1 CHAPTER 10 Solutions. 2 Types of Solutions Solution - homogeneous mixture of 2 or more substances...

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CHAPTER 10

• Solutions

2

Types of Solutions

• Solution - homogeneous mixture of 2 or more substances» solvent - dissolving medium» solute - dissolved species

3

Spontaneity of theDissolution Process

• Assume solvent is a liquid• Major factors that affect dissolution of solutes

» change of energy content, solution

– exothermic favors dissolution

– endothermic does not favor dissolution

» change in disorder, or randomness, Smixing

– increase in disorder favors dissolution

– increase in order does not favor dissolution

• Best conditions for dissolution» exothermic & disordered

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• Disorder in mixing is very common» helps dissolution

• Factors that affect solution

solute-solute attractionsrequires absorption of E to overcome

solvent-solvent attractions requires absorption of E to overcome

solvent-solute attractionsreleases energy

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6

The Solution Process

• A solution is a homogeneous mixture of solute (present in smallest amount) and solvent (present in largest amount).

• In the process of making solutions with condensed phases, intermolecular forces become rearranged.

• Consider NaCl (solute) dissolving in water (solvent):» the water H-bonds have to be interrupted,» NaCl dissociates into Na+ and Cl-,

» ion-dipole forces form: Na+ … -OH2 and Cl- … +H2O.

» We say the ions are solvated by water.» If water is the solvent, we say the ions are hydrated.

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The Solution Process

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There are three energy steps in forming a solution:» separation of solute molecules (H1),

» separation of solvent molecules (H2), andformation of solute-solvent interactions (H3).

• We define the enthalpy change in the solution process as

Hsoln = H1 + H2 + H3.

Hsoln can either be positive or negative depending on the intermolecular forces.

Energy Changes and Energy Changes and Solution FormationSolution Formation

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Energy Changes and Solution Formation

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• Breaking attractive intermolecular forces is always endothermic.

• Forming attractive intermolecular forces is always exothermic.

• To determine whether Hsoln is positive or negative, we consider the strengths of all solute-solute and solute-solvent interactions:H1 and H2 are both positive.

H3 is always negative.

» It is possible to have either H3 > (H1 + H2) or H3 < (H1 + H2).

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Examples: » NaOH added to water has Hsoln = -44.48 kJ/mol.

» NH4NO3 added to water has Hsoln = + 26.4 kJ/mol.

• “Rule”: polar solvents dissolve polar solutes. Non-polar solvents dissolve non-polar solutes. Why?» If Hsoln is too endothermic a solution will not form.

» NaCl in gasoline: the ion-dipole forces are weak because gasoline is non-polar. Therefore, the ion-dipole forces do not compensate for the separation of ions.

» Water in octane: water has strong H-bonds. There are no attractive forces between water and octane to compensate for the H-bonds.

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Solution Formation, Spontaneity, and Disorder

• A spontaneous process occurs without outside intervention.

• When energy of the system decreases (e.g. dropping a book and allowing it to fall to a lower potential energy), the process is spontaneous.

• Some spontaneous processes do not involve the system moving to a lower energy state (e.g. an endothermic reaction).

• If the process leads to a greater state of disorder, then the process is spontaneous.

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• Example: a mixture of CCl4 and C6H14 is less ordered than the two separate liquids. Therefore, they spontaneously mix even though Hsoln is very close to zero.

• There are solutions that form by physical processes and those by chemical processes.

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Solution Formation and Chemical Reactions

Consider:

Ni(s) + 2HCl(aq) NiCl2(aq) + H2(g).• Note the chemical form of the substance being

dissolved has changed (Ni NiCl2).

• When all the water is removed from the solution, no Ni is found only NiCl2.6H2O. Therefore, Ni dissolution in HCl is a chemical process.

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• Example:

NaCl(s) + H2O (l) Na+(aq) + Cl-(aq).

• When the water is removed from the solution, NaCl is found. Therefore, NaCl dissolution is a physical process.

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Dissolution of Solidsin Liquids

• crystal lattice energy - energy absorbed when a mole of formula units of a solid is separated into its constituent ions (molecules or atoms for nonionic solids) in the gas phase

» measure of attractive forces in solid» crystal lattice energy increases as charge density

increases

• energy required to overcome London forces, dipole-dipole or H-bonding

M X (s) crystal lattice energy M (g) X (g)+ - + -

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• dissolution is a competition between1 solute -solute attractions

crystal lattice energy

2 solvent-solvent attractions

H-bonding, etc.

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• energy is released when solute particles are dissolved» energy of solvation» hydration energy (in water)

• look at dissolution of CaCl2.

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CaCl (s) Ca(OH Cl H O

where is approximately 7 or 82

H O2

-2

2 )62 2

x

x

Ca

OH2

OH2

OH2

OH2

H2O

H2O

2+

Cl-

OH H

H OH

HO H

H H O

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• molar energy of hydration - energy absorbed when one mole of formula units becomes hydrated

M (g) + H O M(OH hydration E for M

X H O X(H O) hydration E for X

n+2 2

n n+

y-2 2 n

y-

x

g n

x

y

)

( )

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• hydration energy increases with increasing charge density

kJ/mol 4750- 6.00 0.50 Al

kJ/mol 2160- 2.78 0.72 Cu

kJ/mol 1650- 2.02 0.99 Ca

kJ/mol 351- 0.75 1.33 K

Hydrationofat HeChg/Radius (Å) adius RIon

+3

+2

+2

+

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Dissolution of Liquidsin Liquids (Miscibility)

• Most polar liquids are miscible with other polar liquids

• “like dissolves like” rule» methanol, CH3OH, very

soluble in water

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Dissolution of Liquidsin Liquids (Miscibility)

• nonpolar liquids are miscible with other nonpolar liquids

• “like dissolves like” rule» nonpolar molecules “slide” in between each other

C Cl

Cl

Cl

Cl

C

CC

C

CC

H

H

H

H

H

H

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Rates of Dissolution and Saturation

• Finely divided solids dissolve more rapidly than large crystals• granulated sugar vs. sugar cubes• look at a single cube of NaCl

enormous increase in surface area

dissolves faster

NaCl

Breaks

upmany smaller crystals

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Rates of Dissolution and Saturation

• saturated solutions have established an equilbrium between dissolved and undissolved solutes» air with 100% humidity» liquid solutions with solids

• supersaturated solutions have higher concentrations of dissolved solutes than saturated

equal are rates reverse & forward

XMMX aqaqs

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Saturated Solutions and Solubility

• Dissolve: solute + solvent solution.• Crystallization: solution solute + solvent.• Saturation: crystallization and dissolution are

in equilibrium.• Solubility: amount of solute required to form a

saturated solution.• Supersaturated: a solution formed when more

solute is dissolved than in a saturated solution.

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Factors Affecting SolubilitySolute-Solvent InteractionsSolute-Solvent Interactions• Polar liquids tend to dissolve in polar solvents.• Miscible liquids: mix in any proportions.• Immiscible liquids: do not mix.• Intermolecular forces are important: water and

ethanol are miscible because the broken hydrogen bonds in both pure liquids are re-established in the mixture.

• The number of carbon atoms in a chain affect solubility: the more C atoms the less soluble in water.

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Factors Affecting Solubility

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Factors Affecting Solubility• The number of -OH groups within a molecule

increases solubility in water.

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Factors Affecting Solubility

• Generalization: “like dissolves like”.• The more polar bonds in the molecule, the

better it dissolves in a polar solvent.• The less polar the molecule the less it

dissolves in a polar solvent and the better is dissolves in a non-polar solvent.

• Network solids do not dissolve because the strong intermolecular forces in the solid are not re-established in any solution.

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Effect of Pressure on Solubility

Solubility of a gas in a liquid is a function of the pressure of the gas.

• The higher the pressure, the more molecules of gas are close to the solvent and the greater the chance of a gas molecule striking the surface and entering the solution.» Therefore, the higher the pressure, the greater the

solubility.» The lower the pressure, the fewer molecules of gas

are close to the solvent and the lower the solubility.

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Pressure Effects

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Pressure Effects• Pressure changes have little or no effect on solubility

of liquids and solids in liquids• Pressure changes have large effects on the solubility

of gases in liquids» why carbonated drinks fizz when opened» cause of several scuba diving related problems including

the “bends”

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Pressure Effects• Phenomenon described by Henry’s Law

M

M

gas gas

gas

gas

k P

where molar concentration of gas

k = Henry's Law constant, unique number for each

gas - liquid combination

P = partial pressure of gas

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Effect of Temperature on Solubility

• ionic solids that dissolve endothermically» dissolution enhanced by heating

• ionic solids that dissolve exothermically» dissolution decreased by heating

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Temperature Effects

• Experience tells us that sugar dissolves better in warm water than cold.

• As temperature increases, solubility of solids generally increases.

• Sometimes, solubility decreases as temperature increases (e.g. Ce2(SO4)3).

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Temperature Effects

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Temperature Effects

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Temperature Effects

• Experience tells us that carbonated beverages go flat as they get warm.

• Gases are less soluble at higher temperatures.• Thermal pollution: if lakes get too warm, CO2 and

O2 become less soluble and are not available for plants or animals.

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Ways of Expressing Concentration

• All methods involve quantifying amount of solute per amount of solvent (or solution).

• Generally amounts or measures are masses, moles or liters.

• Qualitatively solutions are dilute or concentrated.• Definitions:

100soln of mass total

soln incomponent of masscomponent of % Mass

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Mole Fraction, Molarity, and MolalityMole Fraction, Molarity, and Molality• Recall mass can be converted to moles using

the molar mass.• Recall

• Recall

components all of moles totalcomponent of moles

component of fraction Mole

solution of literssolute of moles

Molarity

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We define

• Converting between molarity (M) and molality (m) requires density.

• In dilute solutions, molarity and molality are nearly equal.

solvent of kilograms

solute of molesMolality

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Molality & Mole Fraction• Calculate the molarity and the molality of an aqueous

solution that is 10.0% glucose, C6H12O6. The density of the solution is 1.04 g/mL. 10.0% glucose solution has several medical uses. 1 mol C6H12O6 = 180 g

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Molality & Mole Fraction• Calculate the molality and the molarity of an aqueous


? mol C H O

kg H O

g C H O

90.0 g H O

g H O

1 kg H O

mol C H O

180 g C H O C H O

6 12 6

2

6 12 6

2

2

2

6 12 6

6 12 66 12 6

10 0 1000

10 617

.

. m

this is the concentration in molality

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? mol C H O

L H O

g C H O

100.0 g sol'n

g sol'n

mL sol'n

mL

1 L

mol C H O

180 g C H O C H O

6 12 6

2

6 12 6

6 12 6

6 12 66 12 6

10 0 104

1000 10 578

. .

. M

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Molality & Mole Fraction

• Calculate the molality of a solution that contains 7.25 g of benzoic acid C6H5COOH, in 200 mL of benzene, C6H6. The density of benzene is 0.879 g/mL. 1 mol C6H5COOH = 122 g

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Molality & Mole Fraction• Calculate the molality of a solution that contains 7.25

g of benzoic acid C6H5COOH, in 200 mL of benzene, C6H6. The density of benzene is 0.879 g/mL. 1 mol C6H5COOH = 122 g

? mol C H COOH

kg C H

g C H COOH

200.0 mL C H

mL C H

0.879 g C H

g C H

1 kg C H

mol C H COOH

122 g C H COOH C H COOH

6 5

6 6

6 5

6 6

6 6

6 6

6 6

6 6

6 5

6 56 5

7 25 1

1000 10 338

.

. m

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Molality & Mole Fraction• Mole fraction

» number of moles of one component per moles of all the components of the solution

» literally is a fraction using moles as the numerator and denominator » in a 2 component solution

• Mole fraction of component A - XA

X A # moles of A

# moles A + # moles B

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Molality & Mole Fraction• Mole fraction of component B - XB

X

X X

B

B

#moles of B

# moles A + # moles Bnote that

sum of mole fractions must equal 1A 1

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Molality & Mole Fraction• What are the mole fractions of glucose and water in

a 10.0% glucose solution?

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in 100 g of solution there are

10.0 g of glucose & 90.0 g of water

? mol C H O g C H O

1 mol C H O

180 g C H O mol C H O

6 12 6 6 12 6

6 12 6

6 12 66 12 6

10 0

0 0556

.

.

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? mol H O g H O

1 mol H O

18 g H O mol H O

2 2

2

22

90 0

5 00

.

.

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Molality & Mole Fraction• now let’s calculate the mole fractions

X

X

H O2

2 6 12 6

C H O6 12 6

2 6 12 6

2

6 12 6

mol H O

mol H O + 0.0556 mol C H O

mol C H O

mol H O + 0.0556 mol C H O

5 00

5 00

0 989

0 0556

5 00

0 011

100 0 989 0 011

.

.

.

.

.

.

. . .

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Colligative Properties of Solutions

• Colligative properties» solution properties that depend solely on the number of

particles dissoved in the solution and not the kinds of particles dissolved

» physical property of solutions» four common types of colligative properties

vapor pressure lowering freezing point depression boiling point elevation osmotic pressure

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Lowering the Vapor Pressure

Non-volatile solvents reduce the ability of the surface solvent molecules to escape the liquid.

• Therefore, vapor pressure is lowered.• The amount of vapor pressure lowering

depends on the amount of solute.

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Lowering the Vapor Pressure

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Lowering of Vapor Pressure & Raoult’s Law

• Addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution» simply due to fewer solvent molecules at surface» solute molecules occupy some of the spaces

• Raoult’s Law describes this effect in ideal solutionsP P

where P vapor pressure of solvent

P vapor pressure of pure solvent

mole fraction of solvent

solvent solvent solvent0

solvent

solvent0

solvent

X

in solution

X in solution

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• Lowering of vapor pressure, Psolvent, is defined as:

P P P

P - ( P

)P

solvent solvent0

solvent

solvent0

solvent solvent0

solvent solvent0

X

X

)( )

(1

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• Since Xsolvent + Xsolute = 1, we can derive

X X

X

solute solvent

solvent solute solvent0

1 -

P P

which is Raoult' s Law

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Raoult’s Law

Ideal solution: one that obeys Raoult’s law.• Raoult’s law breaks down when the solvent-

solvent and solute-solute intermolecular forces are greater than solute-solvent intermolecular forces.

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Boiling Point Elevation• Addition of a nonvolatile solute to a solution

raises the boiling point of the solution above that of the pure solvent.» vapor pressure is lowered - Raoult’s Law» T must be raised to make vapor pressure equal to

atmospheric pressure

• Amount that T is elevated is determined by the number of moles of solute dissolved in solution.

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Boiling Point Elevation

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• Boiling point elevation relationship is:

T K

where: T boiling point elevation

molal concentration of solution

K molal boiling point elevation constant

for the solvent

b b

b

b

m

m

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What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution?

T Kb b m

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What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution?

T K

T C /

T C

B.P. = 100 C + C = 101.28 C

b b

b0

b0

0 0 0

m

m m( . )( . )

.

.

0 512 2 50

128

128

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Freezing Point Depression• Addition of a nonvolatile solute to a solution

lowers the freezing point of the solution relative to the pure solvent.

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Freezing Point Depression

• Relationship for freezing point depression is:

T K

where: T freezing point depression of solvent

molal concentration of soltuion

K freezing point depression constant for solvent

f f

f

f

m

m

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• Notice the similarity of the two relationships for freezing point depression & boiling point elevation.

• Fundamentally, it is the same phenomenon.» differences are the size of the effect

» reflected in the sizes of the constants, Kf & Kb

• Easily seen on a phase diagram for a solution.

T K vs. T Kf f b b m m

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Freezing Point Depression• Calculate the freezing point of a 2.50 m aqueous

glucose solution.

T K

T (1.86 C /

T C

F.P.= 0.0 C - 4.65 C = - 4.65 C

f f

f0

f

0 0 0

m

m m)( . )

.

2 50

4 650

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• Calculate the freezing point of a solution that contains 8.50 g of benzoic acid (C6H5COOH, MW = 122) in 75.0 g of benzene, C6H6.

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C0.72=C4.76-C5.48 =F.P.

C76.4)929.0)(C/12.5(T

KT

solution. for this depression theCalculate .2

929.0COOHHC g 122

COOHHC mol 1

HC kg 0.0750

COOHHC g 50.8

HC kg

COOHHC mol ?

molality! Calculate .1

000

00f

ff

56

56

66

56

66

56

mm

m

m

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Determination of M.W. by Freezing Point Depression

• Freezing point depression depends on 2 thingssize of Kf for a given solvent

– many of these values are well known

molal concentration– # of moles of solute– kg of solvent

• If Kf and kg of solvent are known, as is often the case in an experiment, then we can determine # of moles of solute and use it to determine the molecular weight.

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• A 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 200 g of water. The resulting solution froze at -5.580C. What is the molecular weight of the compound?

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• A 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 200 g of water. The resulting solution froze at -5.580C. What is the molecular weight of the compound?

T KT

K

C

1.86 C

means that there are 3.00 mol in 1 kg water

? g cmpd. in 1 kg H O =37.0 g cmpd.

0.200 kg H O g cmpd.

185 g

3.00 mol g / mol

f ff

f0

22

m m m

m

558300

300

185

617

0..

.

.

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Osmosis

• Semipermeable membrane: permits passage of some components of a solution.

• Osmosis: the movement of a solvent from low solute concentration to high solute concentration.

• There is movement in both directions across a semipermeable membrane.

• As solvent moves across the membrane, the fluid levels in the arms becomes uneven.

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Osmosis

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Osmosis

• Osmotic pressure, , is the pressure required to stop osmosis:

MRT

RTVn

nRTV

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Osmosis

• Isotonic solutions: two solutions with the same separated by a semipermeable membrane.

• Hypotonic solutions: a solution of lower than a hypertonic solution.

• Osmosis is spontaneous.• Red blood cells are surrounded by

semipermeable membranes.

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Group Question• Medicines that are injected into humans,

intravenous fluids and/or shots, must be at the same concentration as the existing chemical compounds in blood. For example, if the medicine contains potassium ions, they must be at the same concentration as the potassium ions in our blood. Such solutions are called isotonic. Why must medicines be formulated in this fashion?

1 CHAPTER 10 Solutions. 2 Types of Solutions Solution - homogeneous mixture of 2 or more substances...

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Transcript of 1 CHAPTER 10 Solutions. 2 Types of Solutions Solution - homogeneous mixture of 2 or more substances...