Materials Included In Kit

Additional Materials Required

Safety Precautions

Boiling water poses a serious burn risk. Have students use caution when handling hot beakers, hot oil and hot plates. Calcium chloride is slightly toxic by ingestion. Sodium chloride and glucose are not considered hazardous; however, the chemicals provided are for laboratory use only and are not intended for human consumption. Wear chemical splash goggles and chemical-resistant gloves and apron. Have students wash hands thoroughly with soap and water before leaving the laboratory. Please review current Safety Data Sheets for additional safety, handling and disposal information. 

Disposal

Please consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. All of the waste solutions may be disposed of according to Flinn Suggested Disposal Method #26b.

Lab Hints

• The experiment will take two 50-minute lab periods to complete.
• The use of digital thermometers is encouraged. They are fast, accurate, safe and very easy to read.

Part 1.

• Osmosis is observed by placing a solution of low concentration into dialysis tubing and placing this into a beaker containing a solution of higher concentration. This causes the water to exit the tubing in an attempt to equalize the concentration |Of sugar water on each side of the semi-permeable membrane.
• The boiling point elevation in Part 1 is used to demonstrate relative molality values of final solutions. Students will get a dramatic example of the concentration effect on the boiling point.
• Review with students the proper techniques for sealing the dialysis tubing.

Part 2.

• The lab works significantly better with crushed ice rather than with ice cubes. If crushed ice is not available, place some ice cubes in two layers of zipper-lock bags. Using caution, pound the ice with a hammer or other hard object to crush it into small pieces.
• Have each student group use the same thermometer throughout the lab to reduce errors resulting from inaccurate thermometers.
• Distilled or deionized water will produce the best results; however, tap water can also be used as the source of water and ice in this lab. Any ions in the water will be present in every mixture, including the control, and thus will not affect comparative results. The concentration of ions in the local tap water, however, may cause the results to vary slightly from the sample data provided.
• Remind students to use the wooden stirrer to stir the solution and the thermometer strictly to measure the temperature. Also caution students not to leave thermometers standing in beakers unattended as this is a common cause of broken thermometers and spilled solutions.
• Have students hold the thermometer slightly off the bottom of the beaker when reading temperature so that the thermometer is surrounded on all sides by the mixture.
• For accurate temperature measurements, both water and ice must be present in the beaker, indicating that the mixture is at the melting/freezing point.
• This lab is designed as an introduction to colligative properties and their applications. The high concentrations of solutes used in this experiment would be expected to show fairly large deviations from ideality. The high concentrations demonstrate to students, however, just how much the freezing point of water can be lowered using simple salts. Most students are surprised to discover that they can lower the freezing point of water to almost –20 °C using just table salt. In a more rigorous advanced lab, the freezing point of a dilute solution would typically be determined by graphical analysis of a cooling curve and then used to calculate the molar mass of an unknown solute.

Part 3.

• When the heating is stopped and the heating bath cools down, the temperature of the sample will also cool. When the heating bath and the sample reach the precise temperature of its boiling point, the vapor pressure of the liquid inside the capillary tube matches the atmospheric pressure above the boiling tube and the stream of bubbles will stop. Since the vapor pressure of the liquid now equals the pressure of the atmo sphere, the boiling point has been reached. The thermometer can now be read to determine this characteristic temperature.
• It is best to determine the boiling point of the sample as it is cooling rather than heating. Sometimes a sample can superheat and not boil until the temperature is above its boiling point. This is especially true when using new micro test tubes. Consistent and accurate boiling points are easier to attain when the sample cools and the boiling stops.
• Do not allow oil to enter the microscale test tube containing the liquid sample.
• The same capillary tube may be used for all three boiling point determinations.

Answers to Prelab Questions

1. The molality of a solution can be determined using Equation 1 if the boiling point elevation is measured and the boiling point constant (K_b) of the solvent is known. The value of K_b for water is 0.512 kg•K/mol.

ΔT = m x 0.512 kg•K/mol

ΔT = Boiling point of solution – Boiling point of pure water

A solution was prepared by dissolving 12.00 g glucose in 100.0 g water. The resulting solution was found to have a boiling point of 100.34 °C. Calculate the molar mass of glucose. Glucose is a molecular solid that exists as discrete molecules in solution.

ΔT = 0.34 °C, K_b = 0.512 kg•K/mol

2. A solution containing 20 g of sodium chloride, NaCl, is dissolved in 200 mL of water. A second solution has 40 g of sodium chloride dissolved in 200 mL of water.

1. If the boiling points of each solution are determined, which solution has the highest boiling point?

   The solution with 40 g of sodium chloride dissolved in 200 mL of water. The greater the solute concentration, the greater the boiling point elevation, ΔT_b.

2. 50 mL of the solution with 20 g of NaCl dissolved in 200 mL of water is placed in a section of semipermeable dialysis tubing and then sealed. The tube is placed in the 200 mL solution containing 40 g of sodium chloride. After a period of time, the boiling point of the solutions is determined.
   
   Is the boiling point greater than, equal to, or less than each solution’s original boiling point? Explain.
   
   Initially the solution in the beaker is twice as concentrated as the solution in the dialysis tubing. The water in the tubing moves across the tubing membrane to equalize, if possible, the concentration of sodium and chloride ions on each side of the membrane. This results in the solution in the beaker being less concentrated than initially, and the solution in the tubing being more concentrated.
   
   The beaker solution, now being less concentrated in solute, would have a lower boiling point. The solution in the tubing, with an increase in solute concentration, would have a higher boiling point.

3. The freezing point depression of a solution is dependent on the number of particles in solution, not the type. If 0.1 moles of each of the following soluble ionic compounds, sodium phosphate, Na₃PO₄, and sodium chloride, NaCl, are added to separate beakers containing 1-L of distilled water, which would have the lower freezing point? Explain.

When sodium phosphate dissolves in water, it produces four ions in solution

Na₃PO₄(s) → 3Na⁺(aq) + PO₄^3–(aq)

When dissolved, sodium chloride produces only two ions in solution.

NaCl(s) → Na⁺(aq) + Cl^–(aq)

Because 0.1 moles of sodium phosphate would produce twice as many ions in solution as sodium chloride, in equal volumes of water, the sodium phosphate solution would have a higher boiling point.

Sample Data

Part 1. Osmosis

Part 2. Freezing Point Depression
Part 3. Boiling Point Elevation

Answers to Questions

Calculations

Part 1. Osmosis

1. Calculate the initial molality of beakers 1 and 2 and beakers 3 and 4. (molality = moles solute/kg solvent) (Molar mass of glucose = 180.16 g/mole. Assume the mass of water is 0.1000 kg.)

a. Beakers 1 and 2.

b. Beakers 3 and 4.

2. For boiling point elevation:

K_b for water equals 0.512 kg•K/mol

1. Calculate the concentration of glucose in beakers 1 and 3 from their boiling point data and compare these calculated values to the actual concentrations.
   {13113_Answers_Equation_8}

   For beaker 1 the calculated value of 0.98 mol/kg compares favorably with the actual value of 0.95 mol/kg. For beaker 3, the calculated value of 1.76 mol/kg is still close to the actual value of 1.90 mol/kg.

2. The initial concentration of the solution in beaker 2 was the same as that in beaker 1. Compare the boiling point of the solution in beaker 2 after osmosis (step 18) with the boiling point of the solution in beaker 1. Explain the difference.

   The boiling point of the solution in beaker 2 after osmosis was 100.5 °C, compared to 100.7 °C for the solution in beaker 1. The lower boiling point indicates a lowering of the concentration of the solution in beaker 2 after osmosis. Because the solution in the dialysis tubing was more dilute than the solution in beaker 2, a net movement of water molecules in the tubing solution occurred through the membrane until the concentrations of glucose in the tubing and beaker were balanced. This left the final molality of the glucose solution in beaker 2 lower than its original concentration.

Part 2. Freezing Point Depression

1. Using the same mass of additive (i.e., 20 g), which additive to the ice water lowered the freezing point the most (with the greatest ΔT_f)?

The NaCl lowered the freezing point the most.

2. Using the same mass of additive (i.e., 20 g), which additive to the ice water lowered the freezing point the least (with the lowest ΔT_f)?

The glucose lowered the freezing point the least.

3. Which additive has the greatest freezing point depression per mole? Which has the least? Is this what would be expected? Explain.

Calcium chloride has the greatest freezing point depression per mole and glucose has the lowest. This is expected because each mole of calcium chloride dissociates into three particles (ions) in solution while glucose remains as one single particle when in solution, and freezing point depression (a colligative property) depends on the number of particles or ions in solution.

4. What should be true about the freezing point depression per particle or ion? Does your data verify this?

The values for freezing point depression per particle or ion should be the same for any solute in a given solvent. Yes, the data verifies this within experimental error.

5. What factors were held constant in this experiment?

Same thermometer throughout; same amount of ice–water mixture; same mass of solute; solutions were stirred equally throughout.

6. Given the following sample cost data, which deicing chemical would you recommend as the most effective and most cost effective agent for preventing road icing? Explain.

For a given amount of solvent, the freezing point depression depends on the number of solute particles. Therefore, the solute that produces the most particles (moles) for a given cost is the most cost effective.

Sodium chloride is the most cost effective and aluminum chloride, although producing the most particles per compound, is the next to least cost effective.

Part 3. Boiling Point Elevation

1. From the values of ΔT_b (boiling point elevation) for each of the sodium chloride solutions, estimate the boiling point elevation constant. Substitute the molarity values for molality in Equation 1.

ΔT_b = K_bm
K_b ≈ ΔT_b/M

For 1 M NaCl

K_b ≈ (100.2 – 98.0) °C/1.0 M ≈ 2.2 K•kg/mol

For 4 M NaCl

K_b ≈ 104.0 – 98.0) °C/4.0 M ≈ 1.5 K•kg/mol

2. How does using the molarity in place of molality affect the estimate of K_b?

For NaCl solutions

The density of pure water is 1.00 kg/L at STP. This value is greater than 1.00 for solutions of sodium chloride and increases as the concentration of sodium chloride increases. Therefore, the numerical value for molality is always less than the molarity of any water solution, and K_b calculated using molarity values is always less than the true value.

Introduction

Why does adding antifreeze to water keep it from boiling at 100 °C or freezing at 0 °C? Why do cucumbers shrivel up to pickles when placed in a salt solution? How are nutrients transported in plants and animals? All of these observations are examples of the colligative properties of solutions.

Concepts

• Colligative properties
• Osmosis
• Molality
• Boiling point elevation
• Freezing point depression

Background

The properties of a solution are different from those of a pure solvent. Properties of solutions that depend solely on the number of particles dissolved and not on their chemical identity are called colligative properties. The colligative properties are vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.

When a solute is added to a solvent, the solute molecules occupy some positions formerly occupied by solvent molecules. In a sense the solvent is diluted and its molecules become more separated from each other. For example, in a sugar–water solution, sugar molecules occupy a portion of the surface area that would be occupied by water molecules if the sugar were not present.

As expected, the vapor pressure of the water is lowered by the presence of the sugar molecules on the surface in a sugar–water solution. The vapor pressure lowering observed in a solution compared to a pure solvent is the key to understanding the origin of the other colligative properties, namely, boiling point elevation and freezing point depression. The normal boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to the atmospheric pressure (1 atm).

As a result of this vapor pressure lowering, the vapor pressure of a sugar–water solution will be less than one atmosphere when the temperature is equal to the normal boiling point of the solvent. The solution temperature must be elevated above the normal boiling point for the vapor pressure of the solution to equal one atmosphere. The greater the sugar concentration, the larger the increase in boiling point temperature. In equation form, the change in boiling point temperature is:

where

ΔT equals the increase in boiling point temperature (°C).
K_b is a proportionality constant, called the boiling point constant. K_b is unique for each solvent.
m is a concentration unit called molality and is equal to the number of moles of solute per kilogram of solvent.

The freezing point of a solvent is the temperature at which the vapor pressure of the solid is equal to the vapor pressure of the liquid. When a solute is dissolved in the water, the vapor pressure of the solution at 0 °C is less than that of pure water at 0 °C. Ice at 0 °C has a larger vapor pressure than the solution and therefore no ice forms. Only at a lower temperature, when the solid–liquid vapor pressures are equal, will ice form (see Figure 1). The more solute dissolved in water, the lower the freezing point.

Certain substances (solutes) will lower the freezing point of a solution more than other substances. When a molecular substance such as sucrose (C₁₂H₂₂O₁₁) is placed into water, the molecule does not dissociate and remains as just one particle. Ionic solutes, on the other hand, dissociate into ions when put into water. That is, one unit of an ionic salt, such as sodium chloride (NaCl), dissociates in water to produce two particles—one sodium ion (Na⁺) and one chloride ion (Cl^–). One unit of calcium chlo ride (CaCl₂) when placed in water dissociates into three particles—one calcium ion (Ca²⁺) and two chloride ions (Cl^–). Looking at Equation 1, it can be seen that the boiling point elevation depends on the number of particles in solution—the more particles in solution, the greater the change in boiling point.

The equation for the lowering of the freezing point, or freezing point depression, is similar to that of the boiling point elevation.

where

ΔT = decrease in freezing point
K_f = the freezing point constant
m_solute = moles of solute particles per kilogram of solvent

Osmotic pressure is a colligative property that results when two solutions of different solute concentrations are separated by a semipermeable membrane. This membrane allows solvent, but not solute, molecules to pass through.

Since the solvent concentration is greater for the solution containing less solute, the rate of movement of solvent molecules across the membrane is greater from the more dilute side than from the concentrated side of the membrane (see Figure 2a). The osmotic pressure is caused by the increase in solution volume on the concentrated solution side of the membrane. When the pressure becomes sufficient, the rate of solvent transfer is equal for both sides. Volume changes cease and the system is in equilibrium (see Figure 2b). The net result is the dilute solution is now more concentrated in solute and the concentrated solution is less concentrated in solute.

The excess pressure due to the unequal liquid levels is called the osmotic pressure and has the symbol Π. Osmotic pressure increases as solute concentration increases and can be represented by the equation

Π = MRT

where

Π = Osmotic pressure in atmosphere
M = Molarity of the solution
R = 0.08206 L•atm/K•mol
T = Temperature in Kelvin

Experiment Overview

The purpose of this experiment is to observe the colligative properties of osmosis, boiling point elevation and freezing point depression. Boiling point elevation is used to observe the change in the molarity of solutions before and after osmosis experiments are performed. The relative changes in freezing points are observed when two ionic compounds and a molecular compound are added to water. In the final section, students use a microscale procedure to verify that increasing the concentration of sodium chloride in solution results in a comparable increase in the boiling point of the solution.

Materials

Prelab Questions

1. The molality of a solution can be determined using Equation 1 if the boiling point elevation is measured and the boiling point constant (K_b) of the solvent is known. The value of K_b for water is 0.512 kg•K/mol.

ΔT = m x 0.512 kg•K/mol

ΔT = Boiling point of solution – Boiling point of pure water

A solution was prepared by dissolving 12.00 g glucose in 100.0 g water. The resulting solution was found to have a boiling point of 100.34 °C. Calculate the molar mass of glucose. Note: Glucose is a molecular solid that exists as discrete molecules in solution.

2. Solution A contains 20 g of sodium chloride, NaCl, dissolved in 200 mL of water. Solution B has 40 g of sodium chloride dissolved in 200 mL of water.

1. Which solution will have the highest boiling point?
2. 50 mL of Solution A is placed in a section of semipermeable dialysis tubing and then sealed. The tube is placed in a beaker containing the 200 mL of Solution B. After a period of time, the boiling points of the solutions inside and outside the dialysis tube are determined.

How will the boiling point of each solution have changed from its original boiling point? Explain.

3. The freezing point depression of a solution is dependent on the number of particles in solution, not the type of particles. If 0.1 moles of each of the following soluble ionic compounds, sodium phosphate, Na₃PO₄, and sodium chloride, NaCl, are added to separate beakers containing 1 L of distilled water, which would have the lower freezing point? Explain.

Safety Precautions

Boiling solutions pose a serious burn risk. Use caution when handling hot beakers, hot oil and hot plates. Calcium chloride is slightly toxic by ingestion. Sodium chloride and glucose are not considered hazardous; however, the chemicals provided are for laboratory use only and are not intended for human consumption. Wear chemical splash goggles and chemical-resistant gloves and apron. Wash hands thoroughly with soap and water before leaving the laboratory.

Procedure

Part 1. Osmosis

1. Cut and presoak 8 inches of dialysis tubing in a 250-mL beaker containing deionized or distilled water for approximately 5 to 10 minutes.
2. Label four clean 250-mL beakers 1–4.
3. Using a clean 100-mL graduated cylinder, carefully pour 100.0 mL of distilled water into each of the four beakers.
4. Mass 34.2 grams of glucose in each of two weighing dishes.
5. Add 34.2 grams of glucose each to beakers 1 and 2. Record the mass of glucose added to each beaker in the Data Table. Dissolve the glucose in the water.
6. Measure 17.1 grams of glucose in each of two weighing dishes.
7. Add 17.1 grams of glucose each to beakers 3 and 4. Record the mass of glucose added to each beaker in the Data Table. Dissolve the glucose in the water.
8. Clamp off one end of the dialysis tubing with a dialysis tubing clamp. Measure and pour 25 mL of the solution from beaker 4 into the dialysis tubing.
9. Squeeze the tubing gently to remove any air and then clamp off the second end of the tubing.
10. Place the dialysis tube containing beaker 4 solution into beaker 2 (see Figure 3).

11. Cover beaker 2 with Parafilm and let it sit overnight. Cut a square section of Parafilm with sides about 1 inch larger than the beaker lip diameter. Place the Parafilm square over the beaker opening. With both hands, stretch the Parafilm in all directions over the beaker until the lip of the beaker is entirely sealed.
12. Place beaker 1 on a hot plate.
13. Attach a thermometer clamp to the ring stand and move the stand so that the clamp is over the solution.
14. Carefully place the thermometer in the clamp so that the bulb is in the solution (see Figure 4).

15. Turn on the hot plate and heat the solution.
16. Record the boiling point temperature for the solution in beaker 1 in the Data Table. The boiling point occurs at the point after the first bubbles appear when the temperature of the solution stabilizes.
17. Repeat steps 12–16 for the solution in beaker 3.
18. Add 100.0 mL of distilled or deionized water to a clean 250-mL beaker. Repeat steps 12–16 to determine the boiling point of pure water.
19. On the following day, remove the dialysis tubing from beaker 2. Record observations in the Data Table.
20. Repeat steps 12–16 to determine the boiling point of the solution in beaker 2.

Part 2. Freezing Point Depression

1. Label three 250-mL beakers 1–3.
2. Place beaker 1 on the balance and tare the balance, if the balance is electronic.
3. Add 100 grams of ice water to beaker 1:

1. First add approximately 70–80 grams of crushed ice.
2. Then add enough water so the total mass of ice plus water is 100.0 g. Record the precise mass of the ice-water mixture to the nearest tenth of a gram in the Data Table. (Note: If the balance does not have a large enough capacity, weigh the ice in a small weighing dish and then place it in the beaker. Measure the remaining water using a balance or graduated cylinder, assuming d = 1.0 g/mL for water.)

4. Stir the ice–water mixture with a wood stirrer.
5. Carefully insert a thermometer into the ice–water mixture. Wait for the temperature reading to stabilize. Record the temperature of the pure ice–water mixture in °C to the nearest tenth of a degree in the Data Table.

Beaker 1—Sodium Chloride

6. In a weighing dish, weigh out 20.0 g of sodium chloride. Record the mass of the sodium chloride in the Data Table.
7. Add the sodium chloride to the ice–water mixture in beaker 1.
8. Stir the contents of the beaker with a wood stirrer until the mixture has a slushy appearance.
9. Carefully insert a thermometer into the mixture and measure the temperature. Continue to stir the mixture with the wood stirrer. Record the lowest temperature that the mixture reaches before rising again in the Data Table. Record temperature in °C to the nearest tenth of a degree. Note: This may take some time, as the salt does not immediately dissolve in the ice water.

Beaker 2—Calcium Chloride

10. Repeat steps 2–9 for beaker 2, using 100.0 grams of ice–water and 20.0 grams of calcium chloride. Remember to use a clean thermometer, weighing dish and wood stirrer.
11. Record precise masses and temperature readings to the nearest tenth of a degree in the Data Table.

Beaker 3—Glucose

12. Repeat steps 2–9 for beaker 3, using 100.0 grams of ice water and 20.0 grams of glucose. Remember to use a clean thermometer, weighing dish and wood stirrer.
13. Record precise masses and temperature readings to the nearest tenth of a degree in the Data Table.
14. Dispose of the solutions by pouring the mixtures down the drain with plenty of water. Rinse the beakers with tap water.

Part 3. Boiling Point Elevation

1. Using gloves, carefully break a closed-end capillary tube in half and place it in a microscale test tube with the closed end up.
2. Place about 0.5 mL (2 cm) of distilled or deionized water in the micro scale test tube using a pipet.
3. Attach the test tube with two small rubber bands to the thermometer so that the bottom of the test tube is even with the bottom of the thermometer (see Figure 5).

4. Fill the 50-mL beaker with 30 mL of oil and place on a hot plate/ring stand assembly.
5. Attach the thermometer–sample holder to a thermometer clamp and immerse into the oil bath. The sample must be submerged, but the neck of the sample holder must rise at least 5 cm above the surface of the bath. Make sure the rubber band is above the oil surface (see Figure 5).
6. Begin heating the oil bath. When a rapid stream of bubbles begins to come out of the capillary tube, turn off the heat and allow the oil bath to cool. If too much sample boils off, remove the thermometer/test tube setup from the bath and add more sample.
7. Gently stir the oil bath with a glass stir rod to help the oil bath cool evenly. Observe the sample until the steady stream of bubbles coming from the open end of the capillary tube ceases. Record this temperature in the Data Table. This is the boiling point of distilled water.
8. Refill the microscale test tube with the 1.0 M sodium chloride solution and repeat the boiling point procedure (steps 2–7).
9. Repeat steps 2–7 with the 4.0 M sodium chloride solution.
10. When the sample holder and oil bath have cooled, dispose of the sodium chloride solution in the test tube and the oil in the beaker as directed by the instructor.

Student Worksheet PDF

13113_Student1.pdf