Chemistry

Learning objectives

At the end of this unit , students will be able to:

• Define partition coefficient and solvent extraction;

• State the Raoult’s law;

• State and explain the advantages of carrying out distillation processes under reduced pressures;#

• Discuss the chemical principles upon which simple distillation and steam dis-tillation are based;

• State examples of the applications of the distillation methods used in various industries;

• Describe the effect of the solute on vapor pressure, boiling and freezing points of the solvent;

• Explain colligative processes;

• Carry out separation experiments based on the solute partitioning between two immiscible solvents;

• Calculate the amount of the solute extracted from the solvent;

• Interpret boiling point and vapor pressure composition curves of both ideal and non-ideal mixture;

• Calculate the molecular mass of substances using steam distillation;

• Calculate molecular mass of polymers using colligative properties and steam distillation;

• Interpret the boiling point composition curves of azeotropic mixtures;

• Apply Raoult’s Lawto calculate vapor pressure of given solutions and mole fractions;

• Carry out experiments to explain colligative properties;

• Apply colligative properties to determine the molecular mass of the solute in a solution.

Introductory activity

1. Describe the phase diagram for water.

2. Have you ever heard the term “solvent extraction”? If yes can you tell what it is about and give an example of a solvent  extraction?

3. Describe the process of producing banana juice? Is there any application of solvent extraction?

4. Take a tea bag and put it in water (cold or hot); what do you observe? Explain your observation

5. You are provided with the following materials: separating funnel, 1,1,1-trichloroethane, iodine, water, retort stand and its accessories:

Procedure:

Put water into the separating funnel and add 1,1,1-trichloroethane in the water and shake well.

Add iodine in the mixture and shake well, after standing, observe well.

What happens to the iodine? How do you compare the coloration of the two phases? What explanation can you give?

When traditional healers crush leaves of a healing plant, mix with water, filter and give you the filtrated mixture as medicine, they have applied solvent extraction technique, using water as solvent, to extract the medicinal ingredients.

When you prepare your cup of tea, putting a tea bag in the hot water, the color of water change to brown black; you have solvent extracted a certain number of substances present in the tea leaves by hot water, and left other in the tea residue.

Solvent extraction is the separation of a particular substance from a mixture by dissolving that substance in a solvent that will dissolve it, but which will not dissolve any other substance in the mixture.

Let us start with two solvent systems of water and an organic liquid, two immiscible solvents.

When water and an organic liquid are shaken together, and then allowed to stand, the liquids separate into a two phase system with the more dense liquid on the bottom.

If a solute is added to a system of two liquid layers, made up of two immiscible components, then the solute will distribute itself between the two layers so that the ratio of the concentration in one solvent to the concentration in the second solvent remains constant at constant temperature. A quantitative measure of how a component will distribute between the two phases is called the distribution or partition coefficient.

7.1. Definition of partition coefficient and solvent extraction

The ability of a solute (inorganic or organic) to distribute itself between an aqueous solution and an immiscible organic solvent has long been applied to separation and purification of solutes either by extraction into the organic phase, leaving undesirable substances in the aqueous phase; or by extraction of the undesirable substances into the organic phase, leaving the desirable solute in the aqueous phase. Species that prefer the organic phase (e.g., most organic compounds) are said to be lipophilic (“liking fat”) or hydrophobic (“disliking water”), while the species that prefer water are said to be hydrophilic (“ liking water”) or lipophobic (“disliking fat”).

7.1.1. Partition coefficient

Partition means”divided into two parts with a boundary or an interface”.When a solute is added to two immiscible solvents in contact with each other at constant temperature, the solute gets distributed or partitioned between the two solvents with different equilibrium concentrations.

For example, when iodine is added to water and carbon tetrachloride, it distributes in such a way that the equilibrium ratio of the concentrations of iodine in the two solvents is constant at any given temperature. If and are the concentrations of iodine in water and carbon tetrachloride respectively, then

The constant is called the distribution or partition coefficient of the solute between the two solvents at a given temperature. The value of depends on the nature of the solute and the solvent pair. The equation above is the mathematical form of the Nernst Distribution Law (or Nernst’s Partition law) or simply Distribution law or Partition law.

This law states that at constant temperature, when different quantities of a solute are allowed to distribute between two immiscible solvents in contact with each other then at equilibrium the ratio of the concentration of the solute in two

layers is constant. The solutions where this law applies are called “ideal solutions”. However, if the amount of the solute added is sufficiently small, then the distribution coefficient is relatively independent of the concentration.

7.1.2. Solvent extraction

The term solvent extraction refers to the distribution of a solute between two immiscible liquid phases in contact with each other, i.e., a two-phase distribution of solute.

Solvent extraction has become a very powerful method of separation for various reasons. One amongst them is, it is very simple, rapid, selective and sensitive. This method does no need any kind of sophisticated instrument apart from a separating funnel.

It is the partial removal of a solute from one liquid (usually water) into another immiscible liquid (e.g. ether). This is a method of separating one component from a mixture. It is based on finding a solvent which dissolves the desired component much better than it does for any of the others. Solvent extraction is also called liquid-liquid extraction and partitioning.

For the solvent extraction to succeed, the following considerations are recommended:

• the solvent to be used must be a good solvent for the solute and chemically unreactive with the solute

• the solvent must be immiscible with water or any other solvent involved

• the solvent must be volatile, to facilitate the separation with the solute at the end of the operation; example, the most used solvent is ether that fulfills those requirements,

Activity 7.1:

In an experiment, 100 of 0.10mold aqueous propanoic acid was shaken with 50 of organic solvent which is immiscible with water, until equilibrium was reached. 10.0 of aqueous layer required 12.0 of 0.02mold NaOH for reaction in a titration.

Questions:

   a. Calculate the concentration of propanoic acid in aqueous layer and in organic layer.

   b. Calculate the equilibrium ratio between the concentration of propanoic acid in organic layer and in aqueous layer;

  c. What is the name of that ratio?

Multiple extractions

The process of extraction when carried out with the total amount of the given solvent in a single operation is referred to as simple extraction. But to recover the maximum amount of the substance from a solution, the extraction is made in two or more successive operations using small portions of the solvent provided. This is called multiple extractions. In such a process the aqueous solution is first extracted with a portion of the solvent in a separatory funnel. The aqueous layer from which some substance has been removed is then transferred to another funnel. This is shaken with a second portion of the solvent. Similarly, the aqueous layer from the second extraction is treated with a third portion of solvent, and so on.

Example:

The partition coefficient of a substance X between methylbenzene and water is 8.0, X being more soluble in methylbenzene than water.

The mass of X extracted from an aqueous solution containing 6.0 g of X may be calculated as follows:

If shaken with one portion of 100cm3 of methylbenzene:

Let x be the mass of X extracted.

Mass of X remaining in aqueous solution= (6.0-x)g

If shaken with two successive portions of 50 cm3 of methylbenzene:

Mass of X extracted in (i) is less than that in (ii). This shows that it is more efficient to use a solvent in several smaller portions than all at once when extracting a solute from another solvent.

Limitation of Nernst Distribution Law

The law is valid only when the molecular state of the solute is the same in both the solvents. If the solute undergoes dissociation or association in any one of the solvents, then in such cases the distribution law no longer holds.

7.2. Raoult’s law and colligative properties

7.2.1 Raoult’s law and ideal solutions

  

Raoult’s law describes the behavior of ideal solutions of completely miscible and volatile liquids. Ideal solution is defined as the one in which each component obeys Raoult’s law over the entire concentration range. Besides obeying Raoult’s law, an ideal

solution shows two more properties:

       a. When such solution is prepared, no heat is evolved or absorbed, i.e.

          b. The volume of such a solution is the sum of the volumes of its components,

In fact, liquid-pairs rarely form ideal solutions. Benzene-toluene, ethene chloride-ethene bromide and carbon tetrachloride-silicon tetrachloride are few examples of liquid pairs which form very nearly ideal solutions.

A binary solution of components A and B will behave ideally when intermolecular force between a molecule of A and a molecule of B (A-B interactions) is same as intermolecular force between two molecules of A (A-A interactions) or the force acting between two molecules of B (B-B interactions). This means that one component likes itself as much as it likes the other component. This is due to similarity in their structure that their solutions behave ideally.

At any fixed temperature, the vapour phase is always richer in the more volatile component compared to the solution phase. In other words, mole fraction of the more volatile component is always greater in the vapour phase than in the solution phase. The composition of vapour phase in equilibrium with the solution is determined by the partial pressure of components.

i. Calculate the vapour pressure of the solution prepared by mixing 25.5 g of chloroform and 40g of dichloromethane at 25°C.

ii. Calculate the mole fraction of each component in vapour phase.

Answer:

  i. The number of moles in the solution:

Total number of moles in the solution

Mole fractions in the solution:

The vapour pressure of the solution is 347.9 mm Hg.

                   ii. The vapour pressure of each component in vapour phase:

Consider and the mole fractions of components A and B respectively in the vapour phase

The mole fraction in vapour phase is given by the following equation:

Since dichloromethane is more volatile than chloroform and the vapour phase is also richer in dichloromethane, it may be concluded that at equilibrium, vapour pressure will be always rich in the component which is more volatile.

Another type of solutions is solids in liquid solution, in which we take the solid as the solute and the liquid as the solvent. Generally, the solute is nonvolatile in nature and the vapour pressure is less than the pure vapour pressure of the solution.

Examples:

               • Solution of sugar, salt or glucose and water

               • Solution of iodine and sulfur in carbon disulphide

Vapour pressure produced by the solution of nonvolatile solute and solvent results in the vapour pressure of the solution solely from the solvent. This vapour pressure is lower than the vapour pressure of the pure solvent at the same temperature.

The decrease in vapour pressure is due to:

The surface area of the solution is occupied by both nonvolatile solute and the pure solvent particles which results in the reduction of the surface for the solvent particles. As evaporation is a surface phenomenon, the more the surface area is, the greater the evaporation and hence more the vapour pressure. In a pure solvent, there is more surface area available for the particles to vaporize, therefore have more vapour pressure. On the other hand, when we add a nonvolatile solute, the solvent particles get less surface area to escape and hence exert low vapour pressure.

The number of particles escaping the surface is much greater in pure solvent than that of solution containing nonvolatile solute.

Calculation of the vapour pressure of Solutions of Solids in Liquids

Let’s A be the solvent and B a solute; according to Raoult’s law, the partial pressure of individual component is directly proportional to its mole fraction.

Example:

An aqueous solution of glucose is made by dissolving 10 g of glucose, (C6H12O6), in 90 g of water at 30oC. If the vapour pressure of pure water at 30oC is 32.8 mmHg, what would be the vapour pressure of the solution?

Data:

Mass of glucose: 10g

Molar mass of glucose= 180g/mol

Mass of water: 90g

Molar mass of water: 18g/mol

Vapour pressure of pure water: 32.8 mmHg

Calculation of moles of water and glucose211 Mass of water:

Note: The vapour pressure of nonvolatile component is negligeable.

Non-ideal Solutions

Activity 7.2 (b):

You are given the following solutions:

Chloroform-acetone, water-propanol, n-hexane-n-heptane, ethanol-chloroform, ethanol-cyclohexane, ethanol-acetone, phenol-aniline, benzene-toluene.

Choose the solutions which obey Raoult’s law and explain why others do not obey it.

Most of liquid-liquid solutions do not obey Raoult’s law over the entire range of concentrations. Such solutions are called non-ideal solutions. Besides Raoult’s law, non- ideal solutions do not obey the other two conditions: their formation is accompanied by changes of heat and volume. When non-ideal solutions are prepared, some energy is either released or absorbed and the volume of the solution is different from the sum of volumes of the components. The non-ideal behavior of most of liquid-liquid solutions is due to the intermolecular forces between molecules. Vapour pressure of non-ideal solutions is either higher or lower than what predicted by Raoult’s law. Based upon this, there are two types of non-ideal solutions namely, the non-ideal solutions with positive deviations and those with negative deviations.

Most of the real mixtures are non- ideal.

Non-ideal solutions with negative deviations

In some liquid solutions, the observed vapour pressure of the solution is lower than the ideal value as expected from Raoult’s law. Such deviation is called negative deviation.

Examples:

i. Acetic acid-pyridine

ii. Phenol-aniline

iii. chloroform-acetone

iv. water-nitric acid

v. water- sulphuric acid

vi. water-hydrochloric acid

In a mixture of phenol and aniline, the intermolecular hydrogen bond between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bond between similar molecules; the molecules in the mixture are more strongly bonded than those in the pure liquids. Similarly, propanone-trichloromethane mixture forms a solution with negative deviation from Raoult’s law. When the liquids are mixed, hydrogen bonds are formed between the propanone and trichloromethane molecules (figure 7.7). There is no hydrogen bonding in pure liquid propanone or in pure liquid trichloromethane.

Non-ideal solutions with positive deviations

The solutions whose vapour pressure is higher than that expected from Raoult’s law are said to show positive deviations.

Examples of non-ideal solutions showing positive deviation from Raoult’s law:

i. Benzene-cyclohexane

ii. n-butane-n-heptane

iii. carbon disulphide-acetone

iv. carbon tetrachloride-benzene

v. ethanol-acetone

vi. ethanol-hexane

vii. ethanol-chloroform

viii. water-ethanol

ix. water-propanol

x. ethanol-cyclohexane

Example:

When hexane and ethanol are mixed, the ethanol molecules are separated. The hydrogen bonds are broken and much weaker Van der Waals forces hold the ethanol and hexane molecules together in the mixture. The ease of escape into the vapour is caused by the hexane molecules, which cannot form hydrogen bonds, coming between the ethanol molecules and disrupting the hydrogen bonding, making it easier for ethanol molecules to escape into the vapour.

Azeotropic mixtures

All non-ideal solutions with large deviations show either maxima or minima in their vapour pressure compositions curves. The solutions with positive deviations show maxima and those with negative deviations show minima.

The solution corresponding to maximum or minimum has a unique property that the vapour formed on their evaporation has the same composition as the the liquid phase of the solution. Such solution boils at constant temperature and can not be separated by fractional distillation. Such a solution is called Azeotropic mixtures or azeotropes. The azeotrope is defined as the mixture of liquids which boils at constant temperature like a pure liquid and possesses the same composition of components in the solution as well as in vapour phase.

The term azeotrope stems from a Greek word meaning boiling without changing. Each azeotrope has a characteristic boiling point.

Azeotropes formed by the solutions showing positive deviations from Raoult’s law are called minimum boiling azeotropes or positive azeotropes; such an azeotrope has the maximum vapour pressure and, therefore the minimum boiling point among all the mixtures formed by these two liquids. These azeotropes always have boiling point lower than either of the components.

Examples: Ethanol-water mixture

The fractional distillation is able to concentrate the alcohol to the best up to 95% by mass of alcohol. Once this composition is reached, the solution and vapour have the same composition, and no additional fractionation occurs. Other methods of separation have to be used for preparation of 100% ethanol.

Azeotropes formed by the solutions showing negative deviations from Raoult’s law are called the maximum boiling azeotropes or negative azeotropes. These azeotropes have boiling points higher than either of the components.

Azeotropes consisting of two constituents are called binary azeotropes. Those consisting of three constituents are called ternary azeotropes. Azeotropes of more than three constituents are also known.

Examples of azeotropes with minimum boiling points and maximum boiling points:

Checking up 7.2 (b)

1. On the basis of information given below answer the following questions: Information:

a. In bromoethane and chloroethane mixture intermolecular interactions of A-A and B-B type are nearly the same as A-B type interactions.

b. In ethanol and acetone mixture A-A or B-B type intermolecular interactions are stronger than A-B type interactions.

c. In chloroform and acetone mixture A-A or B-B type intermolecular interactions are weaker than A-B type interactions.

Questions:

1. a. Which solution will obey the Raoult’s law?

    b. Which solution will deviate from Raoult’s Law and which kind of deviation?

2. On the basis of information given below, mark the correct option. Information: on adding acetone to methanol some of the hydrogen bonds between methanol molecules break.

a. At specific composition methanol-acetone mixture forms minimum boiling azeotrope and will show positive deviation from Raoult’s law.

b. At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult’s law.

c. At specific composition methanol-acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoult’s law.

Application of Raoult’s Law

Activity 7.2(c):

Use search internet or other resources and then discuss about Raoult’s law and its applications then write a report about your findings.

Questions:

1. Have you carried out a distillation? If yes, what the principle that underlies that separation technique?

2. Have you heard about the production of a local and illegal alcohol called Kanyanga? If yes, do you have an idea of how it is made?

3. Distinguish simple distillation from fractional distillation.

Fractional distillation

Distillation is a separation technique that separates liquid components of a solution according their boiling points. A simple distillation uses a simple condenser; it works well for most routine separation and purification procedures for organic liquids with large differences in their boiling points. When the boiling point differences of the components to be separated are not large, fractional distillation must be used. When an ideal solution of two liquids, such as benzene (boiling point 80°C) and toluene (boiling point 110°C), is distilled by simple distillation, the first vapor produced will be enriched in the lower-boiling component (benzene).

However, when that initial vapour is condensed and analyzed, the distillate will not be pure benzene. The boiling point difference of benzene and toluene (30°C) is too small to achieve a complete separation by simple distillation. So, fractional distillation is an application of Raoult’s law illustrated in figure 7.10.

Industrial application of fractional distillation: crude oil refinery.

Crude oil is a complex mixture of hydrocarbons of different chain lengths, from the short one such as , methane, to the very long ones with more than 20 carbon atoms (C-20). Due to their boiling points which are not very different, there are separated, through fractional distillation, into different fractions of hydrocarbons called: kerosene, naphtha, etc... (Fig.7.12) 

In short: Condensation takes place in discs which increases surface area found in fractionating column. Further heating of the mixture produces more vapour, which in turns will reheat the condensed liquid in discs. During the distillation, the temperatures decrease progressively from the bottom to top of column.

Steam distillation

Activity 7.2 (d)

When people suffer from flue, cold or any other respiratory infections, they collect Cyprus, eucalyptus, and other leaves. They boil them in water and expose themselves to the vapour released. Explain how they are cured.

Distillation using water vapor is called steam distillation. Steam distillation can be used when the material to be distilled has a high boiling point and presents the risk that decomposition might occur if direct distillation is employed. The liquid is added to the still, and steam is passed through it. The solubility of the steam in the liquid must be very low. Most essential oils have been obtained by steam distillation, or, in the more general sense, by hydrodistillation. Essential, volatile, or ethereal oils are mixtures composed of a variety of volatile, liquid, or solid compounds that vary widely in concentration and boiling points. They are present in the interstices of vegetable tissues and can be extracted by hydrodistillation.

Considering the manner in which the contact between water and the original matrix is promoted, a terminology that distinguishes three types of hydrodistillation has been developed:

     • Water distillation,

     • Steam distillation, and

     • Direct steam distillation.

When the first method is employed, the material to be distilled comes in direct contact with boiling water.

In the second method, the material is supported on a perforated grid or screen held some distance above the bottom of the still. In this case, low-pressure, saturated, wet steam rises through the material. The typical features of this method are that the steam is always fully saturated, wet, and never superheated, and the material is only in contact with steam, and not with boiling water. The last type of hydrodistillation, direct steam distillation, resembles the preceding type, except that no water is kept in the bottom of the still, but live steam, saturated or superheated, is passed through the sample, and the process is frequently maintained at higher than atmospheric pressures.

Steam distillation is the most commonly used method for collecting essential oils. Use of this method is prevalent not only because it yields exceptionally pure and clean products, but because it allows for collection of temperature-sensitive aromatic compounds. Unlike simple distillation, steam distillation involves a pressurized system. When the system is pressurized, essential oils can be distilled at temperatures well below their normal boiling point; thus protecting the integrity of their delicate and complex chemical nature.

When using steam distillation, it is vital to pay careful attention to the heat source. The temperature of the system must remain within a strict range; too low and the essential oils will not be distilled, too high and there is risk of damaging the essential oils or collecting unwanted, non-aromatic compounds. Temperatures required for optimal steam distillation typically fall between 60° C and 100° C. One benefit of steam distillation is that the temperature can be continuously adjusted and precisely controlled to ensure that the system always remains within the optimal temperature range. Similarly, pressure must also be rigidly controlled. To do so, steam distillation requires the use of a “closed system.” This means that the system is pressurized to a level above that of atmospheric pressure.

Steam distillation is conducted in a distillation still that uses steam water to remove the essential oil from the plant material. Heat is applied to the water, which produces steam.

The steam rises and moves through a chamber holding the plant material. As the steam forces its way through the plant material, it ruptures small glands that hold the essential oil. Since essential oils contain only aromatic compounds that are readily volatile, the essential oil is easily carried by the steam into the condensing tube. Once there, the liquid is condensed and accumulates in the collecting still. After removal from the still, a mixture of essential oil and water will be formed. Because essential oils are not water soluble, the mixture will naturally separate into two layers (figure 7.14). The watery layer is called hydrosol and is often sold as floral water.

Applications

Steam distillation is widely used in the manufacturing of essential oils, for instance perfumes. This method uses a plant material that contains of essential oils. Mainly orange oil is extracted on a large scale in industries using this method. Application of steam distillation can be found in the production of consumer food products and petroleum industries. They are used in separation of fatty acids from mixtures.

Checking up 7.2 (d)

Explain how the essential oils can be obtained from plants?

7.2.2. Colligative properties

Activity 7.3:1. The photos you see here have been taken in Europe or North America during winter.

On the left, someone is pouring a liquid, an antifreeze in the radiator of his/her car.

On the right, workers are spreading salt from a truck on a road covered by ice.

Question: Do you have an idea why those persons are doing that?

2. Take a container and divide it into two compartments with a thin membrane containing microscopic pores large enough to allow water molecules but not solute particles to pass through.Then add a concentrated salt solution to one compartment and a more dilute salt solution to the other. Initially, the two solutions levels start out the same.

a. What is the name of the membrane?

b. Observe well and after a while, write your observations.

c. Name the process observed and explain it and explain how it can be stopped.

d. Draw with labels the phenomenon.

3. Have you ever heard these words “colligative properties”? If yes explain what they mean and give example to illustrate.

The state of water depends on both pressure and temperature. At sea level, pure liquid water freezes to form solid ice at 0°C and boils to form vapor at 100°C.

However, the addition of particles to water can affect its boiling and freezing points. Particles interfere with the ability of the water molecules to vaporize or freeze. The boiling temperature of water varies depending on the solutes it contains.

Liquid solutions have physical properties significantly different from those of the pure solvent, a fact that has great practical importance. The properties of dilute solutions containing nonvolatile solute, which depend upon relative number of solute and solvent particles but no depending upon their nature, are called colligative properties.

Definition of colligative properties of solutions

Colligative properties are properties that depend only on the number of particles present in solution and not in any way on the nature of the solute particles. These properties are bound together by a common origin; they all depend on the number of solute particles present, regardless of whether they are atoms, ions, or molecules.

The colligative properties are Vapour pressure lowering, Freezing point depression, Boiling point elevation and Osmotic pressure.

Lowering of vapour pressure

Molecules can escape from the surface of a liquid into the gas phase by evaporation. In closed container, the pressure exerted by the vapour in the space above the liquid, if the dynamic equilibrium between liquid and vapour is reached, is called the vapour pressure of the substance. The vapour pressure of a liquid is the pressure exerted by its vapour when the liquid and vapour phase are in dynamic equilibrium. A substance with no quantifiable vapour pressure is called nonvolatile and the one that exerts vapour pressure is called volatile.

According to Raoult’s law, the vapour pressure exerted by a component of a solution at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in pure state.

In a solution of nonvolatile solute in a volatile solvent the vapour pressure will be exerted by the solvent molecules only because the solute is non volatile i.e., does not go into vapour state. Therefore, the vapour pressure of its solution is always less than that of the pure solvent.

In the pure solvent, the molecules occupy large surface and they escape easily; in the solution of nonvolatile solute and solvent, the solute molecules reduce the surface of the solvent molecules which lead to the reduction of molecules of solvent escaping. According to Raoult’s law:

Equation (1) can therefore be rewritten as:

The decrease in vapour pressure,P, is directly proportional to the solute concentration (measured in mole fraction).

In the case of solution of two or more volatile components, the total vapour pressure over the solution is the sum of the partial vapour pressures of each volatile component. Consider the ideal solution containing two volatile liquid, 1 and 2.

By Raoult’s law, the partial pressures of component 1 and 2 vapours above the solution are:

Example:

Calculate the vapour pressure of a solution made by dissolving 218 g of glucose (molar mass=180.2g/mol) in 460 mL of water at 30°C. What is the vapour pressure lowering? The vapour pressure of pure water at 30°C is 31.82 mm Hg. Assume the density of the solution is 1.00 g/mL.

Answer:

The vapour pressure of the solution:

Boiling point elevation

Boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure. As the vapour pressure of a solution containing a nonvolatile solute is lower than that of the pure solvent, its boiling point is higher than that of the pure solvent, because to reach atmospheric pressure at which the solution should boil we need to raise the temperature. The solution has lower vapour pressure than the solvent at the same temperature, figure 7.15. The increase in boiling point is known as elevation in boiling point,

Thus, the molal freezing point depression constant is equal to the depression in the freezing point produced when one mole of solute is dissolved in 1 kg of the solvent.

Example:

Ethylene glycol is common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p 197°C). Calculate the freezing point and boiling point elevation of a solution containing 651 g of this substance in 2505 g of water. The molal mass of ethylene glycol (EG) is

Answer:

Calculate the molality of ethylene glycol (m):

Osmotic pressure

Osmosisis the phenomenon of spontaneous flow of the solvent molecules through a semipermeable membrane from pure solvent to solution or from a dilute solution to concentrated solution. It was first observed by Abbe Nollet. Some natural semipermeable membranes are animal bladder, cell membrane.

You must have observed that if resins are soaked in water for some time, they swell. This is due to the flow of water into the resins through its skin which acts as a semipermeable membrane (permeable only to the solvent molecules). This phenomenon is also observed when two solutions of different concentrations in the same solvent are separated by a semipermeable membrane. In this case the solvent flows from a solution of lower concentration to a solution of higher concentration. The process continues till the concentrations of the solutions on both sides of the membrane become equal. The spontaneous flow of the solvent from a solution of lower concentration (or pure solvent) to a solution of higher concentration when the two are separated by a semipermeable membrane is known as osmosis. The pressure required to prevent osmosis by pure solvent is called osmotic pressure of solution, π.

Thus, osmotic pressure may be defined as the excess pressure that must be applied to the solution side to just prevent the passage of pure solvent into it when the two are separated by a perfect semipermeable membrane.

This is illustrated in figure 7.16. For dilute solutions, it has been found that osmotic pressure is proportional to the molarity, C of the solution at a given temperature, T.

It two solutions of the same osmotic pressure are separated by a semipermeable membrane, there is no osmosis. The two solutions are called isotonic solution. Hypotonic solution is the less concentrated solution and hypertonic solution the more concentrated one.

The osmotic pressure is a colligative property. It depends on the number of particles of solute present in the solution and not on their nature.

Applications of colligative properties

As seen in the introductory activity (7.4), one of the colligative properties, the freezing point depression is applied to de-ice roads in countries which experience very cold winters.

Colligative properties of non electrolyte solutions provide a means of determination of molar masses and molecular formulae of dissolved solutes which are nonvolatile. Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements. In practice, only freezing point depression and osmotic pressure are used because they show the most pronounced changes.

       • Determination of molar mass from relative lowering of vapour pressure

Relative lowering of vapour pressure is the ratio of lowering in vapour pressure to vapour pressure of pure solvent. The relative lowering in vapour pressure of solution containing a nonvolatile solute is equal to the mole fraction of solute in the solution.

The above expression is used to determine the molecular mass of the solute B, provided the relative lowering of vapour pressure of a solution of known concentration and molecular mass of solvent are known. Where, mass of Solute and solvent respectively. molecular weight of solute and solvent respectively. However, the determination of molecular mass by this method is often difficult because the accurate determination of lowering of vapour pressure is difficult. Ostwald and Walker method is used to determine the relative lowering of vapour pressure.

Examples

1. The relative lowering of vapour pressure produced by dissolving 7.2 g of a substance in 100g of water is 0.00715. What is the molecular mass of the substance?

Substituting the values we get

Molecular mass of the substance is 181.26 amu

2. The vapour pressure of 5% aqueous solution of a nonvolatile organic substance at 373 K is 745 mm Hg. Calculate the molar mass of the substance. Solution

                       • Determination of molar mass from elevation of boiling point

The molar mass of a non electrolyte can be calculated from the elevation of boiling point:

where, is the mass of solute ( in grams), and is the molar mass of the solute (in g/mol) and, W1 is the mass of the solvent in Kg units.

This gives

The constant is called the molal elevation constant or ebullioscopic constant for the solvent. may be defined as the elevation in boiling point when one mole of a solute is dissolved in one kilogram of the solvent. BKis expressed in degree per molality.

Examples

1. A solution containing 4.2 grams of an organic compound in 50 grams of acetone shows an elevation of boiling point by 1.8 K. Determine the molar mass of the organic compound. of acetone=1.71 K kg /mol.Solution

Therefore, molar mass of the solute is 58g/mol.

                                        

where, is the mass of solute ( in grams), and is the molar mass of the solute (in g/mol) and, W1 is the mass of the solvent in Kg units.