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2601 Freezing Point Depression

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    1.1 Objectives

    After completing this experiment, the student will be able to:

    • describe colligative properties.
    • find the freezing point depression of a solution.
    • determine the molar mass of an unknown solute using freezing point depression data.

    1.2 Background

    Colligative properties of solutions can be determined by comparing the properties of the pure solvent with the properties of the solvent upon addition of various solute particles. Properties such as boiling point and freezing point depend on the intermolecular attractive forces between solvent molecules. Changes in these physical properties occur when the presence of a solute disrupts these interactions.

    When a small amount of nonvolatile solute is dissolved in a solvent, the vapor pressure of the solution will be less than the vapor pressure of the pure solvent at the same temperature. As a result, the boiling point of the solution, Tb, is higher than the boiling point of the pure solvent, Tb°. The amount by which the boiling point of the solution exceeds the boiling point of the pure solvent, ∆Tb = Tb – Tb°, is called the boiling point elevation. Similarly, because of the reduction in vapor pressure over the solution, the freezing point of the solution, Tf, is lower than the freezing point of the pure solvent, Tf°. The amount by which the freezing point of the solution is decreased from that of the pure solvent, ∆Tf = Tf° – Tf, is called the freezing point depression. A sample vapor pressure diagram for a pure solvent and its solution is shown below. (Figure 1)


    Figure 1. Vapor pressure curve

    Focusing on freezing point depression, the magnitude of the freezing point depression attributed to the added solute is proportional to its molality.

    ∆Tf = Kf m (Equation 1)

    Molality, m, has units of moles of solute per kilogram of solvent, and is more useful for the calculation of this colligative property because, unlike molarity, it is independent of temperature. Kf is known as the freezing point depression constant and depends on the solvent used.

    In this experiment you will determine the molar mass of an unknown solute by adding it into a solvent and measuring the resulting freezing point depression of the solvent. From the measured ∆Tf and the known Kf value of the solvent, you can then determine the value of m using the above equation.

    Solutes, however, dissolve in solvent differently. Most molecular substances such as ethanol, C2H5OH, dissolve in water and remain as discrete solvated molecules, C2H5OH(aq). Other solutes, like ionic substances, dissociate or break apart in solution to form ions. An example is table salt, NaCl, which dissolves in water to form Na+(aq) and Cl-(aq) ions.

    Since the colligative properties of solution depend on the concentration of dissolved solute particles, an additional factor, i, (also called the van’t Hoff factor) is included in the colligative property equations to account for the formation of multiple particles as electrolytes dissociate:

    ΔTf =iKfm (Equation 2)

    The solvent that will be used in this experiment is laboratory water that has a Kf value of 4.5 °C·kg·mol-1. Since only non-dissociating solutes will be used in this experiment, the value of i for your unknown solute will be considered equal to 1.

    To determine the freezing point of this pure solvent, you will measure the temperature as a function of time as the liquid cools in an insulating jacket. At first, the temperature will fall quite rapidly. When the freezing point is reached, solid will begin to form, and the temperature will tend to hold steady until the sample is all solid. This behavior is shown graphically in Figure 2. The freezing point of the pure liquid is the constant temperature observed while the liquid is freezing to a solid.

    The cooling behavior of a solution is somewhat different from that of a pure liquid, also shown in Figure 2. As discussed earlier, the temperature at which a solution freezes is lower than that for the pure solvent. In addition, there is a slow gradual fall in temperature as freezing proceeds. The best value for the freezing point of the solution is obtained by drawing two straight lines connecting the points on the temperature-time graph. The first line connects points where the solution is all liquid. The second line connects points where the solid and liquid coexist. The point where the two lines intersect is the freezing point of the solution.


    Figure 2. Cooling curve for a pure solvent (top) and for a solution (bottom)

    Note that when the solid first appears the temperature may fall below the freezing point, but then it comes back up as more of the solid forms. This effect is called supercooling, a phenomenon that may occur with both the pure liquid and the solution. When drawing the straight line in the solid-liquid region of the graph, ignore points where supercooling is observed. To establish the proper straight line in the solid-liquid region it is necessary to record the temperature until the trend with time is smooth and clearly established.


    3.0 CHEMICALS AND SolutionS

    3.0 CHEMICALS AND SolutionS

    3.0 CHEMICALS AND SolutionS



    Approximate Amount


    Stearic acid


    15 g


    Lauric acid (unknown)


    4 g


    Myristic acid (unknown)


    4 g


    Palmitic acid (unknown)


    4 g



    50 mL beaker

    Hot plate

    100 mL beaker


    150 mL beaker or small Pyrex bowl (70x50)

    Crucible tong


    Paper towel (cut-to-fit)

    Digital thermometer



    5.1 Determining the Freezing Point of Pure Stearic Acid (a Fatty Acid)

    1. Fill a 150 mL beaker with about 30 mL tap water and heat on a hot plate. This will be your hot water bath.
    2. Prepare an insulating jacket by wrapping a piece of paper towel around a clean dry 50 mL beaker and fitting it in a 100 mL beaker. Remove the 50 mL beaker and reserve the 100 mL beaker and the paper towel as the insulating jacket (see Figure 3). The insulating jacket prevents premature cooling due to contact with the skin or other surface.

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    Figure 3. Set-up for an Insulating Jacket

    1. Weigh the 50 mL beaker using an electronic balance and record its mass. Transfer about 15 grams (half-full) of fatty acid to the beaker. Reweigh and record the mass of the beaker with the fatty acid. Calculate the mass of stearic acid by difference.
    2. Place the beaker containing fatty acid into the hot water bath to melt it. After the fatty acid has completely melted, place the thermometer into the ‘melt’ and heat until the temperature reaches 85oC. From this point on, the thermometer is not removed from the sample to prevent loss of material and contamination of bench tops with fatty acids. Remove the beaker from the water bath and dry the outside.
    3. Place the beaker containing fatty acid in the previously prepared insulating jacket. Stir the fatty acid slowly but continuously to help minimize supercooling. Record the initial temperature of the sample and then every 30 seconds for 8-10 minutes. Temperatures are recorded until the temperature of the sample remains constant, changing by 0.1oC per reading, for 3 minutes (6 readings).
    4. Perform a second trial by putting back this beaker containing the solidified fatty acid into the hot water bath and repeating steps 4 and 5.
    1. Determining the Freezing Point of Solutions (Stearic Acid + Unknown)
    1. Remove the 50 mL beaker containing fatty acid from the insulating jacket and cool to room temperature. Carefully place and rest the thermometer on a clean paper towel on your bench top; making sure that no solids sticking on the tip of the thermometer are dislodged.
    2. (Solution I) To the fatty acid sample used above, add approximately 1.8 g of an unknown sample (a different fatty acid). Exactly 1.8 g is not needed, but make sure to record its mass to the nearest 0.1 mg. You may use the ‘tare’ function of the balance in this step.
    3. Repeat steps 4 to 5 above. If time permits, a second trial (step 6) can be performed.
    4. (Solution II) To the same fatty acid mixture, an additional 1.8 g of the same unknown sample is transferred. Make sure to cool the beaker to room temperature before placing on the balance. Similarly, exactly 1.8 g is not needed but make sure to record its mass to the nearest 0.1 mg.
    5. Repeat steps 4 to 5 above. If time permits, a second trial (step 6) can be performed.

    5.3 CLEAN UP

    1. The solidified solutions can be carefully scraped off the beaker using a spatula. All solids can go down the trash bin. Discard any remaining residue from the beaker and thermometer directly into the trash bin.
    2. Return the digital thermometer to the cart/stockroom.

    6.0 DATA RECORDING SHEET Show the equations you used and the calculations you performed below.

    Last Name

    First Name


    Partner Name(s)


    Table 1. Mass Measurements

    Unknown ID


    5.1 Freezing Point of Pure Stearic Acid

    Mass of 50 mL beaker and stearic acid


    Mass of empty 50 mL beaker


    Mass of stearic acid (by difference)


    5.2 Freezing Point of Solution

    Identification code for unknown


    Mass of unknown sample (step 8; for solution I)


    Mass of additional unknown sample (step 10)


    Total mass of unknown sample (for solution II)


    Table 2. Temperature Measurements. Record the temperature every 30 seconds as the pure solvent and solutions are cooled. Note the temperature at which any crystal first appears. (Use additional sheet(s), if necessary.)

    Time Elapsed


    Temperature (oC)

    Pure Solvent

    Solution I

    Solution II












































    7.1 Graphical Analysis of Data

    1. Use Excel to create three (or four) separate graphs of “Temperature versus Time” for the pure solvent and the two solutions studied. Each graph should have an appropriate title and labeled axes with an appropriate scale. Add two trendlines to the data points of each graph. You can do this by hand with a ruler or by using Excel. The first line is applied to data points that correspond to the cooling of the liquid state – these are the points on the steep part of the graph. The second line is applied to data points that correspond to the co-existence of both the solid and liquid (freezing) – these are the points on the part of the graph where the temperature levels out. Extrapolate the two trendlines towards each other until they intersect. The temperature at the point of intersection is the solvent freezing point and should be clearly shown on each graph.
    1. Attach your three graphs to this report.
    1. Record the freezing point temperatures obtained from the graphs below:

    Pure stearic acid: _________ °C

    Solution I: _________ °C

    Solution II: ______________ °C

    7.2 Calculation of Molar Mass

    Complete the table below with the results of your calculations. Be sure to include all units. Note that Kf stearic acid = 4.5 °C·kg·mol-1.


    Solution I

    Solution II

    1. Mass of stearic acid


    2. Total mass of unknown added


    3. Freezing point of pure stearic acid


    4. Freezing point of solution


    5. Freezing point depression, ΔTf


    6. Molality of solution


    7. Moles of unknown in solution


    8. Molar Mass of unknown


    Unknown ID: ______________________ has an average molar mass of ____________.


    1. Using the known freezing point of stearic acid (69oC), determine the percentage error in your experimentally measured freezing point. Show your calculation below and be sure to report your answer to the correct number of significant figures.

    1. Obtain the actual molar mass of your unknown solute (from your instructor) and determine the percentage error in your experimental average value. What factors could have affected such deviation from the actual molar mass of your unknown solute?

    1. The commercial antifreeze used in cars is typically 50% v/v water and ethylene glycol. Discuss why this solution is useful for protecting car engines from both extreme summer and winter temperatures.

    1. Hydrogen chloride (HCl) is soluble in both water and in benzene (C6H6). The freezing points and Kf values are given below for each solvent.


    FP (oC)

    Kf (oC/·kg·mol-1)







    1. For a 0.01 m HCl(aq) solution, the freezing point depression is about 0.04°C. Calculate the van 't Hoff factor for HCl in water. Is this van 't Hoff factor what you would expect for HCl dissolved in water? Explain why or why not.

    1. For a 0.01 m HCl dissolved in benzene solution, the freezing point depression is about 0.05°C. Calculate the van 't Hoff factor for HCl in benzene.

    1. Compare the van 't Hoff factor for HCl when water is the solvent (a) to the van 't Hoff factor for HCl when benzene is the solvent (b). Is there any difference between these two values? Account for the difference in terms of intermolecular forces between HCl and each of the solvents.

    2601 Freezing Point Depression is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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