Chapter 16

Chapter 16
Properties of Solutions

Solution Formation
Solutions are homogeneous mixtures that may
be solid, liquid, or gaseous.
The compositions of the solvent and the solute
determine whether a substance will dissolve.
Stirring (agitation), temperature, and the surface
area of the dissolving particles determine how
fast the substance will dissolve.
These three factors involve the contact of the
solute with the solvent.

Stirring & Solution Formation
Stirring speeds up the process of dissolving because
fresh solvent is continually brought into contact with
the surface of the solute
Stirring affects only the rate at which a solid solute
dissolves. It does not influence the amount of solute
that will dissolve.
An insoluble substance remains undissolved
regardless of how vigorously or for how long the
solvent/solute system is agitated.

Temperature & Solution Formation

At higher temperatures, the kinetic energy of the
solvent molecules is greater than at lower
temperatures so they move faster.
The more rapid motion of the solvent molecules
leads to an increase in the frequency and the force
of the collisions between the solvent molecules and
the surfaces of the solute molecules.

Particle Size & Solution Formation
A spoonful of granulated sugar dissolves more
quickly than a sugar cube because the smaller
particles in granulated sugar expose a much
greater surface area to the colliding solvent
molecules.
The more surface of the solute that is exposed,
the faster the rate of dissolving.

Solubility
If you add 36.0 g of NaCl to 100 g H2O at 25ºC, all of
the 36.0 g of salt dissolves.
If you add one more gram of salt and stir (no matter
how vigorously or for how long) only 0.2 g of the last
portion will dissolve.
According to the kinetic theory, water molecules are
in continuous motion.
They should continue to bombard the excess solid,
removing and solvating the ions.

Solubility
As ions are solvated, they dissolve in the water.
Based on this information, you might expect all of the
salt to dissolve eventually. That does not happen,
however, because an exchange process is
occurring.
New particles from the solid are solvated and enter
into solution. At the same time an equal number of
already dissolved particles crystallize.
These particles come out of solution and are
deposited as a solid. The mass of undissolved
crystals remains constant.

Solubility
In a saturated solution, a state of dynamic equilibrium
exists between the solution and the excess solute.
The rate of solvation (dissolving) equals the rate of
crystallization, so the total amount of dissolved
solute remains constant.
The system will remain the same as long as the
temperature remains constant.
Saturated solution – contains the maximum amount
of solute for a given quantity of solvent at a constant
temperature and pressure.

Solubility
Example: 36.2 g of salt dissolved in 100 g of water is
a saturated solution at 25ºC.
If additional solute is added to this solution, it will not
dissolve.
Solubility of a substance is the amount of solute that
dissolves in a given quantity of a solvent at a
specified temperature and pressure to produce a
saturated solution.
Solubility is often expressed in grams of solute per
100 g solvent. (gas sometimes g/L)

Solubility
Unsaturated solution – a solution that contains less
solute than a saturated solution at a given
temperature and pressure.
If additional solute is added to an unsaturated
solution, it will dissolve until the solution is
saturated.
Some liquids are infinitely soluble in each other. Any
amount will dissolve in a given volume.
Two liquids are miscible if they dissolve in each
other in all proportions (water and ethanol)

Factors Affecting Solubility
Temperature affects the solubility of a solid, liquid
and gaseous solutes in a solvent.
Both temperature and pressure affect the solubility of
gaseous solutes.
The solubility of most solid substances increases as
the temperature of the solvent increases.
Mineral deposits form around the edges of hot
springs because the hot water is saturated with
minerals. As the water cools, some of the minerals
crystallize because they are less soluble at the
lower temperature.

Factors Affecting Solubility
For a few substances, solubility decreases with
temperature.
Supersaturated solution – contains more solute
than it can theoretically hold at a given temperature.

Make a saturated solution of sodium acetate at 30·C
and let the solution stand undisturbed as it cools to
25ºC.
You would expect that solid sodium acetate will
crystallize from the solution as the temperature
drops. But no crystals form.

Supersaturated Solutions
The crystallization of a supersaturated solution can
be initiated if a very small crystal, called a seed
crystal, of the solute is added.
Rock candy is another example of crystallization in a
supersaturated solution.
A solution is supersaturated with sugar and seed
crystals cause the sugar to crystallize out of
solution.
A supersaturated solution crystallized rapidly when
disturbed.

Temperature and Gas Solubility
The solubilities of most gases are greater in cold
water than in hot.
Thermal pollution happens when an industrial plant
takes water from a lake for cooling and then dumps
the heated water back into the lake.
The temperature of the lake increases which lowers
the concentration of dissolved oxygen in the lake
water affecting aquatic animal and plant life.

Pressure and Solubility
Changes in pressure have little affect on the solubility
of solids and liquids, but pressure strongly
influences the solubility of gases.
Carbonated beverages contain large amounts of
carbon dioxide dissolved in water. Dissolved CO2
makes the drink fizz.
The drinks are bottle under higher pressure of CO 2
gas, which forces large amounts of the gas into
solution.
When opened, the partial pressure of CO2 above the
liquid decreases.

Immediately, bubbles of CO2 form in the liquid and
escape from the bottle and the concentration of
dissolved CO2 decrease.

If the drink is left open, it becomes “flat” as it loses its
CO2.

Henry’s Law – sated that at a given temperature, the
solubility (S) of a gas in a liquid is directly
proportional to the pressure (P) of the gas above the
liquid.
As the pressure of the gas above the liquid
increases, the solubility of the gas increases.

Henry’s Law
S1 = S2

P1 P2

Question
The solubility of a gas in water is 0.16 g/L at 104 kPa.
What is the solubility when the pressure of the gas
ins increased to 288 kPa. Assume the temperature
remains constant.
S1 = S2

P1 P2
(288 kPa) ( 0.16g/L) = 4.4 x 10-1 g/L
(104 kPa)

Questions
What determines whether a substance will dissolve?
Chemical composition of the solute and solvent.
What determines how fast a substance will dissolve?
Agitation, temperature and particle size of the solute
What units are usually used to express the solubility
of a solute?
g of solute per 100 g solvent

Questions
What would you do to change a saturated solid/liquid
solution to an unsaturated solution?
Add solvent
What would you do to change a saturated gas/liquid
solution to an unsaturated solution?
Increase the pressure
What are two conditions that determine the mass of
solute that will dissolve in a given mass of solvent?
Temperature and pressure (if the solute is a gas)

Concentration
Concentration of a solution is a measure of the
amount of solute that is dissolved in a given quantity
of solvent.
Dilute solution is one that contains a small amount
of solute.
Concentrated solution – contains a large amount of
solute.
In chemistry the most important unit of concentration
is molarity.

Molarity
Molarity (M) is the number of moles of solute
dissolved in one liter of solution
Molarity (M) = moles of solute / liters of solution.
Note that the volume involved is the total volume of
the resulting solution, not the volume of the solvent
alone.
3 M NaCl is read as “three molar sodium chloride”

Molarity Questions
A solution has a volume of 2.0 L and contains 36.0 g
of glucose (C6H12O6). If the molar mass of glucose is
180 g/mol, what is the molarity of the solution?
M = moles of solute / L of solution
M = 36.0 g glucose 1 mol glucose
180 g glucose 2.0 L
M = 0.1mol/L or 0.1M C6H12O6

Molarity Questions
A solution has a volume of 250 mL and contains 0.70
mol NaCl. What is its molarity?
M = moles of solute / L of solution
M = 0.70 mol NaCl
0.250L solution (convert to L)
M = 2.8 mol/L or 2.8M NaCl

Molarity Questions
How many moles are in each quantity of the following
substances?
12.0 g NaCl
(12.0 g NaCl) (1 mole NaCl / 58.5 g NaCl) =
0.205 mole NaCL
53.8 g KNO3

(53.8 g KNO3) ( 1 mol KNO3 / 101.1g KNO3) =
0.532 mol KNO3

Molarity Questions
Find the mass in grams of each of these amounts of
substances?
1.5 mol NaOH
(1.5 mol NaOH) (40 g NaOH / 1 mol NaOH) =
60 g of NaOH
0.575 mol NaHCO3

(0.575 mol NaHCO3)( 84g NaHCO3 / 1 mol NaHCO3)

48.3 g NaHCO3

Molarity Questions

Which contains more molecules 1.00 mol SO2 or 1.00
mol SO3?

They both contain the same number of molecules
Which contains more mass 1.00 mol SO2 or 1.00 mol

1 mole of SO3 (80.1g) – 1 mole of SO2 (64.1g)

Molarity

Sometimes you may need to determine the number
of moles of solute dissolved in a given volume of
solution.
How many moles are in 2.00 L of 2.5M lithium
chloride (LiCl)?
Moles of solute = molarity (M) x liters of solution (V)
Moles of solute = (2.5 moles/L) ( 2.00L)
Moles of solute = 5.0 mol

Molarity Questions
How many moles of ammonium nitrate are in 335 mL
of 0.425 M NH4NO3?

mol NH4NO3 = M x L of solution
= (0.425 mol/L) (0.335L)
= 0.142 mol NH4NO3

How many moles of solute are in 250 mL of 2.0M
CaCl2? How many grams of CaCl2 is that?

mol CaCl2 = (2.0 mole/L) (0.250L) = 0.50 mol

0.50 mol CaCl2 (111.11 g CaCl2 / 1 mol CaCl2 ) = 56 g

Making Dilutions
Diluting - To make less concentrated by adding
solvent.
Diluting a solution reduces the number of moles of
solute per unit volume, but the total number of
moles of solute in solution does not change.
Moles of solute before dilution = moles of solute after
dilution
moles of solute = M x L of solution and total number
of moles of solute remains unchanged upon dilution.
M1V1 = M2V2

Making Dilutions
M1V1 = M2V2

molarity & volume molarity and volume
of original solution of diluted solution

Volumes can be L or mL as long as the same units
are used for both V1 and V2

Making Dilutions
A student is preparing a 100 mL of 0.40M MgSO 4
from a stock solution of 2.0 M MgSO4. How would
she do this?
M1V1 / M2 = V2
V2 = 20 ml – She would measure 20 mL of the stock
solution (2.0 M MgSO4) and transfer it to a
volumetric flask.
Then she would add water to the flask to make 100
mL of solution.
Try the class activity on page 482

Questions
How many milliliters of a solution of 4.0 M KI are
needed to prepare a 250.0 mL of 0.760 M KI?
V1 = (0.760M)(250.0 mL) / (4.0 M) = 47.5 mL

How could you prepare 250 mL of 0.20M NaCl using
on a solution of 1.0M NaCl and water?
V1 = (0.20M) ( 250 mL) / ( 1.0 M) = 50 mL

Use a pipet to transfer 50 mL of the 1.0M solution to
a 250 mL flask. Then add distilled water up to the
mark.

Percent Solutions (v / v)
The concentration of a solution in percent can be
expressed in two ways:
As the ratio of the volume of the solute to the volume
of the solution or as the ratio of the mass of the
solute to the mass of the solution
Percent by volume (% (v/v)) = volume of solute x 100%
volume of solution
How many milliliters of isopropyl alcohol are in 100
mL of 91% alcohol?

Question
If 10mL of acetone (C3H6O) is diluted with water to a
total solution volume of 200mL, what is the percent
by volume of acetone in the solution?
volume of solution
% by volume of acetone = 10 mL / 200 mL =5.0% v/v

Question
A bottle of the antiseptic hydrogen peroxide is labeled
3.0% (v/v). How many mL hydrogen peroxide are in
a 400.0 mL bottle of this solution?
volume of solution
0.03 = x mL / 400.0 mL
(0.03) (400.0 mL) = x
12 mL = x

Percent Solutions (mass/mass)
Another way to express the concentration of a
solution is as a percent (mass/mass), which is the
number of grams of solute in 100 g of solution.
A solution containing 7 g of NaCl in 100 g of solution
is 7% (m/m)
Percent by mass (% (m/m) = mass of solute x 100%
mass of solution

Percent Solutions (mass/mass)
You want to make 2000g of a solution of glucose in
water that has a 2.8% (m/m) concentration of
glucose. How much glucose should you use?
Percent by mass (% (m/m) = mass of solute x 100%
mass of solution
2000 g solution=(2.8g glucose/100 g solution) = 56 g glucose
How much solvent should be used? The mass of the solvent
equals the mass of the solution minus the mass of the solute.
(2000 g – 56 g ) = 1944 g of solvent
Thus a 2.8% (m/m) glucose solution contains 56 g of glucose
dissolved in 1944 g of water.

Questions
How do you calculate the molarity of a solution?
Molarity (M) = moles of solute / volume of solution
Compare the number of moles of solute before dilution
with the number of moles of solute after dilution.
The total number of moles of solute in solution does not
change (only the number of moles of solute per unit
volume)
What are two ways of expressing the concentration of a
solution as a percent?
Volume solute Mass of solute
Volume of solution Mass of solution

Colligative Properties of Solutions
The physical properties of a solution differ from those of
the pure solvent used to make the solution.
Some of these differences in properties have little to do
with the specific identity of the solute.
They depend upon the number of solute particles in the
solution.
Colligative Property – a property that depends only
upon the number of solute particles, and not upon their
identity.

Three important colligative properties of solutions are:
1. Vapor pressure lowering
2. Boiling point elevation
3. Freezing point depression
Vapor pressure is the pressure exerted by a vapor that is
in dynamic equilibrium with its liquid in a closed
system.
A solution that contains a solute that is nonvolatile (not
easily vaporized) always has a lower vapor pressure
than the pure solvent.

In a pure solvent, equilibrium is established between the
liquid and the vapor.
In a solution, solute particles reduce the number of free
solvent particles able to escape the liquid. Equilibrium
is established at a lower vapor pressure.
Ionic solutes that dissociate, such as sodium chloride and
calcium chloride, have greater effects on the vapor
pressure than does a non-dissociating solute such as
glucose.
Each formula unit of CaCl2 produces three particles in
solution, a calcium ion and two chloride ions.

The decrease in a solution’s vapor pressure is
proportional to the number of particles the solute
makes in solution.
3 moles of NaCl dissolved in H2O produce 6 mol of
particles - each formula unit dissociates into 2 ions
3 moles of CaCl2 dissolved in H2O produce 9 mol of
particles - each formula unit dissociated into 3 ions

3 moles of glucose dissolved in water produce 3 mol
of particles – glucose does not dissociate.


The vapor pressure lowering caused by 0.1 mol of NaCl
in 1000 g of water is twice that caused by 0.1 mol of
glucose in the same quantity of water.
The vapor pressure lowering caused by 0.1 mol of CaCl2
in 1000 g of water is three times that caused by 0.1 mol
of glucose in the same quantity of water.

The decrease in a solution’s vapor pressure is
proportional to the number of particles the solute

Freezing-Point Depression
When a substance freezes, the particles of the solid take
on an orderly pattern.
The presence of a solute in water disrupts the formation
of this pattern because of the shells of water of
solvation. (water molecules surround the ions of the
solute)

As a result, more KE must be withdrawn from a
solution than from the pure solvent to cause the
solution to solidify.
The freezing point of a solution is lower than the
freezing point of the pure solvent.

Freezing-Point Depression – the difference in
temperature between the freezing point of a solution
and the freezing point of the pure solvent.
Freezing-point depression is another colligative property.

The magnitude of the freezing-point depression is
proportional to the number of solute particles
dissolved in the solvent and does not depend upon
their identity.
The addition of 1 mol of solute particles to 1000 g of
water lowers the freezing point by 1.86ºC.

If you add 1 mole (180g) of glucose to 1000 g of water,
the solution freezes at -1.86ºC.
If you add 1 mol (58.5g) of NaCl to 1000 g of water, the
solution freezes at -3.72ºC, double the change for
glucose.
This is because 1 mol NaCl produces 2
mol particles and doubles the freezing
point depression.
Salting icy surfaces forms a solution with
the melted ice that has a lower freezing
point than water. (antifreeze also)

Reminders
Ionic compounds and certain molecular compounds,
such as HCl, produce two or more particles when they
dissolve in water.
Most molecular compounds, such as glucose, do not
dissociate when they dissolve in water.

Colligative properties do not depend on the kind of
particles, but on their concentration.
Which produces a greater change in colligative
properties – an ionic solid or a molecular solid?
An ionic solid produces a greater change because it will
produce 2 or more mole of ions for every mol of solid that
dissolves.

Boiling-Point Elevation
Boiling Point – of a substance is the temperature at
which the vapor pressure of the liquid phase equals
atmospheric pressure.
Adding a nonvolatile solute to a liquid solvent decreases
the vapor pressure of the solvent.
Because of the decrease in vapor pressure, additional KE
must be added to raise the vapor pressure of the liquid
phase of the solution to atmospheric pressure and
initiate boiling.
Thus the boiling point of a solution is higher than the
boiling point of the pure solvent.

Boiling Point Elevation – The difference in temperature
between the boiling point of a solution and the boiling
point of the pure solvent.
The same antifreeze, added to automobile engines to
prevent freeze-ups in winter, protects the engine from
boiling over in summer.
Boiling-point elevation is a colligative property, it depends
on the concentration of particles, not on their identity.
It takes additional KE for the solvent particles to
overcome the attractive forces that keep them in the
liquid.

The magnitude of the boiling-point elevation is
proportional to the number of solute particles dissolved
in the solvent.
The boiling point of water increases by 0.512ºC for every
mole of particles that the solute forms when dissolved
in 1000g of water.
To make fudge, a lot of sugar and some flavoring are
mixed with water and the solution is boiled. As the
water slowly boils away, the concentration of sugar in
the solution increases.
As the concentration increases, the boiling point steadily
rises.

Questions
What are three colligative properties of solutions?
Vapor-pressure lowering, boiling-point elevation, and
freezing-point depression
What factor determines how much the vapor pressure,
freezing point, and boiling point of a solution differ from
those properties of the pure solvent?
The number of solute particles dissolved in the solvent.
Would a dilute or a concentrated sodium fluoride solution
have a higher boiling point? Explain.
Concentrated NaF, - magnitude of the boiling-point elevation is
proportional to the number of solute particles dissolved in
the solvent.

Questions
An equal number of moles of KI and MgF2 are dissolved
in equal volumes of water. Which solution has the
higher boiling point?
MgF2

Higher vapor pressure?
KI solution
Higher freezing point?
KI solution

Molality and Mole Fraction
Unit molality and mole fractions are two additional ways
in which chemists express the concentration of a
solution.
Molality (m) is the number of moles of solute dissolved in
1 kg of solvent.
Molality (m) = moles of solute / kg of solvent
Molarity = moles of solute / L of solution
In the case of water as the solvent, 1 kg = 1000 mL,
1000 g = 1 L

Molality
To prepare a solution that is 1.00 molal (1m) in glucose,
you add 1 mol (180g) of glucose to 1000g of water.
0.500 molal solution in sodium chloride is prepared by
dissolving 0.50 mol (29.3 g) of NaCl in 1.0 kg of water
Molality (m) = moles of solute / kg of solvent
The molality of a solution does not wary with temperature
because the mass of the solvent does not change.
Molarity = moles of solute / L of solution
The molarity of a solution does vary with temperature
because the liquid can expand and contract.

Molality Questions
How many grams of NF are need to prepare a 0.400m
NaF solution that contains 750g water?
750 g H2O 0.400 mol NaF 42g NaF = 13g NaF

1000 g H2O mol NaF

Calculate the molality of a solution prepared by dissolving
10.0g of NaCl in 600 g of water.
10.0 g NaCl 1 mol NaCl 1000 g H2O = 2.85 x 10-1m
600
g H 2O 58.5 g NaCl 1 kg H2O

Mole Fraction
The concentration of a solution also can be expressed as
a mole fraction.
Mole fraction of a solute in a solution is the ratio of the
moles of the solute to the total number o moles of
solvent and solute.
In a solution containing nA mole of solute A and nB mole
of solvent B, the mole fraction of solute A and the mole
fraction of solvent B can be expressed as follows.
XA = nA XB = nB
nA + nB nA + n B

Mole Fraction Questions
What is the mole fraction of each component in a solution
made by mixing 300 g of ethanol (C2H5OH) and 500 g
water.
Xethanol = nethanol
nethanol + nwater

300 g C2H5OH 1 mol C2H5OH = 6.7 mol C2H5OH
45 g C2H5OH

500 g H2O 1 mol H2O = 27.8 mol H2O
18 g H2O

Xethanol = 6.7 mol C2H5OH = 0.190 m C2H5OH

What is the mole fraction of each component in a solution
made by mixing 300 g of ethanol (C2H5OH) and 500 g
water.
Xwater = nwater
nethanol + nwater

45 g C2H5OH

500 g H2O 1 mol H2O = 27.8 H2O
18 g H2O

Xwater = 27.8 mol C2H5OH = 0.810 H2O

Calculate the mole fraction of each component in a
solution of 42g CH3OH, 35g C2H5OH, and 50 g C3H7OH
XA = nA
nA + nB + n C

42 g CH3OH 1 mol CH3OH = 1.3 mol CH3OH
32 g CH3OH

46 g C2H5OH

60 g C3H7OH

X CH3OH = 1.3 mol
1.3 mol + 0.76 mol + 0.83 mol
X CH3OH = 1.3 mol = 0.45
2.89 mol
X CH3OH = 0.76 mol
1.3 mol + 0.76 mol + 0.83 mol
X CH3OH = 0.76 mol = 0.26
2.89 mol
X CH3OH = 0.83 mol
1.3 mol + 0.76 mol + 0.83 mol
X CH3OH = 0.83mol = 0.29

Freezing Point Depression and
Boiling Point Elevation
The freezing point of a solvent is lowered and its boiling point
is raised by the addition of a nonvolatile solute.
The magnitudes of the freezing point depression (ΔTf) and
the boiling point elevation (ΔTb) of a solution are directly
proportional to the molal concentration (m) when the solute is
molecular, not ionic.
ΔTb Ñ m ΔTf Ñ m
Change in the freezing temperature (ΔTf ) is the difference
between the fp of the solution and the fp of the pure solvent.
Change in boiling temperature (ΔTb ) is the difference
between the bp of the solution and the bp of the pure solvent.

Molal Freezing Point
Depression Constant
With the addition of a constant, the proportionality between
the ΔTf and the molality (m) can be expressed in an equation

ΔTf = Kf x m

The constant, Kf, is the molal freezing-point depression
constant, which is equal to the change in freezing point for a
constant
1 molal solution of a nonvolatile molecular solute.

The value of Kf depends upon the solvent. Its units are ºC/m.

Molal Boiling Point
Elevation Constant
The boiling-point elevation of a solution can also be
expressed as an equation

ΔTb = Kb x m

The constant, Kb, is the molal boiling-point elevation
constant, which is equal to the change in boiling point for a 1
constant
molal solution of a nonvolatile molecular solute.

The value of Kb depends upon the solvent. Its units are ºC/m.

For ionic compounds, both the freezing point depression and
the boiling point elevation depend upon the number of ions
produced by each formula unit

Problems
What is the freezing point depression (and boiling point
elevation) of an aqueous solution of 10.0 g of glucose
(C6H12O6) in 50.0 g H2O?

10.0 g C6H12O6 1 mol = 0.0555 mol C6H12O6
180 g
m = mol solute = 0.055 mol = 1.11 m
kg solvent .0500 kg
ΔTf = Kf x m = (1.86 ºC/m) (1.11m) = 2.06 ºC

ΔTb = Kb x m = (0.512 ºC/m) (1.11 m) = 0.568 ºC

Problems
Calculate the freezing point depression of a benzene
solution containing 400 g of benzene and 200 g of the
molecular compound acetone (C3H6O). Kf for benzene is
5.12 ºC/m
200 g C3H6O 1 mol = 3.45 mol C3H6O
58 g
m = mol solute = 3.45 mol = 8.63 m
kg solvent .400 kg
ΔTf = Kf x m = (5.12 ºC/m) (8.63m) = 44.2 ºC

Problems
Freezing points and boiling points cannot be depressed or
elevated without end.
As the concentration of a solute increases, there comes a
point when the quantity of the solute exceeds the quantity
of the solvent.
At this point, the solute then becomes the solvent because
is present is a higher concentration.

Question
Which solution has a higher boiling point, 1 mol of Al(NO 3)3
in 1000g of water or 1.5 mol of KCl in 1000 g of water?
The solution of Al(NO3)3 has a higher boiling point because
Al(NO3)3 dissociates into a larger number of particles.

Why is it important to distinguish between nonvolatile and
volatile compounds when discussing certain colligative
properties?
Volatile solutes would quickly evaporate at higher
temperatures, which would change the molal concentration
of the solution.

Calculating the Boiling Point of an
Ionic Solution
What is the boiling point of a 1.5m NaCl solution?
Each formula unit of NaCl dissociates into two particles,
Na+ and Cl-, the effective molality is 2 x 1.5m = 3.00m.
Calculate the boiling point elevation and then add it to
100ºC.
ΔTb = Kb x m = (0.512 ºC/m) (3.00m) = 1.54 ºC
Boiling Point = 100ºC + 1.54ºC = 101.54ºC

Ionic Solutions
What is the boiling point of a solution that contains 1.25
mol CaCl2 in 1400 g of water.
ΔTb = Kb x m = (0.512 ºC/m) (2.68m) = 1.37 ºC
Boiling Point = 100ºC + 1.37ºC = 101.37ºC
What mass of NaCl would have to be dissolved in 1.000
kg of water to raise the boiling point by 2.00ºC
ΔTb = Kb x m = (0.512 ºC/m) (?m) = 2.00 ºC
m = 3.91 / 2 = 1.96 (Na+ and Cl-)
1.96 mol NaCl 58.5 g = 115 g NaCl
1 kg solution 1 mol

Questions
What are two ways of expressing the ratio of solute
particles to solvent particles?
Molality and mole fractions
How are freezing point depression and boiling point
elevation related to molality?
The magnitudes of the freezing point depression and the
boiling point elevation of a solution are directly
proportional to the molal concentration when the solute is
molecular, not ionic.

Questions
How many grams of sodium bromide must be dissolved
in 400.0 g of water to produce a 0.500 molal solution?
0.500 mol 102.9 g .4000 kg = 20.6 g NaBr
1 kg mol
Calculate the mole fraction of each component in a solution
of 2.50 mol ethanoic acid in 10.00 mol of water.
Xethanoic acid = 2.5 mol / 12.5 mol = 0.200

X water = 10.0 mol / 12.5 mol = 0.800

Questions
What is the freezing point of a solution of 12.0 g of CCl4
dissolved in 750.0 g of benzene? The freezing point of
benzene is 5.48 ºC; Kf is 5.12 ºC/m

m = 12.0 g CCl4 1 mol = 0.104 m
154 g 0.7500kg
ΔKf = m x Kf = (0.104m) ( 5.12 ºC/m) = 0.53ºC

Freezing point = 5.48ºC – 0.53ºC = 4.95ºC

Chapter 16

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (20)

Andere mochten auch

Andere mochten auch (20)

Ähnlich wie Chapter 16

Ähnlich wie Chapter 16 (20)

Mehr von Malcolm Harrison

Mehr von Malcolm Harrison (20)

Chapter 16