Solutions Page 77

Chapter 4

Solutions

Solutions are homogeneous mixtures of two or more

substances.

Dissolving medium is called the solvent.

Dissolved species are called the solute.

There are three states of matter (solid, liquid, and gas)

which when mixed two at a time gives nine different kinds of

mixtures.

Seven of the possibilities can be homogeneous.

Two of the possibilities must be heterogeneous.

Seven Homogeneous Possibilities Solute Solvent Example Solid Liquid salt water Liquid Liquid mixed drinks Gas Liquid carbonated

beverages Liquid Solid dental amalgams Solid Solid alloys Gas Solid metal pipes Gas Gas air

Solutions Page 78

Two Heterogeneous Possibilities Solid Gas dust in air Liquid Gas clouds, fog

Ways of Expressing Concentration

All methods involve quantifying the amount of solute per

amount of solvent (or solution).

Concentration may be expressed qualitatively or

quantitatively.

The terms dilute and concentrated are qualitative ways to

describe concentration.

A dilute solution has a relatively small concentration of

solute.

A concentrated solution has a relatively high

concentration of solute.

Quantitative expressions of concentration require specific

information regarding such quantities as masses, moles, or

liters of the solute, solvent, or solution.

The solution process: Polar materials dissolve only in

polar solvents (NaCI/H2O), and non - polar substances are

• Concentration and solubility

Solutions Page 79

soluble in non - polar solvents. This is the first rule of

solubility "like dissolves like" e.g: benzene in CCI4

Solution Concentration:

The amount of solute dissolved in a given amount of

solvent or dissolved in a given amount of solution is called

the concentration of the solution. Dilute solutions have

relatively low concentrations; concentrated solutions have

relatively high concentrations. A solution that contains as

much solute as can be dissolved is called a saturated

solution; solutions with lower concentration are called

unsaturated solutions.

Methods for expressing the solution concentration:-

a) Weight to weight expression.

b) Weight to volume expression.

a) Weight to weight expression;

1) Weight percent (wt%)

Weight percent is defined as: Number of grams of

solute which present in 100 gram of solution.

Solutions Page 80

100 x solution of weighttotal

solute of weightmass%

N.B

solution of weighttotal

solute of weightfraction weight

e.g. 10% by weight glucose means:

10 gm glucose + 90 gm H2O = 100 gm solution.

% Solute = (10/100) x 100 = 10 %.

% Solvent = (90/100) x 100 = 90 %

2) Mole fraction (X):

The ratio of the number of moles of one component to the total

number of moles of all components in the mixture solution

Mole fraction is essentially self-defined. In equation form

the mole fraction (usually symbolized by X) is

3) Molality

Molality is similar to but not the same as molarity. Molality,

m, is defined by,

Solutions Page 81

( )

N.B

W solution = W solute + W solvent

W solvent = W solution - W solute

W solvent =dV solution - W solute

Where d = density of the solution V = Volume of the solution

Example 4.1: What is the molality of 12.5 % solution of

glucose C6H12O6, in water? M.wt. of glucose is

180.0

Solution: 1) in 12.5 % solution 12.5 gm C6H12O6 is

dissolved in l00 gm solution.

W solvent = 100 - 12.5 = 87.5 g H2O

2) no. of moles glucose = 12.5/180

Solutions Page 82

)

Example 4.2: What are the mole fractions of solute and

solvent in a 1.0m aqueous solution?

Solution: The molecular weight of H2O is 18.0 we find the

number of moles of water in 1000 gm of H2O.

no of moles of H2O mol 55.6 18

1000 H2O

A 1.0 m aqueous solution contains

n solute =1.0 mol

56.6 n

55.6 n

total

H2O

The mole fractions are

X solute 0.018 mol 56.6

mol 1.0

n

n

total

solute

X water = 0.982 6.56 n

55.6 n

total

H2O

Solutions Page 83

b) Weight to volume expression:

1) Molarity (M):

• h d d solution.

•

• M = (W/M. wt) solute X (1000/V) solution

(Molarity is mol of solute per Liter of solution, not per liter

of solvent. Molarity is defined so that we can always know

how many mols of solute there are in any given amount of

solution.)

In dilute aqueous solution the molarity and molality are

about the same. However, in concentrated water solutions

and in solutions where the solvent is not water the molarity

and molality are very different.

Example 4.3: a) How many grams of concentrated nitric

acid solution should be used to prepare 250 ml of

2.0M HNO3? The concentrated acid is 70.0 %

Solutions Page 84

b) If the density of the concentrated nitric acid

solution is 1.42 g/ml. What volume should be

used? M.wt. (HNO3) =63

Solution: a) 70 gm HNO3 l00 gm solution

250

1000 x

M.Wt

WM

250

1000 x

63

W2

Mass of pure HNO3 1000

250 x 63 x 2

mass of HNO3 solution = gm 45.0 70

100 x 31.5

b) ml cone. NHO3 = (45/1.42) = 31.7 ml cone. HNO3

Example 4.4: An aqueous solution of acetic acid was

prepared by dissolving 164.2 gm of acetic acid in 800 ml

of the solution. If the density of the solution was 1.026

gm/ml. M. wt of acetic acid = 60

Calculate:

a) The molar concentration of the solution

b) The molality

c) The mole fraction of both the solute and the solvent

d) The mole %

e) The weight %.

Solutions Page 85

Solution:

a) 3.4

800 x 60

1000 x 164.2

V

1000 x

M.Wt

WM

mlsolute

b) d = 1.026g/ml V = 800 ml

W solution = V x d = 800 x 1.026 = 820.8 gm

W slvent = 820.8 - 164.2 = 656.6 gm

m 4.17 656.6 x 60

1000 x 164.2

W(Solvent)

1000 x (Solute)

M.Wt

massm

c) no. of acetic acid moles = 164.2 / 60 = 2.737 mole

no. of H2O moles = 656.6 / 18 = 36.44 mole

Mole fraction of acetic acid = 0.0699 36.44 2.737

2.737

Mole fraction of H2O = 0.9299 36.44 2.737

36.44

d) mole % acetic acid = 0.0699 x 100 = 6.99 %

mole % of H2O = 0.9299 x 100 = 92.99 %

e) percentage weight of acetic acid = % 20 820.8

100 x 164.2

percentage weight of H2O= % 80 820.8

100 x 656.6

Solutions Page 86

Try ?

1. Five grams of NaCl is dissolved in 25.0 g of H2O. What is

the mole fraction of NaCl in the solution? (Answer =0.0580)

2. What is the mole percent NaCl in the previous problem 1

(Answer = 5.80 mol %)

3. Ten grams of ascorbic acid (vitamin C), C6H8O6, is

dissolved in enough water to make 125 ml of solution.

What is the molarity of the ascorbic acid? (Answer = 5.80

mol %)

4. What is the molality of NaCl in the solution in the previous

problem 1? (Answer = 3.42 m)

5. What is the mass percent of NaCl in the solution in the

previous problem 1? (Answer = 16.7 %)

Factors Affecting Solubility

The extent to which a solute dissolves in solvent depends

1. The nature of the solute.

2. The nature of the solvent.

• Principles of Solubility

Solutions Page 87

3. The temperature.

4. The pressure (for gases).

In this section we will consider in turn the effect of each of

these factors upon solubility.

1. Intermolecular forces are an important factor in

determining solubility of a solute in a solvent. The

stronger the attraction between solute and solvent

molecules, the greater the solubility. For example,

polar liquids tend to dissolve in polar solvents.

Favorable dipole-dipole interactions exist (solute-

solute, solvent-solvent, and solute-solvent).

2. Pairs of liquids that mix in any proportions are said to

be miscible. Example: Ethanol and water are

miscible liquids. In contrast, immiscible liquids do not

mix significantly. Example: Gasoline and water are

immiscible.

3. Consider the solubility of alcohols in water. Water and

ethanol are miscible because the broken hydrogen

1. Solute-Solvent Interactions:

Solutions Page 88

bonds in both pure liquids are re-established in the

mixture. However, not all alcohols are miscible with

water. Why? The number of carbon atoms in a chain

affects solubility. The greater the number of carbons in

the chain, the more the molecule behaves like a

hydrocarbon. Thus, the more C atoms in the alcohol,

the lower its solubility in water. Increasing the number

of –OH groups within a molecule increases its

solubility in water. The greater the number of –OH

groups along the chain, the more solute-water H-

bonding is possible.

4. Generalization: “Like dissolves like”. Substances

with similar intermolecular attractive forces tend to be

soluble in one another. The more polar bonds in the

molecule, the better it dissolves in a polar solvent. The

less polar the molecule the less likely it is to dissolve in

a polar solvent and the more likely it is to dissolve in a

nonpolar solvent..

Example: Most nonpolar substances have very small water

solubilities. Petroleum, a mixture of hydrocarbons, spreads

out in a thin film on the surface of a body of water rather

Solutions Page 89

than dissolving. The mole fraction of pentane, C5H12 in a

saturated water solution is only 0.00003. Few organic

compounds that dissolve readily in water, most contain - OH

groups. Three familiar examples are methyl alcohol, ethyl

alcohol, and ethylene glycol, all of which are soluble in water in

all proportions.

H

C

H

H OH

H

C

H

H C

H

H

OH

H

C

OH

H C

H

OH

H

methyl alcohol ethyl alcohol ethylene glycol

Experience tells us that sugar dissolves better in warm

water than in cold water.

As temperature increases, solubility of solids generally

increases. Sometimes solubility decreases as

temperature increases (e.g., Ce2(SO4)3).

Gases are less soluble at higher temperatures. An

environmental application of this is thermal pollution.

2. Solubility and Temperatune

Solutions Page 90

Since solubility is an equilibrium concentration, we can apply Le

Chateher"s principle in order to find out what happens when the

temperature of a saturated solution is changed. It is important to

know whether the dissolving process is

exothermic

Solute + solvent → + h or

endothermic

Solute + + h →

The heat of solution is defined as H for the dissolving process,

and so is equal to H solution - (H solute + H solvent). Therefore, for the

exothermic case, ΔH solution is negative, and for the endothermic,

positive.

Consider now a saturated aqueous solution of potassium iodide

with excess KI(s) present. For KI, ΔH solution = 21 kJ mol -1

, so we

know that KI dissolves in water with the absorption of heat and

can write the saturation equilibrium equation as

21 kJ + KI(s) K+ + I

-

If we raise the temperature of the saturated KI solution, we

predict according to Le Chatelier's principle that the above

equilibrium will shift to the right (1) using up some of the

Solutions Page 91

added heat (and some of the excess solid KI) and (2)

increasing the concentration of K+ and I

- ions in solution.

After equilibrium has been reestablished at a higher

temperature, the concentration of dissolved KI has become

higher, that is, the solubility of KI increases with increasing

temperature.

An example of an exothermic process is the dissolving of

lithium iodide (Lil) in water for which ΔH solution 71 kJ mol-1. We

can write the saturation equilibrium equation as

LiI (s) Li+ + I

- + 71 KJ

If we raise the

temperature of a

saturated solution of

LiI, the equilibrium

shifts to the left (1)

using up some of the

added heat (and I-

and Li+ ions in

solution) and , (2)

forming more solid Lil.

Solutions Page 92

(We observe the precipitation of some Lil out of solution.)

After equilibrium has been

reestablished at a higher

temperature, the

concentration of dissolved

Lil is lower, so we can say

that the solubility of lithium

iodide decreases with an

increase in temperature.

Dissolving solids or liquids

in liquids can be either positive or negative. For aqueous

solutions it is more frequently positive; so the majority of

substances have solubilities which increase with temperature.

(This should not be used as a rule, however; there are too

many exceptions.)

When gases dissolve in liquids, ΔH is usually negative; that is,

heat is liberated. (The solvation energy usually exceeds the

energy necessary to separate the molecules in the liquid.)Thus

in the majority of cases the solubility of gas decreases with

temperature. This nearly always is true in water. Boiled water,

for example, tastes "flat," in part because dissolved air (and

Solutions Page 93

chlorine) is less soluble at the boiling point and is removed from

the water.

Pressure changes have little or no effect on solubility of

liquids and solids in liquids.

Liquids and solids are not compressible.

Pressure changes have large effects on the solubility of

gases in liquids.

Sudden pressure change is why carbonated drinks fizz

when opened.

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

pressure of the gas. The higher the pressure, the

greater the solubility.

Cg - solubility of gas

3. Effect of Pressure on Solubility

(solubility of gas in liquid)

Henry’s Law – The solubility of a gas increases in direct proportion to its partial pressure above the solution.

ggkPC

Solutions Page 94

Pg - the partial pressure of the gas

k - H ’ w

An application of Henry's law: preparation of carbonated

soda. Carbonated beverages are bottled under PCO2> 1

atm. As the bottle is opened, PCO2 decreases and the

solubility of CO2 decreases. Therefore, bubbles of CO2

escape from solution.

Problem: At 740 torr and 20°C, nitrogen has solubility in

H2O of 0.018 g /I. At 620 torr and 20°C its solubility is 0.015

g/l.

Do these data show that nitrogen obey Henry's law or not?

Example 4.5: At 25°C oxygen gas collected over water at a

total pressure of 101 kPa is soluble to the

extent of 0.0393 g dm-3

. What would its

solubility be if its partial pressure over water

were 107 kPa? The vapor pressure of water is

3.0 kPa at 25°C.

Solution: P total = PH2O + PO2

Solutions Page 95

PO2 = P total - PH2O = 101-3 = 98 kPa

1- Completely miscible liquids

2- Completely immiscible liquids H2O and aniline, H2O and

chlorobenzene

3- Partially immiscible liquids, H2O and phenol, H2O and

ether

Completely miscible liquids of binary solution

a) Ideal solution

b) Non - ideal solution

a) Ideal Solution:-

e.g: (n - heptane / n - hexane)

(chlorobenzene / brombenzene)

• (Solution of liquids in liquids)

Solutions Page 96

1) The force of attraction between all molecules are

identical i.e. the attraction force is not affected by

addition of other components A - A = B-B = A - B.

2) No heat is evolved or absorbed during mixing i.e. ∆H

soln. = Zero

3) The volume of solution is the sum of volumes of the

two liquids.

4) The solution obeys RaouLt's law.

At constant

temperature the

partial vapor

pressure of liquid

component in

ideal solution is

proportional to the

mole fraction of

this constituent in

solution (Figure 1). Figure (1): Vapor pressure of ideal

solutions

Solutions Page 97

At constant T

Where and

= vapor pressure of pure liquids A

and B

PA and PB = partial vapor pressure of liquids A and B in

solution.

PA = (1)

PB = (2)

Pt = PA + PB

Pt = +

(3)

Where XA = mole fraction of A = n

n

t

A

XB = mole fraction of B= n

n

t

B

XA + XB = 1 XA = 1 – XB

By substituting in (1)

Pt = +

( ) (4)

Solutions Page 98

Or

Pt = (

- )

(5)

N.B Composition of solution % XA and % XB.

% A= XA x 100 % B = XB x 100

Example 4.6: Heptane (C7H16) and octane (C8H18) form

ideal solutions What is the vapor pressure at

40°C of a solution that contains 3.0 mol of

heptane and 5 mol of octane? At 40°C, the

vapor pressure of heptane is 0.121 atm and the

vapor pressure of octane is 0.041 atm.

Solution:

The total number of moles is 8.0. therefore

X heptane = 3.0/8.0 = 0.375

X octane = 5.0/8.0 = 0.625

Total = X heptane . Po heptane + X octane. P

o octane

= 0.375 x 0.12 +0.625 x 0.04

= 0.045 atm + 0.026 atm.

= 0.071 atm.

Solutions Page 99

Example 4.7 : Assuming ideality, calculate the vapor

pressure of 1.0 m solution of a non - volatile, on

dissociating solute in water at 50°C. The vapor

pressure of water 50°C is 0.122 atm.

Solution : From example 2 the mole fraction of water in

1.0m solution is 0.982.

PH2O = XH2O PH2O = 0.982 x 0.122 = 0.120 atm.

Problem: At 140°C, the V.P of C6H5CI is 939.4 torr and

that of C6H5Br is 495.8 torr. Assuming that these

two liquids from an ideal solution. Find the

composition of a mixture of two liquids which boils

at 140°C under 1 atm pressure?

Example 4.8: A solution is prepared by mixing 5.81 g

acetone C3H6O, (M. wt = 58.1 g/mole) 11.9 g

chloroform (CHCI3 M.wt 119.4 g/mole). At 35°C

this solution has a total vapor pressure of 260 torr.

Is this an ideal solution? Comment? The vapor

Solutions Page 100

pressure of pure acetone and pure CHCI3 at 35°C

are 345 and 293 torr, respectively.

Solution:

n acetone = M.Wt

W mole 0.1

58.1

5.81

n CHCl3= mole 0.1 119

11.9

nt = 0.1 + 0.1 = 0.2 mole

Xacetone = 0.5 mol 0.2

mol 0.1

n

n

total

acetone

XCHCl3 = 0.5

Pt = +

= 345 x 0.5 + 293 x 0.5 = 319 torr. .

The observed value = 260 torr

• h h w h h

ideal.

• h b d 6 h h d

= 319 this is a negative deviation from Roault's law.

Solutions Page 101

a) In liquid region (apply Rault's law).

b) Non- ideal solutions (Solutions deviate from ideal

behavior).

Negative deviation Positive deviation

1- The force of attraction increase by mixing A - A, B-B < A-B

The force of attraction decrease by mixing A-A , B-B > A-B

2- The vapor pressure will be lower than that given by Roault's law

The vapor pressure will be higher than that given by Raoult's law.

3- H solution :- Ve (exothermic)

H solution: + Ve (endothermic)

4- Temperature change when solution is formed: increase

Temperature change when solution is formed: decrease.

5- Example: Acetone-water

Fig.2: Vapour pressure of non-ideal solution (-ve deviation)

Ethanol-hexane

Fig.3: Vapour pressure of non-ideal solution (+ve deviation)

Solutions Page 102

Fractional Distillation of Binary Miscible liquids

The separation of mixture of volatile liquids into their

components is called fractional distillation, where the

distillate containing the more volatile component and the

residue the less volatile one.

a) Ideal solutions

If a mixture of 2 liquids (A and B) form a completely

miscible ideal solution and PA > PB result in B.P. of A <

B.P of B thus on boiling:-

1) The Liquid A boils at lower B.P than that of liquid B.

2) The liquid A which is more volatile will be passed

from the fractionating column and the liquid B which

is less volatile returned again to the distallating

flask.

A solution of intermediate b.p. between 2 pure liquid -called

azeotropic solution

b) Non - ideal solutions (solutions that exhibit

deviations from Raoults law)

1) Non - ideal solutions with minimum boiling point:

Solutions Page 103

Fig.4: Boiling point-composition diagram

In fig 5 both the liquid and vapor curves meet at a certain

composition having a minimum boiling point, such solution

at this composition called azeotropic mixture.

• I h h d d

the azeotropic mixture will distill first and the excess of

(A) or (B) will remains in the flask e.g 95 % ethanol and

5 % H2O.

Solutions Page 104

2) Non - ideal solutions with maximum boiling point:

In the B.P diagram of both liquid and vapor curves

meet at a certain composition (M) having a maximum

B.P., such solution at this composition called azeotropic

mixture.

• I h h d d h

execs of acetone or CHCI3 will distill first leaving the

azeotropic mixture in the flask.

Solutions Page 105

• Colligative properties depend on number of solute

particles.

Colligative properties do not depend on the kinds of

particles dissolved

• There are four colligative properties to consider:

• Vapor pressure lowering (Raoult's Law).

• Boiling point elevation.

• Freezing point depression.

• Osmotic pressure.

• Colligative Properties of Solutions

• (Solution of solid in liquids)

Solutions Page 106

Vapor pressure lowering is the key to all four of the

colligative properties.

Addition of a nonvolatile solute to a solution lowers the

vapor pressure of the solution.

The effect is simply due to fewer solvent

h ’

The solute molecules occupy some of the spaces

that would normally be occupied by solvent.

R ’ L w d h ideal solutions.

Ideal solution: one that obeys Ra ’ w

Real solutions show approximately ideal behavior when:

• The solute concentration is low.

• The solute and solvent have similarly sized molecules.

• The solute and solvent have similar types of

intermolecular attractions.

R ’ w b down when the solvent-solvent and

solute-solute intermolecular forces are much greater or

weaker than solute-solvent intermolecular forces.

1. Lowering of Vapor Pressure:

Solutions Page 107

Lowering of vapor pressure, DPsolvent, is defined a

Remember that the sum of the mole fractions must equal 1.

Thus Xsolvent + Xsolute = 1, which we can substitute into our

expression.

Raoult’s Law – The equilibrium vapor pressure of the solvent over the solution is directly proportional to the mole fraction of the solvent in the solution

solutioninX

solutionin

X

solvent offraction mole

solvent pure of pressure vapor P

solvent of pressurevapor P where

PP

solvent

0

solvent

solvent

0

solventsolventsolvent

0solventsolvent

0solventsolvent

0solvent

solvent0solventsolvent

)P(

)P)((- P

PP P

X

X

1

Law sRaoult' is which

P P

- 1

0solventsolutesolvent

solventsolute

X

XX

Solutions Page 108

h h h w h w h ’

changed by the mole fraction of the solute, which is

Raoult’s law

Examples 4.9

The vapor pressure of water is 17.5 torr at 20°C. Imagine

holding the temperature constant while adding glucose,

C6H12O6, to the water so that the resulting solution has XH2O

= 0.80 and XGlu = 0.20.

What is, the vapor pressure of water over the solution

Try ?

1. Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a

density of 1.26 g/mL at 25°C. Calculate the vapor

pressure at 25°C of a solution made by adding 50.0

mL of glycerin to 500.0 mL of water. The vapor

pressure of pure water at 25°C is 23.8 torr

2. The vapor pressure of pure water at 110°C is 1070

torr. A solution of ethylene glycol and water has a

vapor pressure of 1.00 atm at 110°C. Assuming that

torrtorrXPXP AAA 145178000 ..

Solutions Page 109

Raoult's law is obeyed, what is the mole fraction of

ethylene glycol in the solution?

Addition of a nonvolatile solute to a solution raises the

boiling point of the solution above that of the pure solvent

lowered as described by Raoult’s law.

The amount that the temperature is elevated is

determined by the number of moles of solute dissolved

in the solution.

Boiling point elevation relationship is

Example 4.9

2. Boiling point elevation

solvent the for

constant elevation point boiling molal K

solution of ionconcentrat molal

elevation point boiling T :where

KT

b

b

bb

m

m

Solutions Page 110

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

C6H12O6, solution?

The addition of a nonvolatile solute lowers the vapor

pressure of the solution. At any given temperature, the

vapor pressure of the solution is lower than that of the pure

liquid

C101.28=C.+C100.0 = solution the of Point Boiling

C.T

).)(C/ .(T

K T

000

0b

0b

bb

281

281

5025120

mm

m

Solutions Page 111

The increase in boiling point relative to that of the pure

Δ b, is directly proportional to the number of

solute particles per mole of solvent molecules.

Molality expresses the number of moles of solute per

1000 g of solvent, which represents a fixed number of

moles of solvent

Example 4.10

Automotive antifreeze consists of ethylene glycol, C2H6O2,

a nonvolatile nonelectrolyte. Calculate the boiling point of a

25.0 mass percent solution of ethylene glycol in water.

Solution:

Boiling point = (normal b.p of solvent + ∆ T

Solutions Page 112

Addition of a nonvolatile solute to a solution lowers the

freezing point of the solution relative to the pure

solvent.

See previous table for a compilation of boiling point

and freezing point elevation constants.

Relationship for freezing point depression is:

Notice the similarity of the two relationships for freezing

point depression and boiling point elevation.

3. Freezing Point Depression

solvent for constant depression point freezing K

soltuion of ionconcentrat molal

solvent of depression point freezingT :where

KT

f

f

ff

m

m

Solutions Page 113

Fundamentally, freezing point depression and boiling

point elevation are the same phenomenon.

The only differences are the size of the effect

which is reflected in the sizes of the constants, Kf

& Kb.

This is easily seen on a phase diagram for a solution.

Example 4.11: Calculate the freezing point of a 2.50 m

aqueous glucose solution.

The size of the freezing point depression depends on

two things:

1. The size of the Kf for a given solvent, which are well

known.

mmbbff

K Tvs. KT

C4.65 - = C4.65 - C0.00=solution of Point Freezing

C65.4T

)50.2)(C/(1.86T

KT

000

0

f

0

f

ff

mm

m

Solutions Page 114

2. And the molal concentration of the solution which

depends on the number of moles of solute and the 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.

Example 4.12: A 37.0 g sample of a new covalent

compound, a nonelectrolyte, was dissolved in 2.00 x 102 g

of water. The resulting solution froze at -5.58oC. What is

the molecular weight of the compound?

g/mol .mol 0.600

g 37 is mass molar the Thus

compound mol .

kg 0.200 3.00=OH kg 0.200 in compound mol ?

water.of kg 0.200 mL 200

are there problem this In

.C1.86

C.

K

T

the thus KT

2

0f

f

ff

761

6000

003585 0

m

mm

m

Solutions Page 115

Osmosis is the net flow of a solvent between two

solutions separated by a semipermeable membrane.

The solvent passes from the lower concentration

solution into the higher concentration solution.

Examples of semipermeable membranes include:

cellophane and saran wrap

skin

cell membranes

Osmosis is a rate controlled phenomenon.

The solvent is passing from the dilute solution

into the concentrated solution at a faster rate than

in opposite direction, i.e. establishing an

equilibrium.

4. Osmotic Pressure

Solutions Page 116

The osmotic pressure is the pressure exerted by a

column of the solvent in an osmosis experiment.

For very dilute aqueous solutions, molarity and

molality are nearly equal.

M m

Osmotic pressures can be very large.

For example, a 1 M sugar solution has an

osmotic pressure of 22.4 atm or 330 p.s.i.

Since this is a large effect, the osmotic pressure

measurements can be used to determine the

molar masses of very large molecules such as:

Polymers

Biomolecules like

proteins

Ribonucleotides

etemperatur absolute = T

K mol

atm L0.0821 = R

V

n solution of ionconcentrat molar =

atm in pressureosmotic = :where

RT

M

M

nlyolutions o aqueous sfor dilute

mRT

Solutions Page 117

Osmosis plays an important role in plant and animal

physiological processes; the passage of substances

through the semipermeable walls of a living cell, the

action of the kidneys, and the rising of sap in trees are

examples.

Application of osmosis

1) Revesse osmosis: When an external pressure is

applied over the solution, the solvent is forced in a

direction contrary to that normally observed. This

process called reverse osmosis is used to secure pure

water from salt water. This is used in desalination of

seawater to be suitable for drinking.

2) Isotonic solution: In the living cells, the osmotic

pressure of solution is equal to the osmotic pressure of

the cell.

e.g: NaCI (0.9%) has the same osmotic pressure as

blood.

3) Hypertonic solution: A solution of higher osmotic

pressure. In this solution red blood cells shrink. The

cells are called plasmolysed.

Solutions Page 118

4) Hypotonic solution: A solution of lower osmotic

pressure. In this solution red blood cells swells up and

burst. The cell is said to be haemolysed

Example 13: A 1.00 g sample of a biological material was

dissolved in enough water to give 1.00 x 102 mL of solution.

The osmotic pressure of the solution was 2.80 torr at 25oC.

Calculate the molarity and approximate molecular weight of

the material.

Try ?

1. What would be the freezing point and boiling point of a

solution containing 6.50 g of ethylene glycol (C2H6O2) in

200.0 g of H2O? KfH2O = 1.86°C /m, kbH2O = 0.512°C /m

(Answer = 0.977 and 100.267oC)

proteins small of typical

..

L

L 0.100

g 1.00

mol

g ?

.K 0.0821

atm . =

= atm .torr 760

atm 1torr 2.80 = atm?

RT RT

molg

K molatm L

4

4

4

1067610501

1

10501298

003680

003680

M

MM

MM

Solutions Page 119

2. What are the boiling point and freezing point of a

solution prepared 2.40 g of biphenyl (C12H10) in 75.0 g of

benzene? b.p of benzene = 80.1°C f.p. of benzene =

5.5°C. (Answer = 80.626 and 4.4oC)

3. A solution prepared by dissolving 0.30 g of an unknown

nonvolatile solute in 30.0 g of CCI4 has a boiling point that is

0.392°C higher than that of pure CCI4. What is the molecular

weight of the solute? Kb = 5.02°C/m.

4. Find the osmotic pressure of blood at normal body

temperature (37°C) if blood behaves as if it were a

0.296 M solution of a nonionizing solute.

5. : An aqueous solution contains 30.0 g of a protein in l.0

L. The osmotic pressure of the solution is 0.0167 atm at

25°C. What is the approximate molecular weight of the

protein?