Purpose
The purpose of the lab is to determine the freezing point depression of antifreeze solution and the molar mass of ethylene glycol by cooling the different solutions in an ice bath.
Data Table
Freezing point of distilled water (T H2O)
|
1°C
|
Freezing point of Solution 1 (T 1)
|
-2°C
|
Freezing point of Solution 2 (T 2)
|
-3°C
|
Calculations
Freezing point depression of Solution 1:
ΔT = T (Solution 1) - T (H2O)
ΔT = -2°C - 1°C
ΔT = -3°C
Freezing point depression of Solution 2:
ΔT = T (Solution 2) - T (H2O)
ΔT = -3°C - 1°C
ΔT = -4°C
Molality of Solution 1:
ΔT = Kf (m) Kf= -1.86°C x kg / mol
-3°C = -1.86m
m = 1.6
Molality of Solution 2:
ΔT = Kf (m) Kf= -1.86°C x kg / mol
-4°C = -1.86m
m = 2.15
Number of grams of antifreeze per 1000 grams of solvent:
- Solution 1:
50 ml of H2O x 1 g of H2O
--------------- = 50 grams of H2O
1 ml of H2O
5 grams antifreeze X grams of anti freeze
----------------------- = --------------------------
50 grams solvent 1000 grams solvent
50x = 5000
----- --------
50 50
x = 100 grams of antifreeze
- Solution 2:
10 grams of antifreeze x grams of antifreeze
---------------------------- = --------------------------
50 grams of solvent 1000 grams of solvent
50x=10000
----- --------
50 50
x = 200 grams of antifreeze
Molar mass of antifreeze:
- Solution 1:
molar mass = grams/moles
100 grams of antifreeze / 1 kilogram of solvent
Molar mass = ----------------------------------------------------------
molality
Molar mass = 100 grams of antifreeze / 1 kilogram of solvent
----------------------------------------------------------
1.6 moles of antifreeze
-----------------------------
1 kg of solvent
Molar mass of antifreeze = 62.5 grams/mole
- Solution 2:
Molar mass = 200 grams of antifreeze / 1 kg of solvent
---------------------------------------------------
molality
Molar mass = 200 grams of antifreeze/ 1 kg of solvent
--------------------------------------------------
2.15 moles of antifreeze
------------------------------
1 kg of solvent
Molar mass of antifreeze= 93.02 grams/mole
ΔT = T (Solution 1) - T (H2O)
ΔT = -2°C - 1°C
ΔT = -3°C
Freezing point depression of Solution 2:
ΔT = T (Solution 2) - T (H2O)
ΔT = -3°C - 1°C
ΔT = -4°C
Molality of Solution 1:
ΔT = Kf (m) Kf= -1.86°C x kg / mol
-3°C = -1.86m
m = 1.6
Molality of Solution 2:
ΔT = Kf (m) Kf= -1.86°C x kg / mol
-4°C = -1.86m
m = 2.15
Number of grams of antifreeze per 1000 grams of solvent:
- Solution 1:
50 ml of H2O x 1 g of H2O
--------------- = 50 grams of H2O
1 ml of H2O
5 grams antifreeze X grams of anti freeze
----------------------- = --------------------------
50 grams solvent 1000 grams solvent
50x = 5000
----- --------
50 50
x = 100 grams of antifreeze
- Solution 2:
10 grams of antifreeze x grams of antifreeze
---------------------------- = --------------------------
50 grams of solvent 1000 grams of solvent
50x=10000
----- --------
50 50
x = 200 grams of antifreeze
Molar mass of antifreeze:
- Solution 1:
molar mass = grams/moles
100 grams of antifreeze / 1 kilogram of solvent
Molar mass = ----------------------------------------------------------
molality
Molar mass = 100 grams of antifreeze / 1 kilogram of solvent
----------------------------------------------------------
1.6 moles of antifreeze
-----------------------------
1 kg of solvent
Molar mass of antifreeze = 62.5 grams/mole
- Solution 2:
Molar mass = 200 grams of antifreeze / 1 kg of solvent
---------------------------------------------------
molality
Molar mass = 200 grams of antifreeze/ 1 kg of solvent
--------------------------------------------------
2.15 moles of antifreeze
------------------------------
1 kg of solvent
Molar mass of antifreeze= 93.02 grams/mole
Conclusion
Based on the data collected in the experiment , it is concluded that the addition of antifreeze lowers the freezing point of the solution. The freezing point depression of Solution 1 was calculated to be -3°C and the freezing point depression of Solution 2 was -4°C. The freezing point depression was calculated by subtracting the initial freezing point of water from the final temperature that was measured. The molality was calculated using the formula ΔT = iKf (m). The molality for Solution 1 was calculated to be 1.6 mol/kg and for Solution 2 to be 2.15 mol/kg. The molar mass of antifreeze was calculated using the molality obtained from the calculation above.
Discussion of Theory
Antifreeze is known to decrease the freezing point of a solution. The impact of antifreeze was tested in the experiment. Two solutions were prepared using 2 different amounts. The larger the amount of antifreeze used, the higher the freezing point depression resulted in, which lowers the freezing point of the solution. This is beneficial because the solution must reach a colder temperature in order to freeze due to colligative properties. Solution 1 was prepared with 5 grams of antifreeze in 50 ml of water. Solution 2 was prepared by using 10 grams of antifreeze in 50 ml of water. This ratio was used to calculate the number of moles of antifreeze in each solution by using the density of water. Because water has a density of 1g/ml, multiplying 50 ml of water by 1g/ml gives us the amount of water used in grams. Proportion was used to calculate the number of moles of antifreeze in 1000 grams of solvent (or 1kg). Moles over kg is the units for molality, so the molality calculated was divided into the number of moles calculated per 1000 grams. The molality was calculated by using the freezing point depression equation. Freezing point depression is a colligative property, which means it depends on the number of particles instead of the type of substance. The formula for freezing point depression is ΔT = (i)Kf (m). i stand for the Van't Hoff factor. For the experiment, the Van't Hoff factor was one because the solute remains as one particle. The solute was ethylene glycol C2H4((OH)2). The Kf constant, -1.86°C x kg / mol, was given information in the experiment. The constant was used to calculate the molality.
Sources of Error
One source of error is the use of tap water instead of distilled water. Tap water contains particles that has not been filtered out versus distilled water. This affects the data because freezing point depression is dependent upon the number of particles in the solution, known as colligative properties. The Van't Hoff factor for tap water is larger than the Van't Hoff factor of distilled water, which affects the freezing point depression even lower. However, this error did not affect the freezing point depression because the freezing point went higher. The freezing point of water is 0°C, however, the tap water was measured to be 1°C. If the tap water affected the freezing point, the additional particles in the solution would have lowered the freezing point of water lower than the freezing point of pure water.
Between measuring the freezing point of Solution 1 and 2 the same test tube was used. After washing the test tube from Solution 1, water still remained in the test tube. This could have affected Solution 2 by increasing the amount of solvent in the test tube. The calculation may be misleading because the extra water was not accounted for in the calculations.
For Solution 2, the test tube was left without stirring for about 2 minutes and when it was stirred the stirring rod was stuck to the bottom of the test tube. Because the solution was left alone for about 2 minutes this may have resulted in the solution freezing more quickly than if it was stirred.
Between measuring the freezing point of Solution 1 and 2 the same test tube was used. After washing the test tube from Solution 1, water still remained in the test tube. This could have affected Solution 2 by increasing the amount of solvent in the test tube. The calculation may be misleading because the extra water was not accounted for in the calculations.
For Solution 2, the test tube was left without stirring for about 2 minutes and when it was stirred the stirring rod was stuck to the bottom of the test tube. Because the solution was left alone for about 2 minutes this may have resulted in the solution freezing more quickly than if it was stirred.
Critical Thinking: Analysis and Conclusion
1. The molar mass of permanent antifreeze is 62.07. The molar mass was calculated by adding the molar mass of each individual element and multiplying the individual molar mass by the subscripts.
Molar mass = 2(12.01g/mol) + 4(1.008g/mol) +2(16g/mol + 1.008g/mol) = 62.07 g/mol
2.
Solution 1:
% Error = l Experiment - actual l
---------------------------- x 100%
actual
l 62.5g/mol - 62.07g/mol l
% Error = ----------------------------------- x 100% = .693%
62.07 g/mol
Solution 2:
% Error = l Experiment - actual l
---------------------------- x 100%
actual
l 93.02g/mol - 62.07g/mol l
% Error = ----------------------------------- x 100% = 49.86%
62.07 g/mol
3. One of the major sources of error is the use of tap water instead of pure water. This could have been prevent by using distilled water, however, it was not available at the time of the experiment. Another major source of error was the balance beam. The balance beams could have resulted in incorrect measurements, which would affect the calculations. This could be correct by using an electronic scale. The electronic scale would accurately measure because the scale would be zeroed and there would be no human error due to eye level.
Molar mass = 2(12.01g/mol) + 4(1.008g/mol) +2(16g/mol + 1.008g/mol) = 62.07 g/mol
2.
Solution 1:
% Error = l Experiment - actual l
---------------------------- x 100%
actual
l 62.5g/mol - 62.07g/mol l
% Error = ----------------------------------- x 100% = .693%
62.07 g/mol
Solution 2:
% Error = l Experiment - actual l
---------------------------- x 100%
actual
l 93.02g/mol - 62.07g/mol l
% Error = ----------------------------------- x 100% = 49.86%
62.07 g/mol
3. One of the major sources of error is the use of tap water instead of pure water. This could have been prevent by using distilled water, however, it was not available at the time of the experiment. Another major source of error was the balance beam. The balance beams could have resulted in incorrect measurements, which would affect the calculations. This could be correct by using an electronic scale. The electronic scale would accurately measure because the scale would be zeroed and there would be no human error due to eye level.
Critical Thinking: Applications
1. The freezing point depression could not be used for substances not soluble in water because Van't Hoff factor would be zero. The Van't Hoff factor is the number of particles that disassociate in the solution. Since insoluble compounds do not disassociate and remain as one the Van't Hoff factor would remain 1. The insoluble substance will not affect the freezing point and therefore, the freezing point depression would be 0.
2. NH4PO3 would increase the freezing point depression because The substance will disassociate into 4 particles. The Van't Hoff factor for NH4PO3 would be 4 (i=4). When 4 is plugged into the freezing point depression equation ΔT = (4)Kf (m), the freezing point depression will be 4 times greater than without the addition of NH4PO3. It would be greater because the molality of the solution is 1 and because Kf is a constant it will result in a higher freezing point depression and significantly lower the freezing point.
3. The assumption made about the density of distilled water in this investigation is that the density = 1g/ml. The density was used to calculate the number of moles of antifreeze in each solution. If the assumption made about the density of distilled water is incorrect, then the molar mass calculated will be incorrect.
4. This method would be practical for other substances soluble in water because as long the amount of solvent and solute is given, the experimenters can find the moles using the ratio. The Van't Hoff factor may vary depending on the substances because the Van't Hoff factor is the number of particles that disassociate. If this number increases, the freezing point depression increases as well and if the Van't Hoff factor decreases, the freezing point depression decreases. They both directly affect each other.
2. NH4PO3 would increase the freezing point depression because The substance will disassociate into 4 particles. The Van't Hoff factor for NH4PO3 would be 4 (i=4). When 4 is plugged into the freezing point depression equation ΔT = (4)Kf (m), the freezing point depression will be 4 times greater than without the addition of NH4PO3. It would be greater because the molality of the solution is 1 and because Kf is a constant it will result in a higher freezing point depression and significantly lower the freezing point.
3. The assumption made about the density of distilled water in this investigation is that the density = 1g/ml. The density was used to calculate the number of moles of antifreeze in each solution. If the assumption made about the density of distilled water is incorrect, then the molar mass calculated will be incorrect.
4. This method would be practical for other substances soluble in water because as long the amount of solvent and solute is given, the experimenters can find the moles using the ratio. The Van't Hoff factor may vary depending on the substances because the Van't Hoff factor is the number of particles that disassociate. If this number increases, the freezing point depression increases as well and if the Van't Hoff factor decreases, the freezing point depression decreases. They both directly affect each other.