Freezing Point Depression
This week, you will observe the effect of concentration on the freezing point of a solution and utilize this information to determine the molecular weight of a solute.
At a given pressure, the temperature at which a specific pure substance undergoes a particular phase change (freezing, boiling, sublimation, etc.) is always the same. For example, pure water at 1 bar pressure will always freeze at 0oC and boil at 100oC. However, if impurities (solutes) are added to the pure solvent in the liquid phase, the temperatures at which phase changes from the liquid occur will be different than those for the pure solvent. In particular the freezing point of a solution formed by mixing a pure liquid with a small amount of solute will decrease and the boiling point of the solution will increase. These effects are referred to as freezing point depression and boiling point elevation and both are part of a larger class of related phenomena called colligative properties - other examples include the lowering of the vapor pressure of a pure liquid by the addition of solutes and the phenomena of osmotic pressure, which is important in keeping blood cells from collapsing. In our everyday lives, the phenomenon of freezing point depression explains why road ice will melt when salt is poured on it (the freezing point is lowered by the addition of the impurity) and why we put ethylene glycol (antifreeze) in our automobile radiators to protect our cooling systems from freezing in extreme cold. Physically, freezing point depression and boiling point elevation occur because a solution consisting of a solvent (the formerly pure liquid) with a small amount of solute added is thermodynamically more stable than the pure liquid itself - primarily due to the fact that the addition of the solute increases the entropy of the solution. This additional stability increases the range of temperatures at which the liquid is stable - lowering the freezing point and increasing the boiling point. In this lab you will be exploring how the freezing point depression of water depends upon the amount and nature of the added solute.
1. Do not allow dry ice to touch bare skin -- it can cause severe frostbite.
2. You must wear goggles at all times while performing this laboratory.
3. Never ingest chemicals or put any object (including pens and pen caps!) into your mouth in the lab. Chemicals in the lab may be contaminated or poisonous.
Think back to last week's lab, in which you acquired the freezing curve of water. (See Freezing of Water and Solutions)
Water is the coolant used in automobile radiators because it can absorb large amounts of thermal energy (heat) to help keep the engine cool. Water typically freezes at 0° C (32° F) and at temperatures below 0° C water is a solid. A solid coolant in a car will not perform its intended function; and since water expands as it freezes it could expand and damage the radiator and parts of the engine.
Given the potential problems above, why is water used as a coolant in automobile engines – even in cold climates?
How do we keep the water in
the radiator from freezing?
When you have your car “winterized” you can test how low the temperature of the radiator coolant can go before you have a problem.
How is the temperature at which the coolant will freeze determined?
The prelab assignment is to answer the above questions.
Laboratory Experiment Part 1
· How do your explanations in the pre-laboratory compare with one another?
· Describe the explanations that you made. Discuss until you reach consensus on the observations, and write them on the blackboard.
Class discussion/ Instruction: 5 – 10 minutes
· Groups share their explanations with other groups in the laboratory.
InvestigationFor this investigation you will use the Vernier LabPro box and LoggerPro software to record the cooling curve and to determine the freezing point of a water/alcohol mixture. Each group will be assigned a different alcohol to use in the investigation; your group will be assigned one of the following:
|Alcohol||Formula||Molecular Weight (g/mol)||Freezing Point
|1-Propanol (n-propyl alcohol)||CH3CH2CH2OH||60.11||-126.5|
|2-Propanol (isopropyl alcohol)||(CH3)2CHOH||60.11||-89.5|
Consider your experience in previous
experiments -- discuss how your group will determine the freezing point of the
alcohol/water mixture, and how that freezing point differs from the normal
freezing points of pure water and the pure alcohol.
What materials might you use?
How will you set up the experiment?
What data will you collect?
How will you determine the confidence that you have in your results?
Each team shares their results from Part 1 with the class. Record the class data in your notebook. Discuss trends seen in the data from Part 1.
What trends do you see in the data?
How can you account for the trends?
What might be the cause of the trends?
How could you test your assumptions?
How could you use the trends to investigate the molecular weight of an
How could you use the trends to investigate the molecular weight of an unknown?
Laboratory Experiment Part 2
In Part 2 you will determine the molecular weight of an unknown alcohol by a method that utilizes freezing point depression. In Part 1 you determined the normal freezing point of water and the freezing points with several alcohols mixed into the water. You will use that information to calculate the change in the freezing point temperature which can then be used to determine the molecular weight of the unknown alcohol.
What trend did you see in Part 1?
Follow the approach you used in Part 1 with your unknown alcohol (assigned by your TA), and use the data to calculate the molecular weight of the alcohol. (See Related Materials section of this document.)
All alcohols are organic molecules containing carbon, hydrogen, and oxygen. Based on the calculated molecular weight estimate how many carbons are in your unknown alcohol. (Hint: look at the table in Part 1 and see where it might fit in that sequence.)
Your TA will assign your team one of five alcohols to use.
1. Weigh out 10.0 grams of the alcohol that you were assigned. Pour the sample into a clean, dry 250-mL Erlenmeyer flask.
2. Weigh out 100.0 grams of distilled water and pour it into the Erlenmeyer flask containing the alcohol. Mix the liquids.
3. Prepare a dry-ice bath by filling a 600-mL beaker 2/3-full with dry-ice chunks. CAUTION: Be careful handling the dry ice as it can cause severe burns when it comes in contact with unprotected skin.
4. Set up the Temperature Probe and the LoggerPro software for collection of temperature data. You may find it helpful to open “Exp 02” from the Chemistry with Computers experiment files of Logger Pro. Each run should take less than 10 minutes.
5. Pour approximately 30 mL of distilled water into a 50-mL Erlenmeyer flask. (Running the experiment first with pure water allows you to determine a baseline freezing temperature.)
6. Place the temperature probe into the mouth of the flask; the probe should be submerged in the liquid.
7. Place the flask in the dry ice bath. CAUTION: Do not touch the dry ice with bare hands.
8. Begin collecting data. Make certain that you continually swirl the water in the flask to maintain a uniform temperature in your solution.
9. After the freezing point has been reached and is relatively stable, you may stop the data collection.
Repeat steps 6-10 using approximately 30 mL of your alcohol/water solution instead
of pure water.
A trend you may have observed in Part 1 is that as the molecular weight of the alcohol increases, the magnitude of DT decreases. This is because the magnitude of a colligative property is, by definition, based on the number of solute particles in a solution. Each group prepares their sample by adding 10.0 grams of their assigned alcohol to 100.0 grams of water. Thus, the total number of alcohol molecules in a particular sample is inversely related to the molecular weight of the constituent alcohol. For example,10.0 grams of methanol (molecular weight 32.04 g/mole) consists of a greater number of molecules than 10.0 grams of 1-butanol (molecular weight 74.12 g/mole).
The equation relating the freezing point depression and concentration of a solution is
DTf = Kf m
where DTf is the change in freezing point (the difference in the freezing point of a pure solvent and the freezing point of a solution using that solvent), Kf is a constant for the solvent, and m is the concentration of the solution in molal units (moles of solute per kilogram of solvent). We can rearrange the equation to determine the molal concentration:
m = DTf / Kf The Kf of water is 1.86 ( ºC/m)
Next, recognize that molality can be defined mathematically as m = (molessolute / kgsolvent). In this case, because both m and kgsolvent are known, the number of moles of solute in the sample (moles solute) can be readily calculated.
Finally, knowing the weight of the solute you used (we'll call it X in the calculation below), and having determined molessolute (we'll call it Y), you can calculate the molecular weight of the solute:
Molecular Weight = (X grams) /(Y moles) = (X/Y) g / mole