Chem 162 Experiment 6 Molar Mass Determination Using Freezing Point Depression
Pre-Lab Questions
1. Define the term colligative property.
2. Suppose the freezing point of a solution of 2.00 g of an unknown molecular substance in 10.00 g of the solvent benzene is measured. If the solution freezes at a temperature of 6.33 °C lower than pure benzene itself, calculate the molar mass of the unknown substance.
3. For a given number of moles of solute, why do ionic substances have a larger effect on the freezing and boiling points of solvents than do nonionic substances?
Be sure to have all of your data tables prepared in your lab notebook.
Purpose • To study the phenomenon of freezing point depression, a Colligative Property. • To determine the molar mass of an unknown solute.
Introduction When a solute is dissolved in a solvent, the properties of the solvent are changed by the presence of the solute. The magnitude of the change generally is proportional to the amount of solute added. Some properties of the solvent are changed only by the number of solute particles present, without regard to the particular nature of the solute. Such properties are called colligative properties of the solution. Colligative properties include changes in vapor pressure, boiling point, freezing point, and osmotic pressure.
For example, if a nonvolatile, nonionizing solute is added to a volatile solvent (such as water), the amount of solvent that can escape from the surface of the liquid at a given temperature is lowered, relative to the case where only the pure solvent is present. The vapor pressure above such a solution will be lower than the vapor pressure above a sample of the pure solvent under the same conditions. Molecules of nonvolatile solute physically block the surface of the solvent, thereby preventing as many molecules from evaporating at a given temperature. (See Figure 1.) As shown in the figure, if the vapor pressure of the solution is lowered, there is an increase in the boiling point of the solution as well as a decrease in the freezing point.
In this experiment, you will determine the freezing points of a pure solvent (c-hexane) and a solution dissolved in c-hexane .
Figure 1. (a) The presence of a nonvolatile solute lowers the vapor pressure of this solvent by blocking the surface.
Figure 1(b) Plot of vapor pressure vs. temperature for a pure solvent and a solution. Note that the presence of a solute changes the properties of the solvent.
The decrease in freezing point (ΔTf) when a nonvolatile, nonionizing solute is dissolved in a solvent is proportional to the molal concentration (m) of solute present in the solvent:
ΔTf = Kf . ms (1)
Kf is a constant for a given solvent (called the molal freezing point depression constant) and represents by how many degrees the freezing point will change when 1.00 mol of solute is dissolved per kilogram of solvent. For example, Kf for water is 1.86°C kg mol-1, whereas Kf for the solvent benzene is 5.12°C kg mol-1.
The molal concentration of a solution represents how many moles of solute are dissolved per kilogram of the solvent. For example, if 0.50 mol of the sugar sucrose were dissolved in 100 g of water, this would be equivalent to 5.0 mol of sugar per kilogram of solvent, and the solution would be 5.0 molal. The molality (m) of a solution is defined as
The measurement of freezing point lowering is routinely used for the determination of the molar masses of unknown solutes. If Equations (1) and (2) are combined, it can be derived that the molar mass of a solute is related to the freezing point lowering experienced by the solution and to the composition of the solution. Consider the following example:
The molal freezing point depression constant Kf for the solvent benzene is 5.12°C kg mo1-1 A solution of 1.08 g of an unknown in 10.02 g of benzene freezes 4.60 °C lower than pure benzene. Calculate the molar mass of the unknown. The molality of the solution can be obtained using Equation (1): m = ΔTf /Kf = 4.60°C / 5.12°C kg mol-1 = 0.898 mol/kg
Using the definition of molality from Equation (2), the number of moles of unknown present can be calculated: m = 0.898 mol/kg = mol unknown / 0.01002 kg solvent moles of unknown = 0.009002
Thus, 0.009002 mol of the unknown weighs 1.08 g, and the molar mass of the unknown is given by:
In the preceding discussion, we considered the effect of a nonionizing solute on the freezing point of a solution. If the solute does indeed ionize, the effect on the freezing point will be larger. The depression of the freezing point of a solvent is related to the number of particles of solute present in the solvent. If the solute ionizes as it dissolves, the total number of moles of all particles present in the solvent will be larger than the formal concentration indicates. For example, a 0.1 m solution of NaCl is effectively 0.1 m in both Na+ and Cl – ions.
Setup
Plastic tube, stopper with one hole for for thermometer, one small stir bar (for inside the tube), one larger stir bar (for mixing ice water bath), salt, unknowns, beakers, ice.
Procedure
A. Determination of the Freezing Point of water (solvent)
1 . Use a 400-mL beaker to make a salted ice bath (1:1). Add appx 50 g of NaCl to 350 mL of ice water bath. Place a large stir bar in the bath and place on a magnetic stirrer and stir at a medium pace.
2. Place a small stir bar in a clean, dry plastic test tube and record the mass to the nearest 0.001 g. The best method is to stand up the test tube in a tared beaker
3. Measure approximately 13 mL of distilled water and pour into the test tube. Record the mass of the the test tube with the water. . Secure the test tube above the ice water bath with a utility clamp
4. Place a one-hole stopper in the test tube and insert a thermometer through the hole. Adjust the thermometer so that it does not touch the bottom of the test tube.
5. Open the Logger Pro software (desktop) and openVernier file for Freezing Point depression (Exp 15)
6. Lower the test tube into the salted ice bath (1:1). Press Collect. Be sure that the solution is stirring. The temperature will be recorded by the program. The temperature should level off as the water freezes.
7. Continue taking temperature readings until the temperature has dropped about one full degree lower than the temperature where it leveled off. If you need to extend the time on the experiment, go to experiment tab and hit extend time. You need to do this before time runs out. It is actually a good odea to extend the time right away to a very long time, and then you can stop the experiment when you are satisfied you have enough data. Print your cooling curve.
8. Your graph should resemble the curve in Figure 2 below. The flat region represents the freezing point of pure water.
Figure 2. Cooling curve for a pure solvent and of a solution of that solvent. The flat portions of the curves represent freezing.
9. Remelt the water by placing the test tube in a beaker with room temperature water.
10. Repeat if your freezing point deviates significantly from the literature value.
B. Determination of the Freezing point of water + Unknown (Solution)
1. Reweigh the test tube with the water to account for any evaporation of the water
2. Add approximately 2.0g ( record the exact weight) of unknown to the test tube containing the water. Shake test tube until the solid is dissolved. Using the same procedure as in Part A, determine the freezing point of the unknown/water solution. Add more salt to ice bath as needed.
3. Remeember to print copies of your cooling curve. Be sure to label them.
4. Reweigh the test tube again and add an additional 2.0g of the same unknown. Repeat freezing process to determine the freezing point. Print a copy of your cooling curve.
Data
A. Determination of the Freezing Point of water
Mass empty test tube ______
Mass test tube plus water ______
Mass of water ______
Freezing point of water (from graph) ______
B. Determination of the Freezing point of water + Unknown Solution
Unknown number/letter ______
First Addition of Unknown
Mass test tube + water ______
Mass of test tube + water + unknown ______
Mass of water ______
Mass unknown ______
Freezing point of water/unknown solution ______
Second Addition of Unknown
Mass test tube + unknown solution ______
Mass of test tube + unknown solution + 2nd addition of unknown ______
Mass of unknown added (second addition) ______
Total mass of unknown (first and second addition) ______
Freezing point of water/unknown solution ______
Analysis and Calculations
1. Obtain freezing points from the three cooling curves and determine = ΔTf ‘s for the two unknown solute additions in part B. ΔTf = Tf (solution) - Tf (solvent). Show your calculations and clearly indicate these values.
2. From the freezing point depression of the unknown/water solution and the molal freezing point depression constant (Kf) for water (1.86 ºC/m), calculate the molality and molar mass of the unknown for each of the additions of solute. Show your calculations.
Show your calculations and clearly indicate the following
Unknown number/letter ______
Molality (m) for first addition of unknown ______
Molar mass of unknown for first addition of unknown ______
Molality (m) for second addition of unknown ______
Molar mass of unknown for second addition of unknown ______
Average molar mass of unknown ______
Post – Lab Questions.
1. A phenomenon called supercooling is frequently encountered in this experiment. supercooling, a solution momentarily drops below its freezing point, and then warms up again, before solidification begins. What event is likely to give rise to supercooling?
2. The molal freezing point constant, Kf , is a property of the solvent, not the solute. What does this say about the fact that freezing point depends on the amount of solute, rather than on the solute’s nature?
3. A phenomenon that happens sometimes during freezing point depression experiments is that the solute is affected in some manner by the solvent. One common occurrence is for a solute to dimerize, that is, two solute molecules combine to produce a single double molecule (a dimer). What effect would there be on a molar mass determination if the solute were to dimerize?