Freezing point depression for solutions of sugar in water

Objective

To investigate the relationship between the molality and the freezing point of a solution.

Hypothesis

We believe that, when doing the experiment, the solutions with a higher concentration will have a lower freezing point. The reason for this is that the process of freezing requires the particles of the solvent to form a regular structure and, when we add a solute, it becomes more difficult for the liquid to adopt that regular structure (we must use more energy to achieve that solid state; more energy = lower temperature)

DTf  = Kc _·m

DTf = change in freezing point of the solution

Kc = cryoscopic constant

m = molality

From the equation, we could state the same hypothesis; the cryoscopic constant, as its name says, is not going to change so, if you multiply the same number with a higher molality (molality is the concentration of a solution expressed as the number of moles of solute per mass of solvent, therefore, if the mass of the solvent stays constant and there is a higher molality, we must have a bigger number of moles of solute) there will be a higher change in freezing point and, consequently, the freezing point will be lower as the molality increases (inversely proportional relationship)

Method

1. In each of the 6 test tubes, measure 5.0 grams of water.
2. Leaving one of the test tubes, add 0.5, 1.0, 1.5, 2.0 and 2.5 grams of sugar.
3. Stir the solutions to ensure all the salt has dissolved (you may need to warm the water slightly to dissolve all of the sugar.
4. Write the molalities of each sugar solution (*):
1. 0 grams of sugar in 5.0 grams of water: Pure water so, molality = 0 m.
2. 0.5 grams of sugar in 5.0 grams of water: molality = 0.6 m,
3. 1.0 grams of sugar in 5.0 grams of water: molality = 1.1 m.
4. 1.5 grams of sugar in 5.0 grams of water: molality = 1.7 m.
5. 2.0 grams of sugar in 5.0 grams of water: molality= 2.2 m.
6. 2.5 grams of sugar in 5.0 grams of water: molality = 2.7 m. 
7. Place the test tubes in a salt ice mixture and note at which temperature each solution freezes. How will you know it is freezing? We know it is freezing because the outside of the test tube will freeze before the inside, the sugar solution starts to get harder because layers will start to form and the bottom of the test tube will start to form ice.
8. Return the remaining liquid to its container.
9. Repeat the whole procedure for each substance.

(*) To calculate the molality, you divide the mass of the solute (grams) by its molecular mass and that would give you the number of moles. Then, you divide the number of moles by the kilograms of solvent. For instance: 1.5 grams of sugar in 5.0 grams of water: number of moles = 1.5/180; The result of that division is 0.0083 moles. Then, you divide that number by the kilograms of solvent to obtain the molality: molality = 0.0083/0.005 = 1.7 m.

Results

With the data we obtained, we have set up a table to organise the mass of sugar in the solution (and therefore its molality), the different results obtained in the different attempts (and therefore the average too) and finally, the difference in the freezing point of the sugar solution compared to water: 

Freezing point of a glucose solution in depending on its molality

Mass of sugar in solution (g)	Molality (mol/kg)	Attempt 1 - Freezing point (oC)	Attempt 2 - Freezing point (oC)	Average freezing point (oC)	Change in freezing point compared to pure water (oC)
0	0	0	0	0	0
0.5	0.6	-0.6	-0.8	-0.7	0.7
1.0	1.1	-0.3	-0.1	-0.2	0.2
1.5	1.7	-1.7	-1.7	-1.7	1.7
2.0	2.2	-2.0	-2.2	-2.1	2.1
2.5	2.7	-2.9	-2.5	-2.7	2.7

• The mass of sugar in the solution is measured with a balance.
• The freezing point is measured with a thermometer.
• The average freezing point is calculated by the addition of all of the attempts and that number is divided by the number of attempts. For example: the average of the freezing point of the sugar solution of 2.7 mol/kg would be: (2.9+2.5)/2 = 2.7 ºC
• The change in freezing point compared to pure water is calculated by the difference between the average freezing point and the freezing point of water (which is 0 ºC). For instance: if the average freezing of the solution is -2.7 and the freezing point of water is 0ºC: change in freezing point compared to pure water = 0-(-2.7) = 2.7 ºC

Afterwards, with these data, we elaborated a graph in which the dependent variable was the freezing point of the glucose solution (y axis) whereas the independent variable was the molality of the glucose solution (x axis):

(*) The reasons why the glucose solution with a molality of 1.0 has a higher freezing point than the rest can be due to the problems with the method stated in the evaluation.

Conclusions

During this experiment we have learnt the relationship between the molality and the freezing point. Due to the fact that molality is strongly related to how much solute there is (number of moles per kilograms of solvent) we have reached the conclusion that the higher the molality is, the lower the freezing point will be (just like we said in the hypothesis, which has been proved that is right); it's an inversely proportional relationship.

The freezing point decreases for the simple reason that freezing requires particles to form a regular structure and it is easier to form it with the pure solvent molecules whereas if you add a solute, it becomes more difficult to form the structure, so it will need a lower temperature than usual (more energy) to reach the solid state.

In other words, our conclusion from this experiment could be synthesized that what we thought before we did the experiment (the hypothesis) was right.

The usefulness of being able to lower the freezing point of water is, for example, on icy roads in winter, it helps to melt the ice from the roads by lowering the melting point of the ice, and this leads to a lower number of car accidents and the capacity of being able to ride on the road (which would be impossible if there were a high amount of ice)

We believe this experiment functioned well as the data we collected reflects what the theory says should happen: the more solute added, the lowest freezing point. This is, in general lines, what should happen. However, there’s one point at which this doesn’t happen: with 1 gram of sugar in the solution, the freezing point has higher than with 0,5, being the temperature of -0,2ºC, while the previous one was of -0,7ºC. Regarding the other data, all seem to be fine, and this is the only one whose data don’t match what really should be. The other strange temperature is the one of pure water, as it should have been of 0ºC, being the one achieved slightly lower than that, with -0,3ºC. This could have been because there was some solute that fell into the solution and changed the results, though this doesn’t seem likely. Another data that could be considered as strange is the one of 2,5 grams of sugar, as the difference between this and the previous one looks too big. This might be perfectly normal, but we see it as a little bit strange, though it still follows the theory.

Evaluation

Regarding the method, there were several problems that we are going to list below:

1. The first problem the experiment had, was that, due to the fact that it did not state the amount of time we should stir the solution, we stirred until we thought that the sugar had dissolved completely. The problem with this is that in one of the solutions the sugar may have dissolved more in others and, the more homogeneous the mixture is, the less it will affect to the freezing point the presence of the solute (in other words, the freezing point won't be so low). To solve this problem we have thought that the method should clarify or determine a specific amount of time in which we should be stirring the mixture, for example, one minute.
2. One of the problems of the experiment was that, as we had to cool down the water, we had to cover it with ice, which consequently provoked that, as the ice completely covered the water we could see, we had to take out the test tube from the beaker in order to see if the water had or hadn't cooled, we had to take the test tube out every few minutes, causing a change of temperature, interfering with the experiment and, therefore, changing the results. Due to his, sometimes, when we took out the test tube, the water sometimes hadn't frozen or had froze more than the previous test tubes, so the experiment wasn't as accurate as it should be. The solution we have thought for this problem is to use a water bath and put cold water inside, which would allow us to see the inside of the test tube, and therefore, we would know when the freezing process started, without having to take the test tube out.
3. Another problem with the method of the experiment was that we didn't know exactly when we had to take the test tube out of the beaker. This is very related to the problem stated above, since we had to keep taking out the test tube in order to see how much the freezing process had advanced, so, many times we took the test tube when it was about to completely freeze and other times, when it had jut started. If we hadn't had this problem and all of the test tubes had been frozen at the same point, some results may have varied (freezing point higher or lower). The best solution to this problem is the one stated above, since if we can see when every stage of the process is taking place, we will be able to determine if it’s the same that the time before and we will take all of the test tubes at more or less the same time, which will lead to more similar and accurate results. Another solution for this problem would be that the method specified the time each test tube spends inside the beaker, for example, 5 minutes, so that the method (and consequently the results) is more objective.

Bibliography

• Atkins, Peter and de Paula, Julio. Physical Chemistry for the Life Sciences. New York, N.Y.: W. H. Freeman Company, 2006. (124-136).

• Chemwiki.ucdavis.edu,. (2015). Freezing Point Depression - Chemwiki. Retrieved 21 February 2015, from http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Colligative_Properties/Freezing_Point_Depression
• Hyperphysics.phy-astr.gsu.edu,. (2015). Freezing Point Depression in Solutions. Retrieved 21 February 2015, from http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/meltpt.html