Chemistry 1C
Freezing Point Depression
Freezing Point Depression, the van’t Hoff Factor, and Molar Mass
Objectives
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To
To
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To
To
understand colligative properties.
find the freezing point depression of a solution.
determine the van't Hoff factor for acetic acid dissolved in cyclohexane.
deduce the nature of acetic acid dissolved in cyclohexane.
find the molar mass of an unknown organic compound.
Discussion – Freezing Point Depression
The vapor pressure of a solution containing a non-volatile solute is lower than that of the pure solvent.
Consequently, the boiling point of a solution is higher. The freezing point of a solvent is also affected by the
solute; the freezing point of a solution is lower than that of the pure solvent. The freezing point of a solution,
Tf, is
Tf = T°f – ∆Tf
Where T°f is the freezing point of the pure solvent and ∆Tf is the freezing point depression. Freezing point
depression, like boiling point elevation, is termed a colligative property. Such properties depend only on the
concentration of the solute particles in the solution, and not on their nature.
For dilute solutions, ∆Tf = mKf = T°f – Tf
(1)
Where m is the molality of the solution and Kf is the molal freezing point depression constant, a property of
the solvent. Values of Kf for various solvents are given in table I.
Table I
Normal Freezing Point
Pure Solvent, T°f (°C)
0.00
5.50
179.75
?
Consider solutions of electrolytes, such as NaCl, dissolved
in water; in this case the solute dissociates to form ions in
the solution. Thus there are more solute particles in
Water
1.86
solution than for the same molality of a non-electrolyte,
Benzene
5.12
such as sucrose or methanol, dissolved in water. For dilute
Camphor
39.7
solutions of NaCl(aq), the solution contains essentially
Cyclohexane
20.0
completely separated Na+(aq) and Cl–(aq) ions. As a
result, the molality of the solute particles (ions in this case) is twice what it would be for a non-electrolyte,
and ∆Tf is twice as large. On the other hand, if the solute particles associate in solution to form aggregates
then the number of solute particles (aggregates in this case) is reduced, causing ∆Tf to be smaller than if the
solute did not associate.
Solvent
Kf (°C/m)
The degree of dissociation or association of the solute particles is given by the van't Hoff factor, i. This
factor is the ratio of the number of moles of solute particles in solution to the number of moles of solute
dissolved.
i = moles solute particles/moles solute dissolved
If the solute exists as single molecules (neither dissociating or associating) in solution, then i = 1
If the solute dissociates in solution, then i > 1
If the solute associates in solution, then i < 1
The value of the van't Hoff factor depends upon: the nature of the solute AND solvent and the concentration
of the solute in the solution. We can determine the van't Hoff factor experimentally for a given system, that is
for a given solute, solvent and concentration of solute, by comparing the experimentally measured freezing
point depression to the calculated freezing point depression given by equation (1)
i=
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!T f (measured)
!T f (calculated)
1
(2)
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Freezing Point Depression
An alternative form of equation (1) that explicitly takes into account dissociation or association of the solute
is:
∆Tf = imKf
(m is the molality without taking dissociation or association into consideration)
(3)
Equation 3 can also be used to determine the van't Hoff factor for a given system, once the freezing point
depression is measured.
In this experiment, we will focus our attention on the effect that the natures of the solute and solvent have on
the value of a van't Hoff factor; we will not investigate concentration effects. Specifically, we will investigate
the behavior of acetic acid dissolved in a non-polar solvent, cyclohexane. The questions we will address
include:
Does acetic acid dissociate, associate, or exist as single molecules when dissolved in cyclohexane?
Can we explain the behavior of acetic acid in cyclohexane?
Discussion – Molar Mass from Freezing Point Depression Data
The freezing point depression of a solution can also be used to determine the molar mass of the solute. Using
equation (3) we can solve for the molality of a solution (if i is known) once ∆Tf has been measured:
m=
!Tf
iK f
(4)
From the molality and known mass of solvent, the moles of solute can be found. From the moles of solute and
known mass of solute the molar mass of the solute can be determined. In this experiment, you are going to
determine the molar mass of an unknown solid organic compound dissolved in cyclohexane.
Materials
LabPro with link cable
Ice
TI Graphing Calculator
Cyclohexane (50 mL per group of two students)
Vernier Stainless steel temperature probe Glacial acetic acid. (0.2 mL per group of two students)
(1) 600-mL beaker
200-µL automatic pipet with disposable tips
(two or more per lab, locate in hoods)
(2) 250-mL beakers
Waste container labeled “Cyclohexane”, “Acetic acid”.
(1) large test tube
Unknown solid organic compound(s).
(2 bottles of each unknown. ≈ 0.3 g per group of two students)
Procedure
Part 1: Freezing Point of the Pure Solvent, Cyclohexane
1. Weigh a large, dry test tube supported-in a 250-mL beaker. (Test tube MUST BE CLEAN!!!) Remember
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3.
!
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that to obtain the mass of the test tube, you can tare (zero) the balance with the beaker alone on it
before placing the test tube in the beaker.
Put about 15-mL of cyclohexane in the test tube.
Put the test tube in a 600-mL beaker with enough room temperature water to fill the beaker about half
way.
Prepare a LabPro for temperature data collection. See the LabPro Quick Start Guide for details.
Plug the temperature probe into Channel 1 of the LabPro System. If you are using an older temperature
probe you may need to use a DIN-BTA adapter to connect the probe to the LabPro. Use the link cable to
connect the LabPro System to the TI Graphing Calculator. Firmly press in the cable ends.
Place the temperature probe in the cyclohexane.
Set up the calculator and LabPro for data collection. Use TIME GRAPH data entry with 5 seconds between
samples and 60 samples.
• Begin to record the temperature by pressing [START] from the MAIN MENU. A temperature graph is
displayed on the calculator screen.
Begin cooling the water bath to ice temperature by adding enough ice to almost fill the beaker. Stir the
cyclohexane with the temperature probe and also move the test tube around in the water bath. Quick
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Freezing Point Depression
cooling while simultaneously stirring with the temperature probe and moving the test tube around in the
water bath tends to produce better cooling curves.
8. Make a note of the temperature and approximate time when the first appearance of
cyclohexane crystals occurs.
9. When data collection stops a graph of temperature vs. time is displayed on the calculator screen. As you
move the cursor right or left, the time (X) and temperature (Y) values of each data point are displayed
below the graph. As a pure liquid freezes, the temperature remains constant. Be aware that even slight
contamination due to unclean glassware can cause the temperature to slowly decrease instead of
remaining constant as freezing occurs. Examine the graph to find the freezing point of cyclohexane. The
freezing point can be found where the slope of the cooling curve breaks and flattens out. Record the
freezing point (to nearest 0.1°C). Compare this value to the value you recorded when crystals were first
observed. The two values should be very close.
10. Press [ENTER] to return to the MAIN MENU.
11. Remove the cyclohexane from the water bath and allow it to melt. Do not discard the cyclohexane, you
will be using it for Part 2 of this experiment.
Part 2: Freezing Point of the Acetic Acid Solution
1. Dry the outside of the test tube and reweigh the test tube with cyclohexane while being supported-in
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a 250-mL beaker.
IN THE HOOD, use the automatic pipet provided to drip 200 µL (0.2 mL) of glacial (pure) acetic acid into
the cyclohexane, avoiding the side of the test tube. BE CAREFUL NOT TO TOUCH THE
CYCLOHEXANE with the pipet tip! Drip the acetic acid into the cyclohexane by holding the tip just above
the surface. (If the tip contacts the cyclohexane, finish delivery, eject the tip into the waste bottle and
replace with a new tip for the next student.)
Reweigh the test tube with cyclohexane and acetic acid while being supported-in a 250-mL beaker.
After weighing, stir the solution well! (Why wait to stir until after weighing?)
Repeat the freezing process as in Part 1 for this mixture.
Unlike pure cyclohexane, cooling a mixture of acetic acid and cyclohexane results in a gradual drop in
temperature during the time period when freezing takes place. To determine the freezing point of the
acetic acid-cyclohexane solution, you need to determine the temperature at which the mixture first started
to freeze. Examine the data points to locate the freezing point of the solution, as shown in Figure 1. It
may be best to locate the freezing point by transferring the data file into Graphical Analysis, using one of
the computers located in the lab room, and then examining the graph using the interpolate mode of
graphical analysis. (Your instructor can assist you with the transfer.) Record the freezing point (to nearest
0.1°C). Compare this value to the value you recorded when crystals were first observed. The two values
should be very close.
Discard your solution into the appropriate waste beaker.
Figure 1: Determining the freezing point of a solution.
Temperature
Freezing Point
Time
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Freezing Point Depression
Part 3: Freezing Point of a Cyclohexane Solution Containing an Unknown Organic Compound
1. Weigh a large, dry test tube supported-in a 250-mL beaker. (Test tube MUST BE CLEAN!!!)
2. Wipe-clean the temperature probe.
3. Add approximately 13 mL (≈ 10 g) of cyclohexane to the test tube.
4. Weigh the test tube with cyclohexane while being supported-in a 250-mL beaker.
5. Determine the mass of cyclohexane added.
6. Find the solid organic unknown in the lab. Record the unknown ID code, mass of unknown needed per
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gram of cyclohexane, and the van’t Hoff factor for the unknown (all given on the unknown label).
From the label information, calculate the mass of unknown to add to your cyclohexane.
Add the calculated mass of organic unknown (to within ±0.01 g) to your cyclohexane.
Stir the mixture with the temperature probe until the unknown solid has completely dissolved.
Repeat the freezing process as in Part 1 for this mixture.
Examine the graph and record the freezing point (to nearest 0.1°C) as in Part 2.
Discard your solution into the appropriate waste beaker.
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Report Sheet
Freezing Point Depression
Name
Partner
Part 1 – Freezing Point of Cyclohexane
Freezing point of pure cyclohexane
Part 2 – Freezing Point of Acetic Acid in Cyclohexane
Mass of test tube
Mass of test tube and cyclohexane
Mass of test tube, cyclohexane and acetic acid
Freezing point of mixture
Part 3 – Freezing Point of Unknown Solid in Cyclohexane
Mass of test tube
Mass of test tube and cyclohexane
Mass of test tube, cyclohexane and unknown
Unknown ID code (on unknown label)
Mass unknown per gram cyclohexane (on unknown label)
van’t Hoff factor for unknown (on unknown label)
Freezing point of unknown mixture
Calculations
Part 2 – Acetic Acid Solution
From the data recorded:
1) Calculate the experimental freezing point depression of the cyclohexane-acetic acid solution.
a)
Calculate the experimental van't Hoff factor for acetic acid in cyclohexane.
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2) Using the van’t Hoff factor calculated above:
a) Does the acetic acid ionize or associate in cyclohexane? How do you know?
b) Determine either the % ionization of acetic acid (if i is >1) or how many acetic acid molecules are
associated per solute particle formed (if i is <1) in the cyclohexane.
c)
Explain why acetic acid behaves this way in cyclohexane. Use Lewis Structures and give a written
explanation in terms of intermolecular forces.
Part 3 - Unknown Solid Molar Mass
1.
From the data recorded, calculate the molar mass of the unknown solid. Show all your work below in
an orderly fashion. Label all calculated quantities.
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Follow-up Questions
1.
Why does the temperature continually decrease as a solution freezes instead of maintaining a constant
temperature as a pure liquid would do?
2.
When making ice cream, the temperature of the ingredients is decreased to below the freezing point of
pure water by using a surrounding ice bath that contains a large amount of table salt, NaCl.
Assuming that it dissolves completely and an ideal value for the van't Hoff factor, what mass of NaCl is
needed to lower the freezing point of 5.5 kg of water to –5.5°C?
a)
b)
3.
Making the same assumptions, calculate the mass of CaCl2 needed.
Conifer, a sugar (a non-electrolyte that does not associate in water) derivative found in conifers such as fir
trees, has a composition of 56.13% C, 6.48% H and 37.39% O by mass. A 2.216 g sample is dissolved in
48.68 g of water and the solution is found to have a boiling point of 100.068°C. What is the molecular
formula of conifer? (Kb for water is 0.51°C/m.)
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4.
Freezing Point Depression
Hydrogen chloride (HCl) is soluble in both water and in benzene (C6H6).
benzene
a)
For a 0.01 m HCl(aq) solution, the freezing point depression is about 0.04°C. Determine the van't Hoff
factor for HCl in water.
i)
Is this van't Hoff factor what you would expect for HCl dissolved in water? Explain why or why
not.
b) For a 0.01 m HCl dissolved in benzene solution, the freezing point depression is about 0.05°C.
Determine the van't Hoff factor for HCl in benzene.
c)
Compare the van’t Hoff factor for HCl when water is the solvent to the van’t Hoff factor for HCl when
benzene is the solvent. Suggest a reason in terms of intermolecular forces for any difference between
the two values.
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