles H2O 18.02 g H2O 110.99 moles H2O = 0.987 1.5 moles kool aid + 110.99 moles H2O
Raoults Law Example 1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25oC. The vapor pressure of water alone is 23.8 mm Hg at 25oC. What is the new vapor pressure of Kool-Aid? Using the mole fraction from the previous page PKool-Aid= H2OPpure H2O= (.987)(23.8 mm Hg) = 23.5 mm Hg
Liquid-liquid solutions in which both components are volatile (non-ideal solutions) Modified Raoult's Law: P P + = = + 0 0 TOTAL P P P A B A A B B P0 is the vapor pressure of the pure solvent PAand PB are the partial pressures This is a variation on Dalton s Law of Partial Pressures that accounts for Raoult s Law
Modified Raoults Law Example What is the vapor pressure of the solution?
Modified Raoults Law Example What is the vapor pressure of the solution?
Ideal Solutions Liquid-liquid solution that obeys Raoult s Law Like gases, none are ideal, but some are close Negative Deviations Lower than predicted vapor pressure Solute and solvent are similar, strong forces of attraction In a solution of acetone and ethanol, there is a negative deviation due to the hydrogen bonding interactions shown
Ideal Solutions Positive Deviations Higher than predicted vapor pressure Particles easily escape attractions in solution to enter the vapor phase Ethanol and hexane are not attracted to each other due to differences In polarity. As a result, its solution is a positive deviation (higher H) From Raoult s Law.
Colligative Properties Properties dependent on the number of solute particles but not on their identity Boiling-Point elevation Freezing-Point depression Osmotic Pressure
Boiling Point Elevation Each mole of solute particles raises the boiling point of 1 kilogram of water by 0.51 degrees Celsius. K i T = bm solute Kb = 0.51 C kilogram/mol m = molality of the solution i = van t Hofffactor
Freezing Point Depression Each mole of solute particles lowers the freezing point of 1 kilogram of water by 1.86 degrees Celsius. K i T = fm solute Kf = 1.86 C kilogram/mol m = molality of the solution i = van t Hofffactor
Boiling Point Elevation A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C2H6O2, d=1.11 g/mL) and water is used, what will be the new boiling point?
Boiling Point Elevation A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C2H6O2, d=1.11 g/mL) and water is used, what will be the new boiling point? K i T = bm solute Van t Hoff factor = 1 (not an ionic salt) Kb = 0.512 kg/mol 1.11 g C2H6O2 1 mole C2H6O2 3000 mL C2H6O2 = 53.15 moles Molality = moles/kg 1 mL C2H6O2 62.08 g C2H6O2
Boiling Point Elevation A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C2H6O2, d=1.11 g/mL) and water is used, what will be the new boiling point? bm K i T = Van t Hoff factor = 1 (not an ionic salt) Kb = 0.512 kg/mol 3 L water = 3 kg water solute m = moles solute/kg solvent Molality = moles/kg m = 53.16 moles C2H6O2 /3 kg water = 17.72 m T = i . Kb. m= 1 x 0.512 x 17.72 = 9.15 C + 100 = 109.07 degrees C
Freezing Point Depression Give the freezing point depression from adding 10 g ethylene glycol to 100 g water. K i T = fm solute DT = (1) (1.86) (m) 10 g C2H6O2 / 620.8g g/mol = 0.16 moles m = 0.16 moles/0.1 kg = 1.6 m DT = (1)(1.86)(1.6) = 2.88 degrees
Determination of Molar Mass by Freezing Point Depression = fm T i K solute
Determination of Molar Mass by Freezing Point Depression = fm T i K solute Rearrange equation Msolute = T/Kf Msolute = 0.240 C / 5.12 C . kg/mol = 4.69 x 10-2 mol/kg
