LAB 14: COLLIGATIVE PROPERTIES: SURFACE TENSION, MELTING & FREEZING POINTS OF SOLUTIONS

LAB 14: COLLIGATIVE PROPERTIES: SURFACE TENSION, MELTING & FREEZING POINTS OF SOLUTIONS PURPOSE: To demonstrate the effect of solutes on the surface tension of water. To determine the effect of solutes on the boiling point of water. To freeze a mixture of milk, eggs, sugar, and cream by using a salt-ice freezing point depression. SAFETY CONCERNS: Always wear safety goggles. Handle and dispose of broken glass safely. SURFACE TENSION: If you fill a glass up to its rim with water and then carefully add more water drops you will find that instead of overflowing the container the water will adhere to itself, forming a dome that rises above the rim of the glass. This effect is the result of the polarity of water. Throughout the liquid in the glass, water molecules are attracted in all directions by surrounding water molecules. However, the water molecules on the surface are pulled like a skin toward the rest of the water in the glass. As a result, the water molecules on the surface become more tightly packed, a feature called surface tension. Because of surface tension, a needle floats on the top of water, certain water bugs can travel across the surface of a pond or lake, and drops of water are spherical. When compounds called surfactants are added to water, they disrupt the hydrogen bonding between the water molecules. As a result, the surface tension is decreased and the water spreads out rather than forming drops. Soap, detergents, shampoos, and fabric softeners are examples of surfactants we use every day. The lungs produce a surfactant that allows efficient exchange of oxygen and carbon dioxide between the alveoli and the capillaries. When premature babies lack an adequate amount of this surfactant, they have difficulty breathing, a condition known as respiratory distress syndrome (RDS). A surfactant may be placed in the lungs and the oxygen increased until the baby produces a sufficient supply of its own surfactant. COLLIGATIVE PROPERTIES: Solution properties that depend only on the concentration of solute particles in solution are called colligative properties. The identity of the solute does not matter. Vapor pressure, boiling point, and freezing point are colligative properties that depend on concentration of solute present. Vapor Pressure: Substances such as gasoline and alcohol have high vapor pressures and evaporate quickly from open containers. Water, on the other hand, has a much lower vapor pressure at equivalent temperatures and evaporates at a much lower rate. In order to bring water to a boil at standard atmospheric pressure, its temperature must be raised to 100 C. At this temperature, the molecules have sufficient energy to overcome the forces of attraction that keep them together and go into the vapor phase. The vapor pressure of water has been raised to a point where it is now equal to the prevailing atmospheric pressure. Throughout our discussion of vapor pressure and its relationship to boiling points, we have been careful to emphasize we were dealing with pure substances. Pure water boils at 100 C and freezes at 0 C when the pressure is 760 torr (1 atm), but what happens when molecules or ions of a solute are added to our solvent? In essence, you are diluting the water molecules, actually diluting the solvent itself. The solute CH105 Lab 14: Colligative Properties (F15) 37

particles get in the way of the H 2 O s. They can t act like pure water molecules anymore. Hydrogen bonding and dipole attractions to each other are altered. So, for example, when you try to boil water that has a solute in it, the H 2 O molecules can t escape to vapor as easily. There are fewer H 2 O molecules at the surface because some of the surface spaces have been taken up by the solute particles. Many of the H 2 O molecules on their way to the surface are also colliding with the solute particles and bouncing into the depths of the solution. Because of this, the water solution s vapor pressure is lower than that of pure water. Boiling Point: You can get the number of escaping water molecules in a solution to increase by speeding them up. So to get a solution to boil, you have to heat it hotter than if it were just pure water. In other words, the boiling point of the solution will be higher than 100 C, and the more solute particles there are, the higher the boiling point will be. The relationship between the change in boiling point and the concentration of a solution is expressed as: t b = ik b m t b is the boiling point increase as compared to pure solvent i is the number of ions or particles formed in solution from 1 unit of solute (called the van t Hoff factor) K b is the boiling point elevation constant for the solvent (also called the ebullioscopic constant) m is the molality of the solution in mols solute per Kg of solvent. i K b m T b = i K b m = # mol solute particles 1 mol solute C o Kg solvent mols solute 1 mol solute particles 1 Kg solvent = C o Water s boiling point goes up 0.52 C o for every mole of solute particles it contains per kilogram of water so the Kb for water is 0.52. Freezing Point: Water with a solute in it should have a harder time getting its molecules all lined up neatly into the solid crystal arrangement of ice. So we would have to cool it down (to slow down the molecules) an extra amount below 0 C, if we want it to freeze. Water s freezing point actually goes down 1.86 C for every molal s worth of solute particles it contains. In fact, any solute will affect (lower) the freezing point of a solvent by a more significant amount than it will affect (raise) the boiling point. The relationship between the freezing point and the concentration of a solution is expressed as: t f = ik f m t f is the freezing point decrease as compared to pure solvent i is the number of ions or particles formed in solution from 1 unit of solute (called the van t Hoff factor) K f is the freezing point depression constant for the solvent (also called the cryoscopic constant) m is the molality of the solution in mols solute per Kg of solvent. We use the principle of freezing point depression in numerous ways. Solute molecules (called antifreeze) are added to the water in car radiators. Antifreeze both lowers the freezing point and raises the boiling point so it doesn t freeze in the winter and is less likely to boil over in the summer. Any solute molecules or ions would do the job but we use ethylene glycol because it is non-corrosive and non-volatile. It doesn t corrode the cooling system or evaporate away when the engine is hot. 38 CH105 Lab 14: Colligative Properties (F15)

Dissolving salt in water will lower its freezing point. For example, to melt ice on streets and sidewalks, salt is often sprinkled on it. Unless it is extremely cold, there is always a slight film of liquid water on the surface of ice, (at least there is as soon as someone steps on it or a car drives over it). The salt dissolves in this film of water, lowering its freezing point (and that of the ice surface it is in contact with ) below the air s temperature, and the surface layer of the ice melts. This puts salt water in contact with the next layer of ice molecules, and they melt, and so on until either all the ice is melted or the salt water gets too dilute to melt more, considering the temperature of the weather. Anything that dissolves in water could be used. Salt is used because it is cheap. Unfortunately, it s corrosive to steel. Many northern states are facing a problem where their bridges may start falling down because of the tons of salt that have been spread on their roadbeds. When there is a lot of ice melting going on in a short time, there is a lot of heat of fusion being used up. Because of this, near an ice/salt mixture, it gets pretty cold. This rapid absorption of heat is the reason we can enjoy ice cream during the summer months. The following table lists the K b and K f values for a few common solvents. K b (C o Kg solvent) (1 mol solute particles) Benzene 2.53 5.12 Camphor 5.95 40 Carbon tetrachloride 4.95 30 Diethyl ether 2.02 1.79 Water 0.52 1.86 K f (C o Kg solvent) (1 mol solute particles) Examples: 1) Determine the boiling point for a solution made up with 2.5 moles of ethanol (C 2 H 6 O) in 1 Kg of H 2 O. Use the equation t b = ik b m i K b m T b = i K b m = # mol solute particles 1 mol solute C o Kg solvent mols solute 1 mol solute particles 1 Kg solvent = C o i for ethanol = 1 This is because as a covalent compound it does not ionize so gives just 1 mol of particles per mole of ethanol. K b for water = 0.52 m for the solution given is calculated as Now to plug into our equation: 2.5 mol Ethanol 2.5m 1Kg water i K b m T b = i K b m = 1 mol particles 1 mol EtOH 0.52 C o Kg water 1 mol particles 2.5 mols EtOH 1 Kg water = 1.3 C o Therefore, the water should boil at 100 o C + 1.3 C o = 101.3 o C CH105 Lab 14: Colligative Properties (F15) 39

2) Determine the boiling point for a solution made up with 2.5 moles of NaCl in 1 Kg of H 2 O. In the previous example ethanol gave only 1 particle for every molecule dissolved in water but if we make up a solution of 2.5 moles of NaCl in 1 Kg of water each NaCl makes 2 particles that get in the H 2 O s way. NaCl is a strong electrolyte so we know it is forming ions in a water solution. Each dissolved NaCl dissociates completely into a Na 1+ and a Cl 1-. So 2.5 m NaCl is actually 5.0 m (5.0 molal) of particles. A weak electrolyte would give an effect somewhere in between the nonelectrolyte alcohol and the strong electrolyte NaCl. Again we ll use the equation t b = ik b m i K b m T b = i K b m = # mol solute particles 1 mol solute C o Kg solvent mols solute 1 mol solute particles 1 Kg solvent = C o i for NaCl = 2 This is because NaCl ionizes to give 1 mol of particles per mole of NaCl. K b for water = 0.52 m for the solution given is calculated as Now to plug into our equation: 2.5 mol NaCl 2.5m 1Kg water i K b m T b = i K b m = 2 mol particles 1 mol NaCl 0.52 C o Kg water 1 mol particles 2.5 mols EtOH 1 Kg water = 2.6 C o Therefore, the water should boil at 100 o C + 2.6 C o = 102.6 o C Calculations of expected freezing points are done very similarly to the prediction of boiling points. The only differences are 1) we use the K f rather than the K b, and 2) we must remember that our final freezing point is below that of the pure solvent rather than above. 40 CH105 Lab 14: Colligative Properties (F15)

PROCEDURES: ACTIONS: I. SURFACE TENSION 1. Fill a shallow, soap-free 1 dish or pan with water to a depth of about 1-2 cm. 2. Float a small piece of tissue paper (about 3 cm 2 ) on the surface of the water. 3. Rest a soap-free straight pin on the tissue paper. 4. Using soap-free fingers, very carefully submerge each side of the tissue paper. The pin should remain on the surface of the water. 5. Slowly add a drop of liquid dish soap or detergent into the water on the opposite side of the pan from the pin. Watch and wait. Record your observations in table I of the report sheet. 2 II. BOILING POINT 1. Obtain 3 small Erlenmeyer flasks (10 ml size) and label them A, B, and C. 2. Into flask A put 10 mls deionized water. Into flask B put 10 mls of a 2 m solution of sodium chloride (salt = NaCl) Into flask C put 10 mls of a 2 m solution of sucrose (sugar = C 12 H 22 O 11 ). 3. Place each flask (A-C) on a hot plate and gently 4 heat to boiling taking care not to let any water boil away. 4. With a digital thermometer, read and record (Box IIA, B 1, C 1 ) the boiling temperature of each solution to the nearest 0.1 C o. 5. Determine the Experimental T bnacl and T bsug relative to the boiling point of your water sample and record in (Box IIB 2& C 2 ) 6. Using the boiling point elevation constant of water 5, calculate the theoretically expected change in boiling points (Theoretical T b ) of your solutions. 6 (Box II B 3& C 3 ) 7. Compare the theoretically calculated T b s (from Box IIB 3& C 3 ) with the actual experimental boiling point changes 7 (from IIB 2& C 2 ) and determine a percent error. 8 (Box IIB 4& C 4 ) NOTES: 1 There must be no soap residue as that will disrupt the ability of water molecules to hydrogen bond with other water molecules and therefore decrease the surface tension. 2 If nothing happens add another drop of detergent. An alternative demonstration is to sprinkle pepper on a clean surface of fresh water and then, with a drop of soap or detergent on your finger touch the water at the edge of the dish. Observe the result on the floating pepper. 3 Molality (m) is a measure of the number of moles of solute per 1 Kg of solvent. m = mol solute Kg solvent In this case the solute is sucrose or sodium chloride and the solvent is water. 4 Gently heat to boiling. If the boiling is too vigorous then the water will boil away, changing the mass of water present and therefore changing the molality of the solutions you are using. 5 The boiling point elevation constant for water is K b = 0.52 C o mol solute particles 6 Remember that 1 mol of NaCl dissociates into 2 mols of ions. 7 The error is how far off you were from what you expected. The error is the difference between your actual T and the expected T. 8 The percent error is an indicator of the relative magnitude of error you had compared to the magnitude of the measured values used. % error = Error X 100 Theorecital CH105 Lab 14: Colligative Properties (F15) 41

III. ICE CREAM: FREEZING POINT 1. Pour a cooled ice cream mixture 9 into a small (sandwich-size) zip lock bag until it is about 1/3 full. 2. Into a larger ziplock bag (quart or ½ gallon), add ice and salt so that the amount of ice is about 4 times the amount of salt. 3. Place the small ziplock bag of ice cream mixture into the larger bag containing salt and ice, making sure that the bags are well sealed. 10 4. With your partner, take turns kneading 11 the zip lock bags. 12 5. When very little ice remains, you may need to add more. Carefully pour off some of the water and add more ice and salt. 13 6. When a desired ice cream consistency has been achieved, 14 remove the inner bag and spoon out the solidified ice cream into cups. Enjoy your ice cream. Record your observations (Box IIIA) 7. With your thermometer, check and record (Box IIIB) the temperature of the water-ice-salt mixture left in the larger bag. 9 Vanilla Ice Cream Mixture: Combine and mix thoroughly: 3Tbl sugar, 1.5 tsp vanilla, 3 c. heavy cream Mix in: 3 c. milk Add and stir well: fruit or flavor of your choice. Chill. Freeze. 10 If the bags are not sealed tightly you will get salty ice cream 11 We need a cold temperature and continuous motion of the cream. The frozen cream will be smooth (not rocky) if it is in motion as it freezes. The continuous motion prevents the molecules from stacking into large crystals that would give a gritty texture. 12 Your hands will get cold as you continually knead the bags, but it will freeze fast. The ice will melt quickly. 13 You won t need to add salt if there is still some undissolved in the bag. 14 This could take about 10-15 minutes depending on the temperature of your ice/salt mixture. 8. From your ice/salt bath temperature, determine the change in freezing temperature ( T) of the ice caused by the salt. (Box IIIC). 9. From your experimental T, calculate the molality of your salt solution. Show calculations. (Box IIID). 42 CH105 Lab 14: Colligative Properties (F15)

LAB 14: COLLIGATIVE PROPERTIES: PRE LAB EXERCISES: NAME DATE 1. In the preparation of Ice Cream, the small zip lock bag should be A. filled only half full with ice cream mix. B. filled completely with ice cream mix and any excess air pressed out. C. left partially unzipped to allow for expansion of gases. 2. How does kneading or churning ice cream make the ice cream smooth rather than full of noticeable ice crystals? A. Kneading serves only to mix the ingredients. B. Kneading helps the molecules find each other so they stack into a solid formation. C. Kneading keeps the molecules from organized stacking so they can t make large crystals. D. More than one of these is correct. 3. Calculate the boiling point of a solution made by dissolving 29.4 g of calcium chloride (CaCl 2 ) in 100. g of water. Show your calculations and circle your answer. 4. Calculate the freezing point of a solution made by dissolving 171 g of sugar (C 12 H 22 O 11 ), a covalent molecule, in 1.00 kg of water. Show your calculations and circle your answer. 5. Calculate the freezing point of a solution made by dissolving 171 g of NaCl, an ionic compound, in 1.00 kg of water. Show your calculations and circle your answer. 6. Devise an explanation to explain why/how salt lowers the freezing point of water and so causes ice on roads or sidewalks to melt. (Tell why don t just say it does.) CH105 Lab 14: Colligative Properties (F15) 43

44 CH105 Lab 14: Colligative Properties (F15)

LAB 14: COLLIGATIVE PROPERTIES: NAME REPORT: PARTNER DATE II. BOILING POINT: Salt (NaCl) -Water Solution A. Boiling Point of pure Water recorded : Actual T bh2o = B. Boiling Point of Salt solution recorded : Actual T bnacl = B 2. Boiling point change due to NaCl: Actual (Experimental) T bnacl = B 3. Expected (Theorietical) T bnacl of a 2 m Salt sln Show calculation B 4. % Error = Show calculation Error (B3-B2) x 100 = Expected Sugar (Sucrose C 12 H 22 O 11 )-Water Solution A. Boiling Point of pure Water recorded : Actual T bh2o = C. Boiling Point of Sugar solution recorded : Actual T bsug = C 2. Boiling point change due to Sugar: Actual (Experimental) T bsug = C 3. Expected (Theorietical) T bsug of a 2 m Sugar sln Show calculation C 4. % Error = Show calculation Explanation/Analysis: Compare salt and sugar and explain why the same mass of each has different influences on boiling point. Explain any errors or anomalies in your results. What did you expect? CH105 Lab 14: Colligative Properties (F15) 45

I. SURFACE TENSION: Observations: Before addition of Soap Observations: After addition of Soap Explanation/Analysis: What is the effect of solutes on the surface tension of water? Why? How? III. FREEZING POINT: ICE CREAM A. Observations: B. Final temperature of Salt/ice solution recorded Actual T fnacl = C. Change in Temperature T fnacl = D. molality (m) of your salt solution. Remember that T fnacl = ik f m Show calculations Explanation/Analysis: Explain why adding a solute to ice causes the ice cream to get cold enough to freeze. Why did the temperature go down? Why does salt lower the freezing point of water? RELATED EXERCISES: 1. Which would be a more effective solute to use to lower the freezing point of ice? A. 1 mol NaCl B. 1 mol CaCl 2 C. 1 mol Sugar D. all are the same. 2. A. If you were to mix liquid A at 20 o C with an equal amount of liquid B at 0 o C what would be the temperature of the final mixture? A. 20 o C B. 0 o C C. 10 o C D. -10 o C E. Not enough information 3. A brine solution (salt already dissolved in water) that was at 15 C would be at making ice cream than a solid salt-ice-mixture as used in this experiment that reached a final temperature of 15 C. A. more effective B. less effective C. just as effective Explain your answer: 46 CH105 Lab 14: Colligative Properties (F15)