Science 3

What you will doActivity 1.1 Types of mixturesMaterials: sand oil sugar three glasses of water flashlight filter paper (if not available use bond paper)Procedure:1. Mix the following: a. sand and water b. sugar and water c. oil and water2. Stir the contents of each mixture. Do the contents settle at the bottom?3. Observe the three mixtures for homogeneity or heterogeneity. Are the components evenly distributed to one another in all proportions? If yes, it is homogenous. If no, it is heterogenous.4. Allow the flashlight to pass through the contents of each mixture. Do the contents scatter light?5. Fold your bond paper into two parts crosswise and cut it. Make a cone out of your cut- out bond paper. This will serve as improvised filter paper. Let the three mixtures pass through the improvised filter paper. Can the contents of each mixture be filtered?Analysis:1. Tabulate your data as follows: Behavior of Particles Sugar + Oil + Sand + Water Water Water Settle at the bottom Scatter light Can be filtered Homogeneity/ Heterogeneity2. Based from the behavior of the particles, identify the type of mixture in each set-up. State whether the mixture is a solution, colloid or suspension. A colloid is a dispersion of particles of one substance (the dispersed phase)throughout a dispersing medium made of another substance. Thus colloids are classifiedbased on the phases of both the dispersed material and the dispersing medium. Forexample, when you beat an egg, you introduce air into the egg. Here, the egg is thedispersing medium and the air is the dispersed material. In Activity 1.1, which mixture is acolloid? -4-

Table 1.1 below shows the different types of colloid and examples of each type: Table 1.1 Different Types of Colloid and ExamplesDispersed Dispersing Name Example Material MediumLiquid Gas Liquid aerosol FogSolid Gas Solid aerosol SmokeGas Liquid Liquid foam MeringueGas Solid Solid foam StyrofoamSolid Liquid Liquid sol Paint, glueSolid Solid Solid sol Certain alloys such as steelLiquid Solid Gel GelatinLiquid Liquid Liquid emulsion MayonnaiseLiquid Solid Solid emulsion Cheese Let us test how well you understand classifying colloids. Can you classify what typeof colloid each of these substances are? 1. cheese 2. marshmallow 3. ink 4. whipped cream 5. cork 6. mist Did you know? Milk is a colloid. The solid particles in milk are evenlyspread throughout a liquid. Vinegar causes the small dissolvedparticles in milk to clump together, making a solid called curd.The liquid part is called whey. Still remember the nurseryrhyme \"Little Miss Muffet?\" This would be a great time toreview it. -5-

What you will doSelf-Test 1.1Now that you are through with the first lesson, try to answer the following and see foryourself how much you learned.Matching Type. Match the items in column A with the type of colloid in column B. Note:Items in Column B may be used twice. Column A Column B_____ 1. clouds a. liquid emulsion_____ 2. soap in water b. gel_____ 3. jellies c. liquid sol_____ 4. soap suds d. solid sol_____ 5. plastics e. solid foam_____ 6. milk f. liquid foam_____ 7. salad dressing g. liquid aerosol_____ 8. butter h. solid aerosol_____ 9. gemstones i. solid emulsion_____ 10. dust in air Did you encounter any problem? Well, compare your answers with the answer keyand see for yourself the items you missed. Good luck! Key to answers on page 17.Lesson 2. What are the Properties of a Colloid? Colloids are a special type of mixture because they exhibit unique properties. That iswhy many kinds of food products and raw materials are in the form of colloids. Theseunique properties of colloids include the Tyndall effect, Brownian motion, adsorption andelectrical charge effect.Tyndall Effect When a beam of light is made to pass through a colloid, it is scattered by the colloidalparticles which appear as tiny specks of light. This light scattering is called the Tyndalleffect. No such scattering is observed with ordinary solutions because the solute moleculesare too small to interact with visible light. Colloidal particles can scatter light because theyare large enough to catch light and reflect it back. This is the reason why oil and watermixture in Activity 1.1 has the ability to scatter light because it is a colloid. The ability of amixture to scatter light is one way of distinguishing colloids from solutions. Manifestations of -6-

Tyndall effect are observed everyday in nature. For instance, when we wake up on a sunnymorning, our senses are awakened by sunlight scattered by dust or smoke in the air. Did you know? Have you ever wondered why the sea and sky are blue? It is because of the scattering of the blue or shorter wavelength of light by the thick layer of small particles in the atmosphere and in the deep waters. The brilliant colors of sunset are also due to light scattering by colloidal water droplets in the atmosphere. Light scattering is often used in determining particle size. The different colors that wesee are not due to pigments but rather to the scattering of light by colloidal substances inthe iris. Green, brown, and black are due to a combination of light scattering and thepresence of yellowish-brown pigment in front of the iris, causing selective absorption of light.Brownian Movement Observe the movement of dust particles floating in air one sunny day. Could youdescribe their movement? Do they settle on standing? The motion of particles floating in air is random and almost in a zigzag fashion. Thisinteresting property of colloids, called Brownian motion, is due to the constant andcontinuous collision of colloidal particles against each other. This is also the reason whycolloidal particles do not settle on standing. The rate of settling of particles is dependent onthe following: a. size of the colloidal particles b. gravitational force acting on the colloidal particles c. viscosity of the medium (ability to resist flow) Water and gasoline mixture has low viscosity as compared to syrup and oil mixture which has high viscosity. The above factors are also useful in identifying viruses, proteins, plastics and othermacromolecules.Adsorption How is adsorption different from absorption? If you place a few drops of water onto acotton ball, the water droplets are immediately soaked up by the cotton. This is absorption. On the other hand, adsorption is when you use the cotton ball against the -7-

chalkboard. Observe closely what happens to the cotton. Chalk particles and dust adheredto the surface. This phenomenon is called adsorption. Physical and/or chemical forcesmay be involved in adsorption. One property of colloidal particles is that they exhibitadsorption. This is due to the large surface areas of colloidal particles. This interestingproperty makes colloids very useful in everyday life. For example, charcoal is used toremove the bad odor produced by vapors of food in the refrigerator. The network of pores inthe charcoal provides extensive surface area that adsorbs the vapors. The adsorptive ability of colloids is used in dyeing fabrics, in the use of aluminumhydroxide in purifying water, in the use of activated carbon in refining sugar andelectroplating solutions, and in the use of bone black in gas masks to remove toxic gases inthe air.Electrical Charge Effect A colloid may allow ions to be adsorbed on its surface, thereby acquiring an electricalcharge. The electrical charge may either be positive or negative. This electrical charge willprevent the particles from clumping together or coagulating. To understand more about the electrical charge effect of colloids, it will be interestingto do Activity 2.1 which is similar to the “Joy Dishwashing Challenge” you have seen in TVcommercials.What you will doActivity 2.1 Electrical Charge Effect of ColloidsMaterials: plastic jar with lid powdered laundry detergent cooking grease or shortening waterProcedure:1. Fill the jar with water about half full.2. Add some laundry detergent and shake the jar until the solution is soapy and bubbly.3. Drop a small glob of grease and put it in the soapy solution.4. Observe what will happen to the grease when the detergent solution is added.Analysis:1. What happens when the detergent comes in contact with the grease?2. Why is the grease removed when it comes in contact with the detergent?3. How is the electric nature of colloids exemplified in this activity? -8-

What causes the colloidal particles to carry a charge? Figure 2.1Colloidal particles have high adsorptive capacity. Thus, particles Cleansing action ofare adsorbed on their surface ions from water or from solutions ofelectrolytes. Such molecules are called surfactants, because detergentsthey tend to adsorb at the surface of a substance that is in contactwith the solution that contains them. Classic examples ofsurfactants are soap and detergents. They have bothhydrophobic (“water fearing” which refers to nonpolar part ofmolecule not attracted to water) and hydrophilic (“water loving”which refers to polar part of molecule attracted to water) groups intheir molecular structure. The nonpolar part of the soap molecule dissolves grease,while the polar ends dissolve in water. The net result is that thegrease/soap complex is water soluble and gets washed away.This process is called emulsification. You can see it working ifyou add soap to some oil-and-vinegar salad dressing. Thevinegar layer of the dressing gets cloudy because the soap hassurrounded little droplets of oil and prevents them from rejoiningthe oil layer. What you will do Self-Test 2.1Again, try to check how much you have learned from the lesson by answering the followingquestions.Identification. Give the principle behind the following. 1. A colored solution poured through a layer of charcoal becomes colorless. 2. Colloidal gold does not sink noticeably in water. 3. Colored glasses are made by using dispersed particles of metals in glass. 4. Beautiful blue beam of light is produced from the glass roof as you enter a building. 5. Soaps can remove oil and grease. Did you encounter any problem? Well, compare your answers with the answer keyand see for yourself the items you missed. Good luck! Key to answers on page 18.-9-

Lesson 3. How are Colloids Prepared? There are actually two ways of making a given substance disperse to colloidal size –by means of dispersion or condensation. Dispersion is the breaking of large pieces intocolloidal particle size. On the other hand, condensation involves tiny particles (molecules,ions, or atoms) clumping together to form clusters. Several processes make use of the principle of dispersion in making colloids. Forexample, grinding using a colloid mill is used in the preparation of paint pigments or facepowder. Also, the process of beating, stirring and whipping in preparing mayonnaise orcreams is a form of dispersion. You may also use chemicals to break down big particles.For example, sodium hydroxide (NaOH) is used to break up clay, glue, starch and gelatinpeptized in water. The process is known as peptization. On the other hand, condensation is involved in the preparation of carbon black byburning methane in limited air and collecting the soot or carbon atoms on cool surfaces.Carbon black is used as filler for rubber tires and in dispersions such as printer’s ink andIndian ink. Condensation is also involved in formation of clouds, fog and mist.What you will doActivity 3.1 Making MayonnaiseMaterials: two egg yolks ¼ cup vinegarProcedure: measuring cups water ¼ cup oil one small cup or bowl one small clear jar with lid ¼ tsp table salt food coloring (optional) ¼ tsp prepared mustard1. Measure out 2-3 cups of water and pour into the jar. Add about 4-6 drops of food coloring to the water and mix (optional). Next, measure out 2-3 cups of salad oil. Pour into the jar and mix again. Observe what happens. Let the contents of the jar sit for 3 - 5 minutes.2. Get some oil and water to mix. What should we add to mix oil and water together?3. Pour the vinegar into your jar.4. Pour the salad oil into the jar and put on the lid. Shake the contents very hard for 3 - 4 minutes, then let the mixture sit for three to four minutes. Compare the vinegar and oil mixture with the oil and water mixture previously done. What do you observe?5. Put the egg yolks into the small cup or bowl. Beat the yolks until they are liquid. - 10 -

6. Pour the beaten yolks into the jar with the oil and vinegar and close the lid. Shake the contents very hard for 3 - 4 minutes. Observe what happens.7. Add the mustard and the salt to the mixture. Again, shake very hard for three to four minutes. Taste it. You now have your home- made mayonnaise.Analysis:1. What happens to the oil? ________________________________________2. Why is the oil in the top layer? ________________________________________3. What is the vinegar for? ________________________________________4. What happens to the mixture when egg yolk is mixed? __________________ ________________________________________________________________5. How does this mayonnaise differ from the one bought from the store? ________________________________________________________________ The oil and the vinegar (water) mixed much better when the egg yolk was added.The lecithin, which is a protein in the egg yolk, acts as an emulsifying agent. Emulsifyingagents have regions that associate with the oil and regions that associate with the vinegar(water). Thus the emulsifier (the egg yolk) acts as a bridge between the oil and water. Themayonnaise created is an emulsion of oil droplets in water. If colloids are made, they can also be broken for certain purposes. There are severalways of breaking colloids. These include applying heat, adding a reactant or chemical, orpassing an electric current through it.1. Applying heat Have you tried cooking egg for breakfast? An egg white is a colloidal protein. Theparticles of an egg clump together because of the heat applied while the egg is beingcooked. This is one way of breaking colloids. Another application of heat in breakingcolloids is by digestion. This involves slow application of heat causing the colloidalparticles to get bigger which is used to prepare precipitates for filtration in the laboratory.2. Adding a reactant or chemical Have you heard of “cloud seeding”? Rain is made by seeding the clouds with solidcarbon dioxide (dry ice) or with silver iodide crystals. These “seeds” provide nuclei on whichwater vapor condenses. For a clearer visibility of the runways during rainy season, airport fog is removed byusing salt or dry ice. - 11 -

3. Passing through an electric current Smoke and other types of harmful aerosols are destroyedby Cottrell method of electric coagulation. This precipitatorremoves smoke particles by attracting them to the chargedplates. In this way, the charge on the smoke particles areneutralized, causing them to coagulate and settle down. Figure 3.1 Cottrell electric precipitator What you will do Self-Test 3.1Let us try to check how much you have learned from this lesson by answering the followingquestions.Identification. Identify the following._______________1. It is the method that involves the breaking down of big particles to smaller ones._______________2. It is the agent which brings about the formation of an emulsion._______________3. It is a method of allowing electric current to pass through the particle so that it will be attracted to the plates and be neutralized causing the particles to coagulate and settle down._______________4. It is the method where colloidal particles are built from the accumulation of smaller particles._______________5. It is the method utilized in “rain making”. Did you encounter any problem? Well, compare your answers with the answer keyand see for yourself the items you missed. Good luck! Key to answers on page 18. - 12 -

Lesson 4. How are Colloids Utilized in Technology, Human Body and Environment?Colloids in Body Processes Colloid substances are involved in many biological processes. For example, thehuman blood is a colloid. Protein molecules in the blood can be very long and their sizefalls within colloidal size range. Soluble waste products of metabolic processes will becarried by the blood to the kidneys for elimination. Sometimes, toxic substances build up in the blood because the kidneys are notworking efficiently specially in the case of patients suffering from Uremia. In order to cleanthe blood, it is made to flow across a large cellophane in an artificial kidney or dialyzingmachine. This process is known as dialysis. Dialysis is the process of separating thecontaminated ions and other smaller molecules from the colloidal particles of the blood byletting the true solutes to pass through a semi-permeable membrane. Dialysis of the bloodis known as hemodialysis. Figure 4.1In the process of hemodialysis, contaminated ions in the blood is allowed to pass in thedialyzing tube, letting the true solutes to pass through a semi-permeable membrane In hemodialysis, the bloodstream is diverted from its normal course in the body andpumped through a dialyzing tube with the semi-permeable membrane. An aqueous solutioncontaining ions, such as Na+, K+, and Cl- are of the same concentration as that of the bloodon the other side of the dialyzing membrane (isotonic). To prevent blood clotting, an anti-clotting agent is added to the dialyzing solution. It is however important that the dialyzingsolution is isotonic because that ensures that the solute particles pass in and out of theblood at equal rates, resulting in no net removal of essential components of the blood. Inthis regard, only the contaminated ions are removed faster than they are returned to theblood. Within a certain time, the procedure reduces the level of toxic substances in theblood. - 13 -

Colloids in Environment The use of colloids in the environment can be both harmful and beneficial. Somemay be natural while others are caused by human activities. Harmful ones can be madebeneficial if we just learn how to manipulate colloids. Aerosols are one of the unwanted colloids in the environment. These consist of solidor liquid particles that are dispersed in air smaller than 100µm in diameter. These verysmall particles include carbon black, silver iodide, and sea salt. These suspended particlesin air are commonly called particulates or simply particulate matter (PM). Largerparticulates include cement dusts and soil dust. Even larger particulates are raindrops, fogand sulfuric acid mist. In some of these particulates, viruses, bacteria and fungal sporesmay also be present. The presence of these bacteria causing diseases can be a threat tohealth in the community as what happened in the meningococcemia epidemic in Baguio Cityon December 2004. Some toxic air pollutants such as unburned hydrocarbons from motorvehicles can cause respiratory irritations when inhaled. Aside from posing a threat to human health, theseparticulates can also affect the transport industry when roadvisibility is reduced by during fog or mist. In our waterways, the disposal of detergents and otherpollutants can act as protective colloids that stabilize foamformation. When these foams accumulate on the surface ofwater, they can seriously reduce the amount of sunlight that canpenetrate the water surface. In this regard, the photosyntheticactivity of aquatic plants is drastically reduced. Figure 4.2 Foams collected on surface of bodies of water Did you know? Colloid science has evolved in technology and invaded the field of photography. Photographic films and paper have a light-sensitive emulsion coating. This emulsion consists of gelatin and one or more silver halides. In color films, several emulsion layers are separated by filter layers enabling the processing of brilliant colors. The gelatin keeps the silver halide crystals suspended when the emulsion is spread on the plastic film or photographic paper. - 14 -

What you will do Self-Test 4.1Let us check how much you have learned from the lesson by answering the followingquestions.Fill in the blanks. Supply the missing word in each statement. 1. In our waterways, the disposal of detergents and other pollutants can act as protective colloids that stabilize _______________ formation. 2. _______________ is the process of separating the contaminated ions and other smaller molecules from the colloidal particles of the blood by letting the true solutes to pass through a semi-permeable membrane. 3. _______________ is a condition where toxic substances build up in the blood because the kidneys are not working efficiently. 4. Suspended particles in air are commonly called _______________. 5. It is important that the dialyzing solution is _______________ to ensure that solute particles pass in and out of the blood at equal rates, resulting in no net removal of essential components of the blood. Did you encounter any problem? Well, compare your answers with the answer keyand see for yourself the items you missed. Good luck! Key to answers on page 18. Let’s Summarize 1. A colloid is a dispersion of particles of one substance, the dispersed phase, throughout a dispersing medium made of another substance. 2. Colloids are classified according to the phases of the dispersed material and dispersing medium. The types of colloids are sol, gel, emulsion, foam and aerosol. 3. A colloid is distinguished from a regular solution by the Tyndall effect, which is the scattering of visible light by colloidal particles. 4. Colloids exhibit special properties like Tyndall effect, Brownian motion, adsorption and electrical charge effect. 5. Brownian motion is the movement of colloidal particles in a random and zigzag fashion. 6. Since colloids have a large surface area, they exhibit high adsorbing capacity. 7. The ions adsorbed on the surface of a colloid produce an electrical charge. This, along with Brownian motion, prevents colloids from coagulating. - 15 -

8. Dispersion and condensation are two ways of making a given substance disperse to colloidal size. When larger pieces are broken into colloidal size, the process is dispersion. But when tiny particles clump together to form clusters the process is condensation.PosttestI. Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. Which of the following is a colloid? c. sand and water a. oil and water d. alcohol and water b. salt and water2. Pond water appears clear on standing. Thus, pond water is _____.a. a colloid c. an emulsionb. a solution d. a suspension3. A metal alloy belongs to what type of colloid?a. sol c. foamb. gel d. emulsion4. As you enter the building, you observe a beautiful beam of light from the glass roof.What best explains this phenomenon?a. adsorption c. Brownian motionb. Tyndall effect d. electrical charge effect5. “Merengue” is prepared by beating the egg white vigorously. What method is used inmaking this colloid?a. deposition c. adsorptionb. dispersion d. condensation6. Which of the following involves adsorption? a. plating of copper on a steel object b. brown color of the eyes of Filipinos c. adherence of paint to wood surfaces d. removal of odor inside a refrigerator using charcoal7. When egg yolk is added to oil and water with vinegar to make mayonnaise, the egg yolkserves as:a. solvent c. surfactantb. coagulant d. emulsifiying agent - 16 -

8. What is the phase of a dispersed material in emulsion?a. gas c. liquidb. solid d. cannot be determined9. Which of the following colloids is considered harmful?a. cheese c. whipped creamb. hair spray d. black diamond10. What phenomenon clearly distinguishes a colloid from a solution?a. adsorption c. Brownian motionb. Tyndall effect d. electrical charge effectII. Analogy. Fill in the blanks with the correct answer based on the relationship provided.1. clouds : liquid aerosol :: mayonnaise : __________________2. solutions : homogenous :: colloids : __________________3. rain : cloud seeding :: smoke : ____________________4. ____________________ : egg yolk :: soap : surfactants5. ____________________ : non-polar :: hydrophilic : polar Key to answers on page 18.Key to AnswersPretest 6. a Matching Type 7. a 1. c Multiple Choice 8. c 2. f 1. c 9. c 3. b 2. a 10. b 4. a 3. d 5. d 4. a 6. a 5. c 7. a - 17 - 8. iLesson 1 9. d 10. hSelf-Test 1.1 1. g 2. d 3. d 4. f 5. e

Lesson 2 Lesson 3 Lesson 4Self-Test 2.1 Self-Test 3.1 Self-Test 4.11. adsorption 1. dispersion 1. foam2. Brownian motion 2. emulsifying agent 2. dialysis3. Tyndall effect 3. electric coagulation 3. uremia4. Tyndall effect 4. condensation 4. particulate or5. Emulsification 5. cloud seeding particulate matter 5. isotonicPosttestMultiple Choice 6. d Analogy1. a 7. d 1. liquid emulsion2. b 8. c 2. heterogenous3. a 9. b 3. electric coagulation4. b 10. b 4. emulsifying agent5. b 5. hydrophobicReferencesAraneta, F.L., Catris, L.V. & Deauna, M.C. (2002). The world of chemistry: Exploring the natural world series. Philippines: SIBS Publishing Inc.Chang, R.N. (2004). Chemistry. (8th ed.) USA: McGraw-Hill, Inc.Green, J. & Damji, S. (2001). Chemistry. (2nd ed.) Australia: IBID Press, VictoriaMagno, M.C., Tan, M.C. & Punzalan A.E. (2000). Chemistry. (3rd ed.) Manila: DIWA Scholastic Press Inc.Moore, J.W., Stanitski, C.L. & Wood J.L. (1998). The chemical world: Concepts and applications. (2nd ed.) USA: Harcourt Brace & CompanyPetrucci, R.H. & Harwood, W.S. (1998). General chemistry: Principles and modern applications. (7th ed.) New Jersey: Prentice-Hall International - 18 -

Module 6 Colligative Properties of Solution What this module is about When a non-volatile (substances that do not readily form vapors), non-electrolytic(substances that do not form ions and do not conduct electricity when placed in water)solute such as sugar is dissolved in a given volume of solvent to form a sugar solution, itchanges the set of properties of the pure solvent entirely. In this regard, the set ofproperties such as freezing point, boiling point, vapor pressure, and osmotic pressure of asolvent are affected by the presence of the solute particles in the solution. This set ofproperties will depend only on the number of dissolved particles in the solution and not ontheir identity. These properties are collectively known as colligative properties of solution. This module will focus on four colligative properties of electrolyte and non-electrolytesolutions namely, vapor pressure lowering, boiling point elevation, freezing pointdepression, and osmotic pressure. To make the discussion easy for you, the module is divided into four lessons:  Lesson 1 – What is the Difference Between Osmotic Pressure and Vapor Pressure?  Lesson 2 – What is Boiling Point Elevation?  Lesson 3 – What is Freezing Point Lowering?  Lesson 4 – What is the Importance of Colligative Properties in Our Daily Lives? What you are expected to learn After going through this module, you should be able to: 1. differentiate between colligative and non-colligative properties of a solution; 2. differentiate vapor pressure from osmotic pressure; 3. identify boiling point elevation; 4. state and explain freezing point lowering; and 5. apply knowledge of the importance of colligative properties of solutions to daily living.

How to learn from this moduleHere are some pointers to remember as you go over this module.1. Read and follow the instructions carefully.2. Answer the pretest first before reading the content of the module.3. Take down notes and record points for clarification.4. Always aim to get at least 70% of the total number of items given.5. Be sure to answer the posttest at the end of the module.What to do before (Pretest)Take the pretest before proceeding to the lessons. Check your answers against the answerkey at the end of the module.I. Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper.1. A substance whose water solution conducts an electric current is called a(n)a. electrolyte c. polar substanceb. nonelectrolyte d. nonpolar substance2. A substance that vaporizes easily is known asa. volatile c. electrolyteb. nonvolatile d. nonelectrolyte3. Adding a solute such as NaCl to water increases itsa. boiling point c. osmotic pressureb. melting point d. vapor pressure4. Which of the following is NOT a colligative property of solution?a. solubility c. melting pointb. boiling point d. osmotic pressure5. Which of the following statements is TRUE about vapor pressure? a. The vapor pressure of the solvent is less than the solution. b. The vapor pressure of the solution is higher than the pure solvent. c. The vapor pressure of the pure solvent is higher than the solution. d. The vapor pressure of the pure solvent and the solution are the same. -2-

6. The i factor gives the number of particles per formula unit of the solute. What is the ifactor for NaCl?a. 0 c. 2b. 1 d. 37. What happens during osmosis? a. Pure solvent diffuses through a membrane but solutes do not. b. Pure solutes diffuse through a membrane but solvent does not. c. Gases diffuse through a membrane into a solution and build up pressure. d. Pure solvent and a solution both diffuse at the same time through a membrane.8. What is the temperature at which the vapor pressure equals the atmospheric pressure?a. 100 oC c. boiling pointb. freezing point d. melting point9. The freezing point of a solution is always __________ the pure solvent.a. less than c. greater thanb. same as d. greater than or equal to10. If a potato is placed in brine (concentrated NaCl solution), the potato cell is expected to:a. shrink c. become rigidb. swell d. no considerable changeII. TRUE or FALSE. Write TRUE if the statement is correct. Otherwise, write FALSE.__________ 1. Colligative properties will only depend on the number of dissolved particles in the solution and not on their identity.__________ 2. The addition of a solute will increase the boiling point of the solution.__________ 3. The van’t Hoff factor is only applicable for non-electrolyte solutions.__________ 4. The boiling point of the solution is always greater than that of the pure solvent.__________ 5. All biological membranes are regulated via osmosis. Key to answers on page 21.Lesson 1. What is the Difference Between Osmotic Pressure and Vapor Pressure? Before differentiating these two terminologies, let us first understand the differencebetween the two classifications of solutions according to their properties – colligative andnon-colligative properties. Earlier, we have defined that colligative properties will just -3-

depend on the number of dissolved particles in the solution and not on their identity. Non-colligative properties state otherwise. To clearly understand the difference between thesetwo properties, let us do Activity 1.1.What you will doActivity 1.1 Classifying Colligative and Non-Colligative Properties of SolutionMaterials: 1% (w/v) brown sugar solution 0.5 % (w/v) salt solution (NaCl) thermometerProcedure: Prepare the brown sugar and salt solutions of concentration 1% and 0.5%,respectively. Observe the following properties in the two solutions and identify whether theproperty observed is a colligative or non-colligative property based on the definition.Properties of Solution 0.5 % (w/v) salt 1% (w/v) brown sugar solution solutionSolubility (ability of thesolute to be dissolved ina solvent)Viscosity of solution(“viscosity” is a propertyof solution to resist flow)Color of the solutionTaste of solutionTemperature uponboilingTemperature uponfreezingAnalysis:1. Which of the following properties of solution can be considered colligative? Why?2. Which of the following properties of solution can be considered non-colligative? Justify your answer. After finishing the activity, we can clearly distinguish what properties of solution arecolligative and what properties are not. Again, for as long as the number of dissolvedparticles in solution is taken into consideration, then that property of solution (e.g. boiling -4-

point, freezing point, etc.) is a colligative property. On the other hand, if the identity of thedissolved species and solvent is being analyzed (e.g. color, taste, viscosity, solubility), thenit is a non-colligative property. Having learned their difference, we can now describe vaporpressure and osmotic pressure of solutions.Vapor-Pressure Lowering When a nonvolatile (does not turn to vapor easily) solute is added to a liquid to forma solution, the vapor pressure above that solution decreases. To understand why that mightoccur, let us do a simple experiment in Activity 1.2.What you will doActivity 1.2 Vapor-Pressure LoweringMaterials: Water Aqueous sugar solution Two glasses One sealed enclosed containerProcedure:1. Get two glasses placed side by side in a sealed enclosed container. One glass contains pure water, the other an equal volume of an aqueous solution of sugar. Take note of the volume of water and sugar solution. (See figure 1.1a)2. Leave the set-up until the next day. Gradually measure the volume of the sugar solution and that of the pure water. Is there any change in their original volume? (See figure 1.1b) Figure 1.1 Experiment set-up for vapor-pressure loweringAnalysis:1. What can you infer about the change in volume of the sugar solution and that of pure water?2. What has caused this change in volume of the sugar solution and that of pure water? -5-

After finishing the activity, you should now see why the vaporpressure of the solvent decreases upon addition of a non-volatile, non-electrolyte solute (sugar). Figure 1.2 A nonvolatile solute reduces the rate of vaporization of the solvent. The extent to which a nonvolatile solute lowers the vapor pressure is proportional toits concentration. This was discovered by French chemist Francois Raoult (1830-1907).Raoult’s law states that for nonelectrolyte solutions, the partial vapor pressure of a solventover a solution (P1) is equal to the vapor pressure of the pure solvent (Po1) multiplied by themole fraction of the solvent (X1). This law is mathematically expressed as: P1 = X1 Po1 For solutions of electrolytes, the vapor pressure lowering equation can be expressedas: P1= iPo1 X1where i is the van’t Hoff factor, named after Jacobus Henricus van’ Hoff (1852-1911), whowon the very first Nobel Prize in chemistry in 1901 for his work on colligative properties ofsolution. The i factor gives the number of particles per formula unit of the solute. Forexample, NaCl solution dissociates to give one Na+(aq) and one Cl-(aq), the iNaCl = 2, becausethere is one Na+ and one Cl- ion in solution per formula unit of solute. On the other hand, if both components of a solution are volatile (readily evaporates),the vapor pressure of the solution is the sum of the individual partial pressures. The totalpressure is given by Dalton’s law of partial pressure: PT = PA + PB or PT = XA PoA + XB PoB -6-

Sample Problem: What is the vapor pressure of the solution containing 20 g of sugar (C12H22O11) in 1.5kg water at 25oC. Note: Powater at 25oC is 24 torrExplanation:1 mole of sugar (C12H22O11) is 342 g and 1 mole of water (H2O) is 18 g. The numberof moles of each component is computed as:For sugar: 20 g C12H22O11 × 1mol C12H22O11 = 0.06 mol C12H22O11 342 g C12H22O11For water: 1500 g H2O × 1mol C12H22O11 = 83.33 mol H2O 18 g C12H22O11And mole fraction of water (solvent) is computed as: XH2O = 83.33 C12H22O11 = 1.00 83.33 + 0.06 C12H22O11Therefore, the vapor pressure is: PH2O = XH2O × Po H2O = 1.00 × 24 torr PH2 O = 24.0 torr Practice Exercise: Calculate the vapor pressure of salt solution containing 20 g salt (NaCl) in 1.5 kg of water at 25oC? Note: Powater at 25oC is 24 torr and molecular weight of NaCl is 58 g/mol. Note: Remember that NaCl is a strong electrolyte and dissociates to Na+ and Cl- ion respectively. Answer: 48 torr -7-

Osmotic Pressure Osmosis is of prime importance to living organisms because it influences thedistribution of nutrients and the release of metabolic waste products. Living cells of bothplants and animals are enclosed by a semipermeable membrane called the cellmembrane, which regulates the flow of liquids and of dissolved solids and gases into andout of the cell. Did you know? Oceans cover over 72% of the earth and a reservoir for 97.2% of the earth’s water. However, out of the 97.2% of earth’s water, only 2.5% is available as freshwater and less than 1% is available as groundwater or surface water for human use. Scientists have found a new, cheaper and more economical way of making seawater possible for drinking. The process is known as reverse osmosis. Through this process, seawater is made to pass through a semi-permeable membrane and by applying a pressure greater than 30 atm (Note: 30 atm is the pressure that must be applied to saline solution of seawater in order to facilitate osmosis), the osmotic flow would be reversed, and fresh water made to run through the membrane. Now, that’s chemistry in action! The osmotic pressure (Π ) of a solution is the pressure required to stop osmosis.The osmotic pressure of the solution is given by: Π = MRTwhere M is the molarity of solution, R the gas constant (0.0821 L . atm / K . mol), and T theabsolute temperature (in Kelvins). The osmotic pressure is expressed in atm. And since osmotic pressuremeasurements are carried out under constant temperature, molarity is preferred overmolality as concentration. Again, for solutions of electrolytes, the osmotic pressure equationcan be expressed as: Π = iMRTwhere i is the van’t Hoff factor, the number of particles per formula unit of the solute. -8-

Sample Problem: What is the osmotic pressure of the solution containing 0.2 M sugar (C12H22O11)solution at 25oC?Explanation:Π = MRTΠ = 0.2 mol C12H22O11 × 0.0821 L ⋅ atm × 298 K = 4.89 1L solution K ⋅ molΠ = 4.89 atm Practice Exercise: Calculate the osmotic pressure of 0.2 M salt solution (NaCl) at 25 oC? Answer: 9.78 atmWhat you will doSelf-Test 1.1Now that you are through with the first lesson, try to answer the following questions and seefor yourself how much you learned.Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. What happens during osmosis? a. Pure solvent diffuses through a membrane but solutes do not. b. Pure solutes diffuse through a membrane but solvent does not. c. Gases diffuse through a membrane into a solution and build up pressure. d. Pure solvent and a solution both diffuse at the same time through a membrane.2. What is the van’t hoff factor for CaCl2? c. 2 a. 0 d. 3 b. 13. The osmotic pressure found for a 2.50 g sugar solution at 25oC is 1.79 atm in a 1.0 L ofwater. What is its molar mass (g/mol)?a. 342 g/mol c. 3.42 g/molb. 34.2 g/mol d. 7.32 x 10-2 g/mol -9-

4. Cold-blooded animals and fish thrive in sea water. Which of the following situations is true about their condition? a. Their blood has greater osmotic pressure than that of sea water. b. Their blood has less osmotic pressure than that of sea water. c. Their blood has the same osmotic pressure as that of sea water. d. The osmotic pressure of sea water is greater than that of their blood.5. According to Raoult's law, which statement is FALSE? a. The vapor pressure of a solvent over a solution decreases as its mole fraction increases. b. The solubility of a gas increases as the temperature decreases. c. The vapor pressure of a solvent over a solution is less than that of pure solvent. d. The greater the pressure of a gas over a solution the greater its solubility. Key to answers on page 21.Lesson 2. What is Boiling Point Elevation? Remember that boiling point is the temperature at which the vapor of a liquid is equalto the atmospheric pressure. In the presence of a non-volatile, non-electrolytic solute suchas sugar, the pressure that will be exerted by the vapor of the solvent is lesser incomparison with that of the pure solvent at equal volumes. And as a consequence, thesolution will boil at a higher temperature than that of the pure solvent. To clearly understand what boiling point elevation is all about let us do Activity 2.1.What you will doActivity 2.1 Boiling Point ElevationMaterials: three eggs 1 tbsp of salt 1 tbsp of sugar boiling waterProcedure:1. Put the first egg in two cups of water and take note of the time until the water boils. - 10 -

2. Again, using two cups of water, put the second egg in the water and add 1 tbsp of salt. Record the time it will take the water to boil. 3. Repeat step 2, but add 1 tbsp of sugar instead of salt. Again, take note of the time it will take the water to boil. 4. Compare the time it will take for water to boil and cook the hard-boiled egg in step 1 to step 3. Record your observations.Analysis: 1. Which set-up took less time to cook hard-boiled eggs? Why? ________________________________________________ 2. Did the water take more time to boil when electrolyte (NaCl) solute was added? Why? ___________________________________________ 3. Did the water take more time to boil upon addition of non-electrolyte (sugar) solute? Justify your answer. ___________________________________________________________ Since we have already established that the boiling point of the solution is greater thanthat of the pure solvent, the boiling point elevation (∆Tb) is mathematically expressed asfollows: ∆Tb = Kbmwhere (Kb) is boiling point elevation constant, equivalent to 0.52 oC/m for aqueous solutions.This means that, for example, 1 mole of sugar (nonelectrolyte) in 1 kilogram of water willincrease the boiling point from 100oC to 100.52oC. And (m) is the molal concentration ofsolute. It is also important to note that ∆Tb is a positive quantity and should be added to theboiling point of pure solvent (water), which is 100oC. Remember that molality is used here over molarity because we are dealing with asolution whose temperature is not constant and concentration cannot be expressed inmolarity because it changes with temperature. Again, for solutions of electrolytes, the boiling point elevation equation can beexpressed as: ∆Tb = iKbmwhere i is the van’t Hoff factor, the number of particles per formula unit of the solute.Sample Problem: At what temperature will the sugar solution boil if 20 g sucrose (C12H22O11) is added - 11 -

to 1.5 kg of water?Explanation:Remember that sugar is a nonelectrolyte so there will be no need for the van’t Hofffactor. A mole of sugar (C12H22O11) is 342 g. Thus, molality of sugar can becomputed as: 20 g C12H22O11 × 1mol C12H22O11 = 0.04 mol C12H22O11 = 0.04 m 1.5 kg H2O 342 g C12H22O11 kg H2OThe boiling point elevation is calculated as: ∆Tb = Kbm = 0.52 oC x 0.04 m m ∆Tb = 0.02oCThus, the boiling point of the solution is: 100oC + 0.02oC = 100.02oC Practice Exercise: At what temperature will salt solution boil if 20 g salt (NaCl) is added to 1.5 kg of water? (Molecular weight of NaCl is 58 g/mol) Note: Remember that NaCl is a strong electrolyte and dissociates to Na+ and Cl- ion, respectively. Answer: 100.24 oC What you will do Self-Test 2.1Again, try to check how much you have learned from the lesson by answering the followingquestions.Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper. - 12 -

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1. If Solution A contains 5 g NaCl in 100 mL water and Solution B contains 10 g NaCl in 100 mL water, which of the following conditions is true about solution A and B? a. Solution A has greater boiling point than Solution B. b. Solution B has greater boiling point than Solution A. c. Solution A and B have the same boiling point. d. Solution A has less boiling point than Solution B.2. What is the boiling point of a 0.20 m NaCl solution in water? For water, Kb is 0.512oC.kg/mol. a. 100.0oC c. 100.3oC b. 100.2oC d. 100.4oC3. What happens to the boiling point of water when 0.5 m sugar (nonelectolyte) is added toit? a. 100.0oC c. greater than 100.0oC b. less than 100.0oC d. none of the above4. Which of the following statement is TRUE about the boiling point of the solution? a. The boiling point of the solvent is greater than that of the solution. b. The boiling point of the solution is less than that of the pure solvent. c. The boiling point of the solution is the same as that of the pure solvent. d. The boiling point of the solution is greater than that of the pure solvent.5. The boiling point of an impure compound is generally __________ that of pure solid. a. same as c. greater than b. less than d. greater than or equal to Key to answers on page 21.Lesson 3. What is Freezing Point Depression? Have you ever tried sprinkling salt over ice? What have you noticed? Salts such asNaCl and CaCl2 cause ice to melt. This method of thawing depresses the freezing point ofwater. To learn more about freezing point depression, it will help to do Activity 3.1. - 14 -

What you will do?Activity 3.1 Freezing Point DepressionMaterials: one tsp. salt stirrers one glass of water spoon one glass of crushed Ice thermometerProcedure:1. Using one glass of water and one glass of crushed ice, stir the mixture; then using a thermometer, observe and record this temperature.2. Add one tsp. of salt to the water/ice mixture, then observe and record the temperature. You should repeat this procedure until the temperature reaches 10oC. More ice should be added if necessary.Analysis:1. What happens to water and ice when salt is added to this mixture?2. What happens to the temperature when salt is added to the mixture?3. What variables would cause these differences? Water freezes at 0oC and boils at 100oC. Salt water will not freeze until thetemperature is below 0oC. The more salt, the lower the freezing point of the solution. In theabove experiment, energy is lost from the water in the form of heat. This heat is used tomelt the ice. Since heat is lost from the water the temperature of the water goes down.Since there is now salt dissolved in the water it cannot freeze again, hence we observe alower temperature. The freezing point depression (∆Tf) is mathematically expressed as: ∆Tf = Kf mwhere Kf is the freezing point depression constant. equivalent to -1.86 oC/m for aqueoussolutions. Again, for example, 1 mole of sugar (nonelectrolyte) in 1 kilogram of water willdecrease the freezing point from 0oC to -1.86oC. (m) is the molal concentration of solute.And since we have already established that the freezing point of the solution is less thanthat of the pure solvent, then ∆Tf is a negative quantity and should be subtracted from thefreezing point of pure solvent (water), which is 0oC. Again, for solutions of electrolytes, the freezing point lowering equation can beexpressed as: - 15 -

∆Tf = iKfmwhere i is the van’t Hoff factor, the number of particles per formula unit of the solute.Sample Problem: At what temperature will the sugar solution freeze if 20 g sucrose (C12H22O11) isadded to 1.5 kg of water?Explanation:Since sugar is a nonelectrolyte, there will be no need for the van’t Hoff factor. A moleof sugar (C12H22O11) is 342 g. Molality (m) of sugar can be computed as:20 g C12H22O11 × 1mol C12H22O11 = 0.04 mol C12H22O11 = 0.04 m1.5 kg H2O 342 g C12H22O11 kg H2OThe freezing point depression is calculated as:∆Tf = Kf m = -1.86 oC x 0.04 m m∆Tf = -0.07oCThus, the freezing point of the solution is:0oC - 0.07oC = -0.07oC Practice Exercise: At what temperature will salt solution freeze if 20 g salt (NaCl) is added to 1.5 kg of water? (Molecular weight of NaCl is 58 g/mol) Note: Remember that NaCl is a strong electrolyte and dissociates to Na+ and Cl- ion, respectively. Answer: -0.86 oC - 16 -

Did you know? NaCl (salt) is added to ice to make a freezing mixturethat results in a tasty, homemade ice cream. Lowering thefreezing temperature of the ice-water-salt mixture causes theice cream ingredients to freeze more quickly.What you will doSelf-Test 3.1Let us try to check how much you have learned from this lesson by answering the followingquestions.1. The melting point of an impure compound is generally __________ that of the pure solid.a. less than c. greater thanb. the same as d. greater than or equal to2. Adding sodium chloride to water will cause the: a. boiling point to rise and the freezing point to lower. b. boiling point to lower and the freezing point to rise. c. both boiling point and freezing point to rise. d. both boiling point and freezing point to lower.3. What is the freezing point of a 0.20 m sucrose (nonelectrolyte)?Kf = 1.86 oC /m.a. 0.00oC c. 3.7oCb. 0.37oC d. 37oC4. What is the freezing point of a 0.50 m NaCl (strong electrolyte)?Kf = 1.86 oC /m.a. 0.0oC c. 9.3oCb. 0.93oC d. 1.86oC5. Which of the following solutions has the HIGHEST freezing point?a. 0.1 m sugar c. 0.08 m CaCl2b. 0.1 m NaCl d. 0.04 m Na2SO4 Did you encounter any problem? Well, compare your answers with the answer keyand see for yourself the items you missed. Good luck! Key to answers on page 21. - 17 -

Lesson 4. What is the Importance of Colligative Properties in Our Daily Llife? We have already established that addition of salt, such as NaCl, lowers the freezingpoint of ice. Even some organisms have evolved to survive freezing water temperatureswith natural \"antifreeze.\" Some fishes living in the Arctic ocean have blood containing ahigh concentration of a specific protein. This protein behaves like a solute in a solution andlowers the freezing point in their blood enabling them to survive in very cold temperatures. Biologists and biochemists often take advantage of osmotic pressure when theyisolate the components of a cell. They further classify solutions depending on theconcentration of the solutes inside the cell as hypertonic, isotonic or hypotonic solutions.Suppose you cut a potato into cubes and immerse it in brine (concentrated NaCl). What willhappen? The potato will tend to shrink. This method is used for curing meats like countryham. It is done to eliminate the bacteria since the bacteria cannot survive in high saltconcentration. Alternatively, hypotonic means the solution has a lower concentration of particlesthan the cell. Suppose you cut a potato into cubes, but this time immerse it in water. Whatwill you notice? There would be more water moving into the potato cell causing it to swell.When a cell is placed in a solution that has a much smaller ionic strength, water pours intothe cell, and the cell expands until the cell membrane bursts. In the case of red blood cells,the process is known as hemolysis. On the other hand, an isotonic solution has the same osmotic strength on both sidesof the semipermeable membrane. Physiological saline (0.9% NaCl) is isotonic with blood.In this condition, the cell retains its normal shape and function. Did you know? Dialysis uses the same principle as osmosis to remove salt or other ions in the blood. In the dialysis machine, the blood is circulated through a long tube of cellophane in an isotonic solution and then returned to the patient’s vein. The ions can pass out and the large proteins remain. When protein is found in the urine, that means there is some type of damage to the kidney. The isotonic solution has 0.6% NaCl, 0.04% KCl, 0.2% NaHCO3, and 0.72% glucose (w/v). Thus no sodium ions or glucose is lost from the blood. The isotonic solution is changed every 2 hours and the patient stays on the dialysis machine for 4 to 7 hours. - 18 -

What you will do Self-Test 4.1Let us try to check how much you have learned from the lesson by answering the followingquestions.Modified True or False. Write TRUE if the statement is correct otherwise CHANGE theunderlined word to correct the statement.__________ 1. A hypertonic solution is a solution that has a higher concentration of article than the cell.__________ 2. The normal physiological saline concentratrion of the blood is 0.6% NaCl.__________ 3. Cells shrink if the concentration of solute inside the cell is greater than that outside the cell.__________ 4. Hemolysis is the phenomenon that causes the membrane of red blood cells to burst due to a hypotonic environment inside the cell.__________ 5. Some fishes living in the Arctic ocean were able to survive in very cold temperature because their blood contained the same concentration of solutes as the seawater. Did you encounter any problem? Well, compare your answers with the answer keyand see for yourself the items you missed. Good luck! Key to answers on page 21.Let’s SummarizeA. Summary of Key Equations P1 = Po1 X1 PT = PA + PB For nonelectrolytes: Π = MRT ∆Tb = Kbm Raoult’s Law: ∆Tf = Kf m Dalton’s Law of Partial Presssure: Osmotic Pressure: P1 = iPo1 X1 Boiling Point Elevation: PT = iPA + iPB Freezing Point Depression: For electrolytes: Raoult’s Law: Dalton’s Law of Partial Presssure: - 19 -

Osmotic Pressure: Π = iMRTBoiling Point Elevation: ∆Tb = iKbmFreezing Point Depression: ∆Tf = iKf mB. Summary of Facts and Concepts1. Colligative properties (or collective properties) are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.2. The four colligative properties of a solution are vapor pressure, osmotic pressure, boiling point and freezing point.3. The change in vapor pressure where the solute is less volatile than the solvent is regulated by Raoult’s law, which states that the vapor pressure of a solvent over a solution is equal to the mole fraction of the solvent times the vapor pressure of pure solvent.4. The osmotic pressure of a solution is the pressure required to stop osmosis.5. The freezing point of the solution is always less than the freezing point of the pure solvent.6. The boiling point of the solution is always greater than the boiling point of the pure solvent.7. In electrolyte solutions, the interaction between ions leads to the formation of ion pairs. The van’t Hoff factor (i) provides a measure of the extent of dissociation of electrolytes in solution.8. Solutions can be classified as hypertonic, hypotonic or isotonic depending on the concentration of solute inside and outside the cell.9. A hypertonic solution has a higher concentration of particle than the cell causing the cell to shrink.10. A hypotonic solution has a lower concentration of particles than the cell causing the cell to swell.11. An isotonic solution has the same osmotic strength on both sides of the semipermeable membrane. PosttestI. Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper.1. According to Raoult's law, which statement is FALSE? a. The solubility of a gas increases as the temperature decreases. b. The greater the pressure of a gas over a solution the greater its solubility. c. The vapor pressure of a solvent over a solution is less than that of pure solvent. d. The vapor pressure of a solvent over a solution decreases as its mole fraction increases. - 20 -

2. Consider a solution made from a nonvolatile solute and a volatile solvent. Which statement is TRUE? a. The osmotic pressure is the same as vapor pressure of the solution. b. The vapor pressure of the solution is always greater than the vapor pressure of the pure solvent. c. The boiling point of the solution is always greater than the boiling point of the pure solvent. d. The freezing point of the solution is always greater than the freezing point of the pure solvent.3. Dissolving a solute such as NaCl in a solvent such as water results in: a. an increase in the melting point of the liquid b. a decrease in the boiling point of the liquid c. a decrease in the vapor pressure of the liquid d. no change in the boiling point of the liquid4. What is the freezing point of a solution that contains 10.0 g of glucose (C6H12O6) in 100g of H2O? Kf = 1.86 oC/m?a. -0.186 oC c. -0.10 oCb. 0.186 oC d. -1.03 oC5. Which of the following solutions has the highest osmotic pressure at 25oC?a. 0.2 M KBr c. 0.2 M Na2SO4b. 0.2 M C2H5OH d. 0.2 M KCl6. A solution that contains 55.0 g of ascorbic acid (Vitamin C) in 250 g of water freezes at -2.34 C. Calculate the molar mass (g/mol) of the solute. Kf = 1.86 oC/m.a. 1.26 c. 43.6b. 10.9 d. 1757. What mass of ethanol, C2H5OH, a nonelectrolyte, must be added to 10.0 L of water togive a solution that freezes at -10.0oC? Assume the density of water is 1.0 g/mL?a. 85.7 kg c. 5.38 kgb. 24.8 kg d. 2.48 kg8. If solution A contains 5 grams of NaCl in 100 grams of water and solution B contains 10 grams of NaCl in 100 grams of water, which of the following comparisons is true? a. A has a higher boiling point and a higher freezing point than B. b. A has a lower boiling point and a lower freezing point than B. c. A has a lower boiling point and a higher freezing point than B. d. A has a higher boiling point and a lower freezing point than B.9. Which solution has lower concentration of particles than the cell causing the cell toswell?a. hypertonic c. isotonicb. hypotonic d. unsaturated - 21 -

10. In what temperature does the vapor pressure of the liquid equal the atmosphericpressure?a. 100oC c. melting pointb. boiling point d. freezing pointII. Matching Type. Match the correct equation in Column B with the principles in Column A. Column A Column B_____ 1. Raoult’s law a. ∆Tb = Kbm_____ 2. Dalton’s law of partial pressures b. Π = MRT_____ 3. Boiling point elevation c. P1= Po1 X1_____ 4. Freezing point depression d. PT = PA + PB_____ 5. Osmotic pressure e. ∆Tf = Kf m f. ∆P = P2 - P1 Key to answers on page 22.Key to AnswersPretest 6. c II. True or False 7. a 1. True I. Multiple Choice 8. c 2. False 1. a 9. a 3. False 2. a 10. a 4. True 3. a 5. True 4. a 5. cLesson 1 Lesson 2 Lesson 3 Lesson 4Self-Test 1.1 Self-Test 2.1 Self-Test 3.1 Self-Test 4.1 1. a 1. b 1. a 1. True 2. d 2. b 2. a 2. 0.9 % NaCl 3. b 3. b 3. b 3. True 4. c 4. d 4. d 4. Hypertonic 5. a 5. c 5. c 5. greater - 22 -

Posttest 6. d II. Matching Type 7. d 1. c I. Multiple Choice 8. c 2. d 1. d 9. b 3. a 2. c 10. b 4. e 3. c 5. b 4. d 5. cReferencesChang, R.N. (1998). Chemistry. (6th ed.) USA: McGraw-Hill, Inc.Green, J. & Damji, S. (2001). Chemistry. (2nd ed.) Australia: IBID Press, VictoriaMoore, J.W., Stanitski, C.L. & Wood J.L. (1998). The chemical world: Concepts and applications. (2nd ed.) USA: Harcourt Brace & CompanyPetrucci, R.H. & Harwood, W.S. (1998). General chemistry: Principles and modern applications. (7th ed.) New Jersey: Prentice-Hall InternationalRegalado, E.C. (2000). Chemistry. Philippines: St. Augustine Publications, Inc. - 23 -

Module 7 Solutions What this module is about Many of the substances we deal with in daily life such as air, gasoline, vinegar, cornsyrup, coconut oil, seawater, cooking gas, and stainless steel share common characteristics--- they are all solutions. Each of them is made up of more than one component, but eachcomponent is evenly distributed. Solutions of this kind are known as homogenous solutions. In this module, you will learn more about solutions, their nature, properties and theways of expressing their concentration. As you read the module, it will introduce you to theversatile world of solutions. Hopefully, you will learn to appreciate the importance ofsolutions not just in our life but in the world we live in. To make the discussion easy for you, the module is divided into four lessons:  Lesson 1 – What Distinguishes Solutions from Non-Solutions?  Lesson 2 – What Makes a Solute Dissolve Faster?  Lesson 3 – How Much Solute Can Be Dissolved in a Solution?  Lesson 4 – Why Do You Need to Study Solutions? What you are expected to learn After going through this module, you should be able to: 1. identify the types of solutions; 2. describe the changes that occur in the dissolving process of solutions; 3. relate the changes that go with the dissolving process to energy changes and disorderliness; 4. differentiate saturated, unsaturated and supersaturated solutions; 5. identify the factors affecting solubility; 6. express concentration of solution in terms of percent composition, mole fraction, molarity, and molality ; and 7. appreciate the importance of solutions in our daily activities.

How to learn from this moduleHere are some pointers to remember as you go through this module. 1. Read and follow the instructions carefully. 2. Answer the pretest first before reading the content of the module. 3. Take down notes and record points for clarification. 4. Always aim to get at least 70% of the total number of items given. 5. Be sure to answer the posttest at the end of the module.What to do before (Pretest)Take the pretest before proceeding to the lessons. Check your answers against the answerkey at the end of the module.I. Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. All of the following are solutions EXCEPTa. milk c. alloyb. wine d. coffee2. All of the following affects the solubility of a solid in a liquid EXCEPTa. pressure c. surface areab. stirring d. temperature3. A metal solution is called a(n) c. suspension a. alloy d. electrolyte b. colloid4. Why is a solution considered homogenous? a. It is usually liquid. b. It contains a solute and solvent. c. It can be dilute or concentrated. d. Its components are distributed evenly in all proportions.5. All of the following describes a solution EXCEPTa. clear c. cannot pass through filter paperb. homogenous d. can be separated by physical means -2-

6. A solution that contains the maximum amount of dissolved solute is:a. saturated c. concentratedb. unsaturated d. supersaturated7. The following substances are soluble in water EXCEPTa. oil c. sugarb. salt d. alcohol8. What is the maximum amount of solute that can be dissolved in a fixed amount ofsolvent at a given temperature?a. dilution c. solubilityb. molarity d. dissolution9. What is the concentration of a solution expressed in moles of solute per kilogramsolvent?a. molality c. mole fractionb. molarity d. percentage by weight10. The pH of human saliva is 6.5. How would you describe the solution?a. basic c. saturatedb. acidic d. supersaturatedII. True or False. Write the word TRUE if the statement is correct and FALSE if it is incorrect.________ 1. Solutions are homogenous.________ 2. Solutes can exist as solid, liquid or gas.________ 3. Water can dissolve anything that is why it is called the universal solvent.________ 4. When a liquid solute is dissolved in a liquid solvent they are said to be soluble to one another.________ 5. Solubility of solids, liquids and gas increases with increasing temperature. Key to answers on page 25.Lesson 1. What Distinguishes Solutions from Non-Solutions? When you are preparing coffee in the morning, you are actually making a solution. Itis considered a solution because it is a homogenous mixture of two or more substancesevenly distributed in each other. The coffee and sugar that you mix in the hot water are thesolutes or the substances dissolved, and the solvent is the hot water, the liquid material inwhich the solute has dissolved. These are the components of a solution. The solute can bein the form of solid, liquid, or gas. -3-

To know more about the general properties of solutions, try to do Activity 1.1.What you will doActivity 1.1 Solutions vs. non-solutionsMaterials: Sand Soy Sauce Glass of water Spoon Salt Bond papersProcedure:1. Add some sand to a glass of water and stir. To another glass of water, add some salt, stir and taste it. Observe the results. Which of the two set-ups dissolved the solute completely? ____________________________________________________________2. Leave the two set-ups for five minutes and take note of the results. Did the sand dissolve and form a solution? How about the salt? _____________________________3. Prepare another set-up by mixing soy sauce with water. Stir and taste it. Compare the third set-up with the two previous set-ups by observing the three samples in a transparent glass under sunlight, or you may use a flashlight and allow light to pass through the samples. Which of the three set-ups form a clear solution? (Note: A clear solution is not necessarily colorless, but is transparent to light.) ____________________4. Add more salt, sand, or soy sauce to each corresponding set-up and stir. For the second time, taste the salt and soy sauce set-ups. Are they saltier than the first taste test? ___________________________________________________5. Get a piece of bond paper and fold it to form a cone. Allow the three set-ups to pass through this improvised filter paper. Can they be separated by physical means? ______________________________________________________________________Analysis:1. Identify the solute and solvent in each of the three set-ups.2. Which of the three set-ups forms a solution? Why?_____________________________ ______________________________________________________________________3. What property of solution is evident in Step 1? 2? 3? 4? 5? Explain. ______________________________________________________________________ ______________________________________________________________________ You have just explored the different properties that distinguish a solution from a non-solution. There are several ways of classifying solutions. It can be based on the phase ofthe solution and according to the relative amounts of the components of the solution. -4-

Did you know? The oldest way of making salts is through solar evaporation. The collected sea water is made to evaporate under the sun and is allowed to crystallize to form rock salts in the process of crystallization. Crystallization is the process in which dissolved solute comes out of the solution and forms crystals. Crystals form through crystallization in warm solutions that have cooled or evaporated. These rock salts are added to food to give it a salty taste. In terms of phase, a solution can be classified as gaseous, solid or liquid. Air is anexample of a gaseous solution. It is the combination of nitrogen, oxygen, and carbondioxide gases. Certain alloys such as brass (solution of zinc and copper) and coinage silver(solution of copper and silver) are examples of solid solutions. Seawater and blood areexamples of liquid solutions. Can you think of other examples of gaseous, solid or liquidsolutions that you encounter in your daily activities? Solutions are also classified as dilute and concentrated based on the relativeamounts of components of the solution. When the solute is present in small amounts, thesolution is said to be dilute. But when the solute is present in considerably significantamounts, the solution is said to be concentrated. However, solutions must not be confused with suspensions. A basic example ofsuspension is the sand mixed with water as was shown in our activity. As you can see, thesand is not dissolved in water; it is merely suspended in it. The suspension produced is notalso clear; it is actually opaque. And upon standing, you can see that the sand is slowlysettling. In this regard, the composition of suspension is actually changing as the sandsettles in, so it is a heterogenous mixture, unlike that of the solution. Also, when the clayparticles are allowed to pass through the improvised filter paper, the sand is left behind.This is because the particles of a suspension are too large to pass through the filter paper,or through a membrane with finer openings.Energy in the Dissolution Process Get a glass of water and put a pinch of powder dye. Observe the pattern of themovement of dye in water. What you just did best describes the dissolution process whichfascinates chemists for years. The dissolution process is due to intermolecular forces(forces that exist between molecules). When two molecules overcome the attractive forcesthat exist between them, they absorb energy. On the other hand, when the attractive forceis formed between two molecules, they release energy. Thus, we can say that dissolutionprocess involves the absorption (endothermic) and release (exothermic) of energy. Whenthe energy absorbed is greater than the energy released, the dissolution process isendothermic. The opposite scenario is true for exothermic process. -5-

Think? The process of removing stain in shirts involves a process of dissolving stain in water. It is an exothermic process because energy absorbed between stain and water is less than the energy released when blood is hydrated. How can you easily remove the stain using water alone? Solubility is the measure of how much solute will dissolve in a solvent at a specifictemperature. The famous saying “like dissolves like” is important in predicting the solubilityof a substance in a given solvent. This expression means that when two substances ofsame intermolecular forces and magnitude are mixed, they are likely to be soluble to oneanother. Take for example salt (NaCl) and water (H2O) which are both polar. When thesetwo substances are mixed, the attraction between NaCl and H2O molecules has the samemagnitude to the forces between NaCl molecules and between H2O molecules. However,solubility is often confused with miscibility. Miscibility is the term used when two liquids arecompletely soluble with one another in all proportions while solubility is the term used whena solid or gas is completely soluble in liquid in all proportions. For example, alcohol andwater are miscible to one another.What you will doSelf-Test 1.1Now that you are through with the first lesson, try to answer the following and see foryourself how much you learned.Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. What do you call a mixture that is not evenly distributed in all proportions and cannot beseparated by filtration?a. colloid c. suspensionb. solution d. homogenous mixture2. All of the following will exhibit a homogenous mixture EXCEPTa. salt and water c. sugar and waterb. sand and water d. alcohol and water3. When oil is mixed with water, the mixture is said to be __________ to one another.a. soluble c. insolubleb. miscible d. immiscible -6-

4. What is the dissolving substance in a solution?a. water c. solventb. solute d. particles5. Substance A was dissolved in water, and the energy absorbed in overcoming theattractive intermolecular forces between the molecules is 500 kJ. The energy releasedwhen substance A was dissolved in water is 800 kJ. How will you describe thedissolution process?a. basic c. endothermicb. acidic d. exothermic Key to answers on page 25.Lesson 2. What Makes a Solute Dissolve Faster? The solubility of a substance is the maximum amount of solute that can be dissolvedin a certain amount of solvent at a certain temperature. In this case, water is assumed asthe solvent unless otherwise stated. But, there are also other solvents that can be used.One common solvent is alcohol. An alcohol solution used medicinally is called “tincture”.Thus, a tincture of iodine contains iodine dissolved in alcohol. There are actually different factors that contribute to the rate of dissolution of solute inthe solution. To clearly understand these factors, do Activity 2.1.What you will doActivity 2.1 How can we make solutes dissolve faster?Materials: Sugar Oil Glass of hot water Mortar and pestle Glass of cold water Boiling water in a kettleProcedure:1. Place a spoonful of sugar in a glass of hot water and in a glass of cold water. Where did sugar dissolve faster? Why?2. Repeat step 1. But this time, stir the sugar simultaneously in both glasses until it completely dissolves. In which of the two set-ups did sugar dissolve faster?3. Using a mortar and pestle, pound the sugar simultaneously until it is completely -7-

pulverized. Then, add the powdered sugar to the glass of hot water and cold water and stir. Compare the result with Step 1.4. Mix oil in a glass of hot water and in a glass of cold water. Take note of the result. Did the oil dissolve in water?5. Heat water in the kettle until it boils. Observe the bubbles formed on the side of the kettle even before the water boils. What happens to the bubbles as the water “boils out”?Analysis:1. What happens when sugar is stirred?2. What happens to the rate of solubility of sugar when heat is applied?3. What is the effect of particle size on the rate of dissolving?4. What can you infer about the solubility of gases (water vapor) as the temperature increases as in step 5?Factors Affecting Solubility of a SoluteA. Temperature In the previous experiment, you were able to dissolve excess solute upon heating.Thus, temperature affects the solubility of substances. The solubility of a solid in liquidusually increases with increasing temperature since most dissolution process that involves asolid solute over a liquid solvent is endothermic. However, in the case of the solubility ofgas in liquid, increasing the temperature will usually result to decrease in solubility. Did you know? Are you a softdrinks addict? The reason why we love to drink carbonated drinks is because of the biting taste caused by carbon dioxide gas dissolved in the solution. Cleopatra craved for it, too. She used pearls in her beverage. These pearls contained calcium carbonate which could react with water to produce carbon dioxide. But you know how expensive pearls are nowadays. Today, carbon dioxide is dissolved in liquid by decreasing the temperature and increasing the pressure. This is how commercially available carbonated drinks are made.B. Nature of Solvent The ability of a solute to be dissolved in a given solvent is affected by the type of -8-

bond of both the solute and solvent. In general, polar liquids dissolve polar compounds andthe same is true with nonpolar liquids. Thus goes the saying “like dissolves like”. This is theguiding rule in preparing solutions. Since water is a polar liquid, it dissolves polarcompounds such as sugar (C6H12O6) and salt (NaCl). Sometimes, in spite of all attempts, asubstance does not appreciably dissolve in water even though it is polar. These substancesare called insoluble salts (Table 2.1)Table 2.1 Water Solubililty of Common Salts and BasesSoluble Insoluble Sodium salts Carbonates (except sodium, Potassium salts potassium and ammonium) Ammonium salts Phosphates (except sodium, potassium and ammonium) Acetates Sulfides (except sodium, potassium Nitrates Chlorides (except silver, lead, and and ammonium) mercury) Hydroxides (except sodium,Sulfates (except calcium, barium and potassium and ammonium) lead)C. Pressure The effect of pressure on the solubility of a solid or liquid solute is not as noticeableas that of a gaseous solute. The solubility of gases in water usually increases withincreasing pressure. This relationship is first noticed by William Henry (1775-1836) in 1803and its law is called after his name—Henry’s law or Pressure-Solubility Law. This lawstates that in a given temperature, the mass of a gas that dissolves in a given volume ofliquid is directly related to its pressure. An example is the carbon dioxide gas dissolved in carbonated drinks. If the bottle isclosed, the carbon dioxide gas remains dissolved in the beverage due to considerableamount of partial pressure present. But if it is opened, the pressure is released and carbondioxide gas bubbles off the liquid due to a decrease in gas solubility. Did you know? On a hot summer day, an experienced fisherman wouldusually pick a deep spot in the river or lake to cast the baitbecause the oxygen content is greater in the deeper, coolerregion. Alas! most fish will be found there. So next time you tryfishing on a summer day, try this secret technique for a moreenjoyable recreation. -9-

Factors Affecting the Rate of DissolutionA. Surface area Surface area does not affect the amount of solute that will dissolve, but it does affectthe solute’s rate of dissolution. Thus, in order to make a solid solute dissolve faster, wefrequently powder it, thereby increasing the surface area. This is why powdered coffeedissolves faster than granulated coffee even without stirring.B. Rate of Stirring The rate at which a solute dissolves can be increased by stirring the mixture. Thisprocess brings fresh solvent into contact with the solute and so permits a faster rate ofdissolution.C. Temperature Generally, solubility increases with increasing temperature for most cases of solid inliquid. The increase in temperature causes an increase in kinetic energy of the solute,solvent and the solution thus facilitating rapid interaction with one another.Saturated, Unsaturated and Supersaturated Solutions Solutions can also be classified as saturated, unsaturated and supersaturated.When a small amount of sugar is mixed in a glass of water, all the sugar will dissolve. Ifmore and more sugar is added while stirring, a point is reached when some sugar will settleat the bottom of the glass even with continued rapid stirring. This type of solution is said tobe saturated. Thus, a saturated solution is one that contains as much of the solute as it canhold at a given temperature. An unsaturated solution contains less solute than it has thecapacity to dissolve. The third type, the supersaturated solution contains more solute than is present in asaturated solution. This is a rather unstable condition. In this case, the excess solid willeventually separate from the solution as a precipitate in a process known as precipitation oras crystals in a process known as crystallization. Crystals have bigger granules thanprecipitate. Tawas or alum crystals are prepared as supersaturated solutions. What you will do Self-Test 2.1Again, try to check how much you have learned from the lesson by answering the followingquestions. - 10 -

Multiple Choice. Choose the letter of the best answer. Write the chosen letter on aseparate sheet of paper.1. Which of the following affects both the rate of solution and the solubility of a solid in aliquid?a. stirring c. increasing surface area exposedb. increasing pressure d. increased in temperature2. Which solution contains more dissolved solute than a saturated solution contains underthe same conditions?a. saturated c. concentratedb. unsaturated d. supersaturated3. All of the following are soluble in water EXCEPTa. lead acetate c. lead chlorideb. lead sulfate d. lead phosphate4. Which substance usually becomes less soluble with increasing temperature?a. gas c. liquidb. solid d. all of the above5. All of the following affect the solubility of solid in liquid EXCEPTa. pressure c. surface areab. temperature d. nature of solvent Did you encounter any problem? Well, compare your answers with the answer keyand see for yourself the items you missed. Good luck! Key to answers on page 25.Lesson 3. How Much Solute Can Be Dissolved in a Solution? In order to clearly understand solutions, we must know how much of the solute ispresent in the solution and also how to control the amounts of solute used to bring about asaturated, unsaturated or supersaturated solution. The concentration of a solution is the amount of solute present in a given quantityof solvent or solution. For simplicity of the discussion, we will assume the solute is a liquidor solid and the solvent is a liquid. Chemists use several different concentration units, eachof which has advantages as well as limitations. Let us examine four most common units ofconcentration: percent composition, mole fraction, molarity, and molality. - 11 -

Types of concentration unitsA. Percent composition The percent of the solute in the solution is referred to as percent composition. Inexpressing the percent concentration, the units mass or volume are used. The following arethe different ways of expressing percent composition.1. Percent by mass (% m/m) The percent by mass (also called percent by weight or weight percent) is the ratioof the mass of solute to the mass of solution, multiplied by 100 percent: Percent by mass of solute = mass of solute × 100% mass of solution Sometimes, percent by mass is also termed as mass fraction only that thepercentage multiplier is deleted. Mass fraction is just the fraction of the mass of thesolute over the total mass of the solution. Mass fractions of very dilute solutions areoften expressed in parts per million (abbreviated as ppm). One part per million isequivalent to one gram of solute per one million grams of solution, or one milligram ofsolute per thousand grams of solution (1mg/kg). For even smaller mass fractions, partsper billion (ppb) and parts per trillion (ppt) are often used. And as their name implies, wecan express the mass fraction as parts per million by multiplying it by 1 000 000 ppm(106ppm) or 1 000 000 000 ppb (109 ppb) and so on.Sample Problem: A healthy snack claims to contain “low sodium” (Na+). It contains 30 mg of Na+(molecular weight is 23 g/mol) in each 8-oz serving (8 oz = 250 mL). What is theconcentration of sodium in terms of ppm? (1ppm = 1mg/L).Explanation:30 mg of Na+ × 1000 mL = 120 mg/L 250 mL 1LSince 1 ppm is equivalent to 1 mg/L, then sodium content is also 120 ppm.2. Percent by volume (% v/v) On the other hand, the percent by volume is expressed the same as percent bymass only that mass is changed to volume, as given by the equation: - 12 -