CHAPTER 13
SOLUTIONS
CHAPTER 13
Solution Topics
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Solution review
Solution forming process
Solution concentration definitions
Solution Concentration calculations
Solution stoichiometry
Colligative Properties
13.2 SOLUTION REVIEW
Solutions are homogenous mixtures.
 They consist of a larger component called the
solvent and one or more smaller components called
the solutes.
 Can be in the solid, liquid, or gaseous state.
13.2 Solution Examples
•Margarine
•Tap Water
•Steel
•18 Carat Gold
•Air
•Sterling Silver
13.2 Solution Examples
Composition of air: Dry air contains roughly (by volume)
78% nitrogen, 21% oxygen, 0.93% argon,
0.038% carbon dioxide, and small amounts of other gas
Air also contains a variable amount of water vapor,
on average around 1%
What is the solvent in air ?
13.2 Solution Examples
Composition of air: Dry air contains roughly (by volume)
78% nitrogen, 21% oxygen, 0.93% argon,
0.038% carbon dioxide, and small amounts of other gases.
Air also contains a variable amount of water vapor,
on average around 1%
What is the solvent in air ?
Nitrogen, N2
13.2 Solution Examples
Composition of air: Dry air contains roughly (by volume)
78% nitrogen, 21% oxygen, 0.93% argon,
0.038% carbon dioxide, and small amounts of other gases.
Air also contains a variable amount of water vapor,
on average around 1%
What is the solvent in air ?
Nitrogen, N2
What is a solute in air?
13.2 Solution Examples
Composition of air: Dry air contains roughly (by volume)
78% nitrogen, 21% oxygen, 0.93% argon,
0.038% carbon dioxide, and small amounts of other gases.
Air also contains a variable amount of water vapor,
on average around 1%
What is the solvent in air ?
Nitrogen, N2
What is a solute in air?
Oxygen, O2
13.2 Solution Examples
Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper.
What is the solvent in 18 ct gold ?
13.2 Solution Examples
Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper.
What is the solvent in 18 ct gold ?
Gold
13.2 Solution Examples
Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper.
What is the solvent in 18 ct gold ?
Gold
What are the solutes?
13.2 Solution Examples
Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper.
What is the solvent in 18 ct gold ?
Gold
What are the solutes?
Silver and Copper
13.2 Solution Examples
Examples of solutions include:
Salt water:
what is the solvent in salt water ?
13.2 Solution Examples
Examples of solutions include:
Salt water:
what is the solvent in salt water ?
Water, H2O
13.2 Solution Examples
Examples of solutions include:
Salt water:
what is the solvent in salt water ?
Water, H2O
What is a solute in sea water?
13.2 Solution Examples
Examples of solutions include:
Salt water:
what is the solvent in salt water ?
Water, H2O
What is a solute in sea water?
NaCl (salt)
13.2 Solution Properties
Some general properties of solutions include:
 Solutions may be formed between solids, liquids or
gases.
 They are homogenous in composition
 They do not settle under gravity
 They do not scatter light (Called the Tyndall Effect)
Solute particles are too small to scatter light and therefore
light will go right through a solution like is shown on the
next slide.
13.3 Solution Properties
Tyndall Effect
Laser light reflected by a colloid. In a solution you would not
see any red light.
13.3 Solution Properties
Soluble substances are those that can dissolve in a given
solvent.
Insoluble or immiscible substances are those that cannot
dissolve in a given solvent.
Which of the following are soluble in water?
NaCl, sugar, cooking oil, alcohol, gasoline, motor oil
13.3 Solution Properties
Soluble substances are those that can dissolve in a given
solvent.
Insoluble or immiscible substances are those that cannot
dissolve in a given solvent.
Which of the following are soluble in water?
NaCl, sugar, cooking oil, alcohol, gasoline, motor oil
13.3 Solution Properties
Soluble substances are those that can dissolve in a given
solvent.
Insoluble or immiscible substances are those that cannot
dissolve in a given solvent.
Which of the following are soluble in water?
NaCl, sugar, cooking oil, alcohol, gasoline, motor oil
Which of the following are immiscible in cooking oil?
NaCl, sugar, alcohol, gasoline, motor oil, water
13.3 Solution Properties
Soluble substances are those that can dissolve in a given
solvent.
Insoluble or immiscible substances are those that cannot
dissolve in a given solvent.
Which of the following are soluble in water?
NaCl, sugar, cooking oil, alcohol, gasoline, motor oil
Which of the following are immiscible in cooking oil?
NaCl, sugar, alcohol, gasoline, motor oil, water
13.3 Solution Properties
The maximum amount of a given solute a solvent can dissolve
is called the solubility. The solubility is dependent on the
temperature and pressure.
Solubility is often expressed in terms of grams of solute per
100 g of solvent but may have other units.
When a solvent contains the minimum amount of a solute
possible the solutions is said to be unsaturated.
When a solvent contains the maximum amount of a solute
possible the solutions is said to be saturated.
When a solvent contains more than the maximum amount of
a solute possible the solutions is said to be supersaturated.
13.3 Solution Properties
Solutions form when a soluble solute(s) is dissolved in a
solvent.
In biological systems aqueous (solutions where water is the
solvent) are of particular importance.
The solubility of most liquids and solids in water increases
with temperature.
The effect of pressure on the solubility of liquid or solid
solutes in water is negligible.
13.3 Solution Properties
Solubility Curves
13.3 Solution Properties
By forming a solution at a high temperature then slowly
cooling it we can form supersaturated solutions that contain
more solute than in a saturated solution.
These kinds of solutions are very unstable and tend to
separate out the excess solute with the slightest disturbance.
http://www.youtube.com/watch?v=uy6eKm8IRdI&NR=1
http://www.youtube.com/watch?v=aC-KOYQsIvU&feature=related
13.3 Solution Properties
The solubility of gases in water decreases with temperature.
 Are cold carbonated drinks bubblier than warm
carbonated drinks?
The solubility of many gases in water is directly proportional
to the pressure being applied to the solution.
i.e. double the pressure, double the solubility
 What happens when the cork is removed from a
bottle of champagne?
 What is the origin of decompression sickness?
 Anyone heard of hyperbaric therapy?
13.3 Solution Properties
When we place an ionic solid in water there will be
attractive forces between the ions at the surface of the
crystal and the water molecules. These attractive forces
are called ion-dipole forces.
Water molecules orient such that the positive end of the
molecule is oriented towards the negative ions at the
surface and vice versa.
13.3 Solution Properties
How do solutions form?
Why do some substances leave one phase and enter the
solution and others don’t?
How can we use chemistry to predict solubility's?
Lets first look at the formation of a solution between an ionic
solute and a polar solvent such as H2O.
13.3 Solution Properties
Ionic compounds are composed of oppositely charged ions
arranged in a repeating 3-d arrangement.
They are held together
by attractive forces
between oppositely
charged ions.
13.3 Solution Properties
Ionic compounds are composed of oppositely charged ions
arranged in a repeating 3-d arrangement.
They are held together
by attractive forces
between oppositely
charged ions
Why is chloride ion
larger than sodium ion?
13.3 Solution Properties
red is the region
where electrons are
found most often and
blue is
where electrons are
rarely found
13.3 Solution Properties
If the attractive force between the surface ion and the
solvent is greater than the forces between the ion and the
solid then the ion will enter the solution phase.
The ion that has left the solid
H2O and becomes completed
surrounded by water
molecules. It has become
solvated or hydrated. Strong
K+ Ion dipole forces hold the ions
to water.
13.3 Solution Properties
Note the different
orientation of water
molecules around
the oppositely
charged ions.
Positive pole of
water directed to
the negative ions
and the negative
pole directed to the
positive ions
13.3 Solution Properties
In a solution of an ionic compound a solvated ion will
occasionally collide with the surface of the solid.
Sometimes when this happens the ion will “stick” to the
surface and become part of the solid phase again.
This will happen more frequently the more concentrated the
solution is.
13.3 Solution Properties
When the rate of ions leaving the solid equals the rate of ions
going back to the solid the system is at equilibrium and the
solution is saturated.
When a solution is at equilibrium with its solute
macroscopically there will be no change occurring.
However, at the molecular level lots is happening, just in
equal and opposite directions.
13.3 Solution Properties
Supersaturated solutions can form because there are no sites
for solute ions to collide with.
When we place a “seed” crystal in a supersaturated solution
this provides the needed sites and the excess solute
crystallizes very quickly.
13.3 Solution Properties
In the you tube video
we watched you can
just see the tiny seed
crystals on the persons
finger.
13.3 Solution Properties
Polar but non-ionic solutes dissolve in water via a similar
mechanism as for ionic compounds.
13.3 Solution Properties
A solute will be insoluble in a solvent if:
1. Forces between solute particles are greater than the
forces between solute particles and the solvent.
13.3 Solution Properties
A solute will be insoluble in a solvent if:
Forces between the solvent particles and solute particles
are stronger than forces between the solvent and the
solute.
e.g. The only attractive force between oil and water will is
dispersion forces. These are weak compared to hydrogen
bonds between water molecules.
13.3 Solution Properties
In a polar solvent
there will be
attraction between
the oppositely
charged ends of the
molecule.
Hydrogen bonds are represented by dotted lines
between the water molecules. A hydrogen bond is and
intermolecular force between hydrogen of one molecule
and O, N, or F of another molecule. Hydrogen must be
directly attached to O, N, or F in at least one of the two
hydrogen bonded molecules.
13.3 Solution Properties
A good “rule of thumb” that works especially well for nonionic compounds is:
“Like dissolves like”
i.e. Polar solvents dissolve polar solutes well and non-polar
solvents dissolve non-polar solutes well.
13.3 Solution Properties
The rate of dissolution is dependent upon:
1. The surface area of the solute.
i.e. how finely divided it is.
Increasing rate
13.3 Solution Properties
2. How hot the solution is.
i.e. the kinetic energy of solute and solvent.
3. The rate of stirring.
Typically when we are
preparing a solution in the lab
we will both heat and stir.
13.3 Solution Properties
When a solute dissolves in a solvent heat can be released or
absorbed.
When heat is absorbed the process is endothermic and the
solution becomes cooler.
This effect is used in instant cold packs for sporting injuries
and first aid.
13.3 Solution Properties
Endothermic Solution
Solvent temperature 22.2°
Solvent temperature 11.3°
13.3 Solution Properties
Exothermic Solution
More commonly dissolution is an exothermic process and
heat is released when a solute is dissolved.
Sometimes when we make a solution it will get so hot it
boils!!
13.5 SOLUTION CONCENTRATION
The ratio of the amount of solute to amount of solution, or
solvent is defined by the concentration.
Concentration = solute
solution
=
solute
solvent
There are various combinations of units that are used in these
rations.
Ratio
X 102 X 103 X 106 X 109
g solute
g solution
= % (w/w)
g solute
= % (w/v)
mL solution
mL solute
mL solution
=
% (v/v)
ppt (w/w) ppm (w/w) ppb (w/w)
ppt (w/v)
ppm (w/v) ppb (w/v)
ppt (v/v)
ppm (v/v)
ppb (v/v)
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
1. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
33.6g H2O + 25.2 g NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl
100 g solution
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
100 g solution
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
100 g solution
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
= 149 g NaCl
100 g solution
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
= 149 g NaCl
100 g solution
Mass of water?
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
= 149 g NaCl
100 g solution
Mass of water? 333 g solution – 149 g NaCl = 184 g H2O
13.5 SOLUTION CONCENTRATION
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
= 149 g NaCl
100 g solution
Mass of water? 333 g solution – 149 g NaCl = 184 g H2O
13.5 SOLUTION CONCENTRATION
3. How many grams of NaCl are required to dissolve in 88.2 g
of water to make a 29.2% (w/w) solution.
4. A sugar solution is 35.2%(w/v) find the mass of sugar
contained in a 432 mL sample of this sugar solution.
13.5 SOLUTION CONCENTRATION
The solution concentration can also be defined using moles.
The most common example is molarity (M).
The molarity of a solution is defined as:
“The number of moles of solute in 1 L of solution”
and is given the formula:
Molarity (M) =
Moles solute
Liters solution
13.5 SOLUTION CONCENTRATION
1. A student dissolves 25.8 g of NaCl in a 250 mL volumetric
flask. Calculate the molarity of this solution. (picture of
volumetric flask is on the next slide)
2. Find the mass of HCl required to form 2.00 L of a 0.500 M
solution of HCl.
3. A student evaporates the water form a 333 mL sample of a
0.136 M solution of NaCl. What mass of salt remains?
4. Find the molarity of sodium ions in a solution containing
2.35g of in a one liter volumetric flask.
13.5 SOLUTION CONCENTRATION
In the lab we would use a piece of glassware called a volumetric flask to
prepare this solution.
13.5 SOLUTION CONCENTRATION
13.5 SOLUTION CONCENTRATION
13.7 SOLUTION CONCENTRATION
Often we will want to make a dilute solution from a more
concentrated one.
To determine how to do this we use the formula :
C1V1 = C2V2
Where:
C1 = concentration of more concentrated solution
V1 = volume required of more concentrated solution
C2 = concentration of more dilute solution
V2 = volume of more dilute solution
We can use any units in this equation but they must be the
same on both sides.
13.7 SOLUTION CONCENTRATION
How would one prepare 50.0 mL of a 3.00 M solution of
NaOH using a 7.10 M stock solution?
C1V1 = C2V2
(7.10 M)V1 = (3.00 M) (50.0 mL)
(7.10
M)V1 = (3.00 M) (50.0 mL)
(7.10 M)
(7.10 M)
V1 = 21.1 mL
This means that you add 21.1 mL of the concentrated stock solution to a
50.0 mL volumetric flask and add water until the bottom of the meniscus
touches the line on the volumetric flask.
13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3
L solution
13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution
L solution
103 mL
13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution moles AgCl 215.35 g AgCl 33.2 mL
moles AgNO3 moles AgCl
L solution
103 mL
13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution moles AgCl 215.35 g AgCl 33.2 mL
moles AgNO3 moles AgCl
L solution
103 mL
13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution moles AgCl 215.35 g AgCl 33.2 mL
moles AgNO3 moles AgCl
L solution
103 mL
13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution moles AgCl 215.35 g AgCl 33.2 mL
moles AgNO3 moles AgCl
L solution
103 mL
= 0.715 g AgCl
13.8 SOLUTION STOICHIOMETRY
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
2. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and 200.0 mL of a 0.200
M solution of calcium chloride solution.
3. Find the volume of the excess reactant.
13.9 COLLIGATIVE PROPERTIES
Colligative properties are physical properties that depend on
the number of particles, but not the nature of the particles.
We will discuss three colligative properties:
 Boiling point elevation
 Freezing point depression
 Osmotic pressure
13.9 COLLIGATIVE PROPERTIES
The vapor pressure above a solution is lower than that above
the pure solvent.
This has some interesting effects:
• The boiling point of a solution is higher than the pure
solvent (boiling point elevation).
• The freezing point of a solution is lower than the
pure solvent (freezing point depression).
Molality
Molality is another frequently used solution
concentration, specifically for colligative properties.
moles solute
Molality (m) =
Kg solvent
13.9 COLLIGATIVE PROPERTIES
The change in boiling point and freezing point can be
calculated using the following constants:
Boiling Point Elevation = (i) 0.512 °C
m
°C
Freezing Point Depression = - (i) 1.86° m
(i) = moles of ions found in an ionic solid
M = molality = moles of solute
kg solvent
13.9 COLLIGATIVE PROPERTIES
Lets do an example:
What is the boiling point and freezing of a 0.1 m solution of
NaCl? NaCl (s) → Na+ (aq) + Cl- (aq) n = 2 ions
Freezing point:
-1.86 °C
m
2
0.1 m
13.9 COLLIGATIVE PROPERTIES
Lets do an example:
What is the boiling point and freezing of a 0.10 m solution of NaCl?
NaCl (s) → Na+ (aq) + Cl- (aq) n = 2 ions
Freezing point:
-1.86 °C
m
2
0.10 m
= - 0.37 °C ( Amount Lowered)
Boiling point:
0.512 °C
m
2
0.10 m
Lets do an example:
What is the boiling point and freezing of a 0.10 M solution of NaCl?
NaCl (s) → Na+ (aq) + Cl- (aq) n = 2 ions
Freezing point:
-1.86 °C
m
2
0.10 m
= - 0.37 °C ( Amount Lowered)
F.P. = O °C + ( - 0.37 °C) = - 0.37 °C
Boiling point::
0.512 °C
m
2
0.10 m
= 0.102 °C
B.P. = 10O °C + 0.102 °C) = 100.102 °C
Lets do an example:
What is the boiling point and freezing of a 0.10 M solution of NaCl?
NaCl (s) → Na+ (aq) + Cl- (aq) i = 2 ions
Freezing point:
-1.86 °C
m
2
0.10 m
= - 0.37 °C ( Amount Lowered)
F.P. = O °C + ( - 0.37 °C) = - 0.37 °C
Boiling point::
0.512 °C
m
2
0.10 m
= 0.102 °C
B.P. = 10O °C + 0.102 °C) = 100.102 °C
Osmotic Pressure
Sailors adrift at sea, know that drinking sea water will cause
dehydration, do you? Maybe a better question is why? The
answer involves another colligative property called osmosis.
Osmotic Pressure
Osmosis being a colligative property is dependent upon
particles passing through a semipermeable membrane. A
semipermeable membrane will let only smaller particles pass
through. A chain link fence is and example of a
semipermeable membrane. The fence will block baseballs
from passing through, but not air. The roots of a tree contain
a semipermeable membrane, they allow water to flow into a
tree, but not salt or sugar. On the next diagram two different
solutions are separated by a semipermeable membrane.
Osmotic Pressure
mm
The compartment on the right has more water molecules compared to
the one on the left, thus water flows from the right to the left in
increasing the height of the column on the left and decreasing the
height on the right. Measuring the difference in heights gives the
osmotic pressure of the solution. Calculation of osmotic pressure in
atm. can be done using the ideal gas constant:
0.08206 L-atm/mole-K, notice L/mole in the constant.
Osmotic Pressure
Osmotic pressure coupled with capillary action is responsible for
moving water from the ground up to the top of a tree. Osmotic
pressure also affects our blood cells.
Isotonic solution: When the solution concentration outside the
cell is the same as inside. Here the migration of water into and
out of the cell are equal. Thus the cell remains in its normal state
and works normally.
Hypotonic solution: When the solution concentration outside the
cell is lower than inside the cell. In this state more water flows
into the cell than out of the cell. Overall result is the cell expands
and could rupture. Called hemolysis.
Hypertonic solution: When the solution concentration outside
the cell is greater than inside the cell. In this state more water
flows out of the cell than into the cell, causing the cell to shrink.
This is called crenation.
Osmotic Pressure
Crenation
Hemolysis
Now we can explain about drinking sea water. Since the solute
concentration is greater in the stomach (less water molecules), then
water flows from the cells to the stomach, thus causing dehydration.
The End Ch#13
Solutions