Preparation of Ice Bath
1. Fill the large beaker 3/4 full with crushed ice, and cover the ice with 1-2 cm of table salt.
2. Stir the mixture with a stirring rod ensuring that the temperature drops to at least -10°C.
Determination of Freezing Points of Solutions
1. Prepare 100ml of a solution with 5.85g of NaCl.
2. Prepare 10ml of a solution with 3.42g of sucrose.
3. Place a test tube 1/2 full of water in the ice bath.
4. Stir the water in the test tube gently with a thermometer and keep track of the temperature.
5. Record the temperature when the first ice crystals appear on the wall of the test tube. This is
the freezing point of the liquid and the liquid is pure water.
6. Repeat steps 3-5 with the other prepared NaCl and sucrose solutions.
7. Calculate the molalities of the NaCl and sucrose solutions.
8. Determine the value of i for the 2 solutions.
9. From the answers in 7 & 8 above and the the molal elevation constant (Kb) for water is 0.51
calculate the boiling points for the NaCl and sucrose solutions evaluated.
Molecular Mass Determination From Freezing Point Depression - Part 1
Using the 3 unknown solutions:
1. Take the first unknown and dissolve the amount stated on the label in 10 ml of water.
2. Freeze the solution the same way from ‘determination of freezing points’ and record the Tf.
3. Calculate the molecular mass of this unknown solute from its freezing point depression using
the following formula:
Molecular mass of solute = [(Kf) (grams of solute)] ÷ [(T) (kg of solvent)]
4. Repeat this for the 2 other unknown samples.
Molecular Mass Determination From Freezing Point Depression - Part 2
Lauric acid is used as the solvent and benzoic acid is the solute. A number of
repeats will be performed using different concentrations of benzoic acid in the lauric acid.
Determination of the melting point of pure lauric acid.
1. Weigh out 1.0 g of solid lauric acid and put into a test tube.
2. Fill a 600 mL beaker about three-quarters full with water and heat on the hot plate.
3. Place test tube in beaker and heat until the lauric acid is completely melted (stir the lauric acid
carefully with the thermometer)
4. Once the lauric acid is completely melted, carefully remove the lauric acid from the hot water.
(Keep the beaker of water warm)
5. Record the temperature every 30 seconds until the temperature reaches 35°C.
Determination of the freezing point of lauric acid/benzoic acid mixture
1. Weigh out 1.0 g of solid lauric acid and put it into the test tube.
2. Weigh out 0.1 g of benzoic acid and place into the test tube with the lauric acid.
3. Using hot water from the previous procedure with pure lauric acid place the test tube in a
beaker and heat until the lauric acid/benzoic acid mixture is completely melted
4. Once the mixture is completely melted, carefully remove it from the hot water and allow it to
cool.
5. Record the temperature every 30 seconds until the temperature reaches 25°C.
6. Repeat steps 1-5 two more times increasing the amount of benzoic acid by 0.1 g each time.
Record the changes in temperature with time for each.
Results
Freezing Points of Compounds
- Water = 0°C
- NaCl = -4°C
- Sucrose = -2°C
- Unknown A = -4°C
Molalities
The equation used is: Molality = moles of solute / solvent in kg
NaCl: Molality = 0.1/0.1
= 1mol/kg
Sucrose: Molality = 0.01/0.01
= 1mol/kg
i Values
To be able to calculate the i values for the solutions we need to use the equation:
T = Kf im
NaCl: 4 = 1.86 x i x 1
i = 2.15
Sucrose: 2 = 1.86 x i x 1
i = 1.07
Boiling Points
Calculations of NaCl and sucrose boiling points were determined using:
ΔTb = Kbm
Kb = molal elevation constant
M = molality if the solution
NaCl: ΔTb = 0.51 x 1
ΔTb = 0.51°C
Sucrose: ΔTb = 0.51 x 1
ΔTb = 0.51°C
Molecular Mass of Unknown A
Molecular mass of solute = (Kf x solute in grams) / (T x solvent in kg)
Molecular mass of Unknown A = (1.86 x 1) / (4 x 0.01)
Molecular mass of Unknown A = 46.5 kg/mol
Table 1: Showing the temperature decrease as time increases using pure Lauric Acid
Table 2: Showing the temperature decrease to 25°C as the time increased using 1.0g Lauric Acid & 0.1g Benzoic Acid, 1.0g Lauric Acid & 0.2g Benzoic Acid and 1.0g Lauric Acid & 0.2g Benzoic Acid.
Freezing Points for Each Trial of Benzoic Acid
The freezing point is determined from the graph by reading off the temperature at which the graph levels off:
- 1g Lauric Acid (Graph 2): Freezing point = 42°C
- 1g Lauric Acid + 0.1g Benzoic Acid (Graph 1): Freezing point = 37°C
- 1g Lauric Acid + 0.2g Benzoic Acid (Graph 1): Freezing point = 34°C
- 1g Lauric Acid + 0.3g Benzoic Acid (Graph 1): Freezing point = 36°C
Tf Values
To determine the Tf value you subtract the values from the benzoic acid trials from the pure Lauric acid, this is the freezing point depression (the change in 2 freezing points)
-
1g Lauric Acid + 0.1g Benzoic Acid: Tf = 5°C (42 – 37)
-
1g Lauric Acid + 0.2g Benzoic Acid: Tf = 8°C (42 – 34)
-
1g Lauric Acid + 0.3g Benzoic Acid: Tf = 6°C (42 – 36)
Molality of Each Trial
The equation used is: Molality = moles of solute / solvent in kg
No moles 0.1g Benzoic Acid = 0.1 / 122.12 = 8.19x10-4 moles
No moles 0.2g Benzoic Acid = 0.2 / 122.12 = 1.64x10-3 moles
No moles 0.3g Benzoic Acid = 0.3 / 122.12 = 2.46x10-3 moles
1g Lauric Acid + 0.1g Benzoic Acid:
Molality = 8.19x10-4 / 0.001
= 0.82 mol/kg
1g Lauric Acid + 0.2g Benzoic Acid:
Molality = 1.64x10-3 / 0.001
= 1.64 mol/kg
1g Lauric Acid + 0.3g Benzoic Acid:
Molality = 2.46x10-3 / 0.001
= 2.46 mol/kg
Determination of the Molar Mass of Benzoic Acid
Using the equation:
Molecular mass of solute = (Kf x solute in grams) / (T x solvent in kg)
1g Lauric Acid + 0.1g Benzoic Acid:
(3.9 x 0.1) / (5 x 0.001) = 78.0 kg/mol
1g Lauric Acid + 0.2g Benzoic Acid:
(3.9 x 0.2) / (8 x 0.001) = 97.5 kg/mol
1g Lauric Acid + 0.3g Benzoic Acid:
(3.9 x 0.3) / (6 x 0.001) = 195.0 kg/mol
Average of the 3 trials = 123.5 kg/mol
The actual molar mass of benzoic acid is 122.12 kg/mol, therefore we can determine the experimental error:
(123.5 – 122.12) x 100 = 1.13%
122.12
Discussion
Effect of solutes on the freezing point of water
The freezing point of pure water was determined at 0°C and then the freezing points of a water/NaCl solution and water/sucrose solution where determined as -4°C and -2°C respectively. This was done by recording the temperature at which the first ice crystals appeared on the inside wall of the test tubes.
The freezing point is dependent on the number of solute particles in a solution. When sodium chloride dissolves in water it will dissociate into its 2 ions (Na+ and Cl-) therefore is will have the greatest effect on the freezing point of water and reduce it the most from 0°C to -4°C.
When sucrose dissolves in water it doesn’t dissociate into its ions because it’s a non electrolyte and will not have a significant effect on the freezing point of water. However, because it is a solute it will lower the freezing point slightly from 0°C to -2°C. This occurs because the solute can bind to variable amounts of water and will remove some water from the free water that will freeze.
The higher the molality of the solution the greater the effect on the freezing point of water. Both sucrose and NaCl has the same Molalities and therefore backs up that fact, as both lower the freezing point. In carrying out of this experiment the molecular weight of an Unknown Solute A was able to be determined as 46.5 kg/mol.
Accounting for the Experimental Error
The theoretical molar mass of benzoic acid is 122.12 kg/mol and the experimental molar mass was determined as 123.5kg/mol. This is a very slight difference as the low percentage error of 1.13% illustrates. This means that the results are reliable; however this error could be due to the 0.2g of benzoic acid having a lower freezing point of 34°C compared to the 0.3g benzoic acid having a freezing point of 36°C which is not expected. This leads to the very likely chance of a procedural error in weighing out the masses of compounds or carrying out the experiment.
Effect of a strong acid on the freezing point
If a strong acid was used it would’ve completely dissociated into its ions and therefore behave in a similar manner as the NaCl did on the freezing point of water; ie, it would lower the freezing point. This occurs because the dissociation means there is an increase in the number of solute particles and leads to the depression of freezing point.
Conclusion
There are therefore 3 things that can affect the freezing point:
- addition of a solute (eg; sucrose) – the greater the amount the greater the depression of freezing point
- addition of a strong acid - dissociation of ions increase the number of solute particles which cause the greatest decrease in freezing point (eg; NaCl)
- high molality solution will have greater effect on freezing point
References
1: wikipedia: Wikimedia foundation inc [2001]: Colligative properties [Available online] http://en.wikipedia.org/wiki/Colligative_properties [accessed 26 March 2007]
2: Activity 2: The Freezing Point Depression of Lauric Acid [Avaliable online] http://intro.chem.okstate.edu/ChemSource/Solutions/solutionslab2teach.html [ accessed 26 March 2007]
3: Patrick J Sinko:[2005]Martins Physical Pharmacy 5th Edition: Lippincott Williams & Wilkins
