Purpose
To study how molality affect the freezing point by comparing two molar freezing point of anti-freeze solutions.
Data Table
Freezing Point of Distilled Water (T H2O)
Freezing Point of Solution 1 (T1) Freezing Point of Solution 2 (T2) |
1*C
-2*C -6*C |
Calculations
1. Determine the freezing point depression of solution 1.
Freezing point depression = temperature change between freezing point of antifreeze solution (mostly C2H4(OH)2) and of H2O
Freezing point depression = freezing point of solution 1 - freezing point of distilled water
Freezing point of solution 1 = -2*C - 1*C = -3*C
2. Determine the freezing point depression of solution 2.
Freezing point depression = freezing point of solution 2 - freezing point of distilled water
Freezing point of solution 2 = -6*C - 1*C = -7*C
3. Determine the molality of solution 1. (Kf = -1.86*C*kg/mol for water)
Freezing point depression of solution 1 = kf * m1
-3*C = (-1.86*C*kg/mol)(m1)
m1 = 1.61 mol/kg
4. Determine the molality of solution 2.
Freezing point depression of solution 2 = kf * m2
-7*C = (-1.86*C*kg/mol)(m2)
m2 = 3.76 mol/kg
5. To find the molar mass of antifreeze, you first need to calculate the number of grams of antifreeze per 1000 grams of solvent for the solutions.
Solution 1
given: 5 g of antifreeze
50 mL of water
50 mL l 1 g l = 50 g
l 1 mL l
5 g of antifreeze = Xg of antifreeze
50g of water 1000 g of water
Xg = 100 g antifreeze
Solution 2
given: 10 g of antifreeze
50 mL of water
50 mL l 1 g l = 50 g
l 1 mL l
10 g of antifreeze = Xg of antifreeze
50g of water 1000 g of water
Xg = 200 g antifreeze
6. Find the molar mass of antifreeze.
Solution 1
100 g antifreeze/1000 g water = 100 g antifreeze/1 kg water
molality = 1.61 mol/kg
molar mass = g/mol
(100 g antifreeze/1 kg water) / (1.61 mol/kg)
62.11 g/mol
Solution 2
200 g antifreeze/1000 g water = 200 g antifreeze / 1 kg water
molality = 3.76 mol/kg
molar mass = g/mol
(200 g antifreeze/1 kg water) / (3.76 mol/kg)
53.19 g/mol
Freezing point depression = temperature change between freezing point of antifreeze solution (mostly C2H4(OH)2) and of H2O
Freezing point depression = freezing point of solution 1 - freezing point of distilled water
Freezing point of solution 1 = -2*C - 1*C = -3*C
2. Determine the freezing point depression of solution 2.
Freezing point depression = freezing point of solution 2 - freezing point of distilled water
Freezing point of solution 2 = -6*C - 1*C = -7*C
3. Determine the molality of solution 1. (Kf = -1.86*C*kg/mol for water)
Freezing point depression of solution 1 = kf * m1
-3*C = (-1.86*C*kg/mol)(m1)
m1 = 1.61 mol/kg
4. Determine the molality of solution 2.
Freezing point depression of solution 2 = kf * m2
-7*C = (-1.86*C*kg/mol)(m2)
m2 = 3.76 mol/kg
5. To find the molar mass of antifreeze, you first need to calculate the number of grams of antifreeze per 1000 grams of solvent for the solutions.
Solution 1
given: 5 g of antifreeze
50 mL of water
50 mL l 1 g l = 50 g
l 1 mL l
5 g of antifreeze = Xg of antifreeze
50g of water 1000 g of water
Xg = 100 g antifreeze
Solution 2
given: 10 g of antifreeze
50 mL of water
50 mL l 1 g l = 50 g
l 1 mL l
10 g of antifreeze = Xg of antifreeze
50g of water 1000 g of water
Xg = 200 g antifreeze
6. Find the molar mass of antifreeze.
Solution 1
100 g antifreeze/1000 g water = 100 g antifreeze/1 kg water
molality = 1.61 mol/kg
molar mass = g/mol
(100 g antifreeze/1 kg water) / (1.61 mol/kg)
62.11 g/mol
Solution 2
200 g antifreeze/1000 g water = 200 g antifreeze / 1 kg water
molality = 3.76 mol/kg
molar mass = g/mol
(200 g antifreeze/1 kg water) / (3.76 mol/kg)
53.19 g/mol
Conclusion
According to the information from the antifreeze lab, one can assume that increased concentration of antifreeze in the solution would further lower the freezing point. The antifreeze solution with only 5 grams of antifreeze lowered the freezing point of the solution from 1*C to -2*C, but the antifreeze solution with 10 grams of antifreeze lowered the freezing point of the solution further, from 1*C to -6*C. The relationship between the concentration of the solution and the freezing point of solution shows how antifreeze prevent engine failures in cars during wintertime. Antifreeze does so by lowering the freezing point.
Discussion of Theory
Antifreeze is used to lower the freezing point, and its colligative property allows it to happen. Colligative properties are boiling point elevation, freezing point elevation, osmotic pressure, and lowering of vapor pressure. It solely depends on the number of the solute particle in a solution, which allows the colligative properties to be a tool for finding the nature of a solute and determining molar masses of substances. One of the important colligative property used in this lab was the freezing-point depression, which lowers the freezing point of a solution as more and more solute are dissolved in a solvent. This occurs because of the splitting ions called Van't Hoff Factor, or i. The i number increases as ions split, and increasing i number lowers the freezing point. Ice will not form at 0*C because the vapor pressure of the water was lowered by the splitted ions, having lower pressure than that of ice. The equation for freezing-point depression is the following:
Difference in T = Kf * m
where Kf represents the freezing point and m represents molality.
Difference in T = Kf * m
where Kf represents the freezing point and m represents molality.
Sources of Error
There are several possibilities of error in the antifreeze lab.
1. The water used was distilled water from the sink. The water contained other impurities that may have distorted the data. It may have been washed with flourine and other chemicals to be cleansed. If more accurate data is required, pure water, such as Fiji water, is recommended.
2. When measuring the freezing point of the antifreeze solution, the observer may have failed to record the exact temperature in which the first ice crystal has began to form. Pinpointing the exact point where the solution starts to freeze is very difficult for humans to do, and this may have affected some of the calculations.
3. The measuring equipments might not have been the most accurate. The triple-beam-balance and the thermometer might have defects and could have affected the data and the results.
1. The water used was distilled water from the sink. The water contained other impurities that may have distorted the data. It may have been washed with flourine and other chemicals to be cleansed. If more accurate data is required, pure water, such as Fiji water, is recommended.
2. When measuring the freezing point of the antifreeze solution, the observer may have failed to record the exact temperature in which the first ice crystal has began to form. Pinpointing the exact point where the solution starts to freeze is very difficult for humans to do, and this may have affected some of the calculations.
3. The measuring equipments might not have been the most accurate. The triple-beam-balance and the thermometer might have defects and could have affected the data and the results.
Critical Thinking: Analysis and Conclusions
1. Permanent antifreeze is almost 100% ethylene glycol (1, 2 ethanediol, C2H4(OH)2). Calculate its molar mass.
2(14g) + 2(1g) + 2(16g) + 2(1g) = 62 g
Molar mass of permanent antifreeze is approximately 62 grams/mol.
2. Calculate the percent error in both trials.
(Experimental - Theoretical) x 100% = Percent Error
Solution 1
(62.11 - 62.00) x 100% = .1774%
62.00
Solution 2
(53.19 - 62.00) x 100% = -14.21%
62.00
3. What do you think are the major sources of error in this investigation? How might some of them be reduced?
1. The water used was distilled water from the sink. The water contained other impurities that may have distorted the data. It may have been washed with flourine and other chemicals to be cleansed.
Solution: Pure water could be used to reduce the sources of error.
2. When measuring the freezing point of the antifreeze solution, the observer may have failed to record the exact temperature in which the first ice crystal has began to form. Pinpointing the exact point where the solution starts to freeze is very difficult for humans to do, and this may have affected some of the calculations.
3. The measuring equipments might not have been the most accurate. The triple-beam-balance and the thermometer might have defects and could have affected the data and the results.
Solution: Use of digital scales and trusted thermometer will help to minimize error.
2(14g) + 2(1g) + 2(16g) + 2(1g) = 62 g
Molar mass of permanent antifreeze is approximately 62 grams/mol.
2. Calculate the percent error in both trials.
(Experimental - Theoretical) x 100% = Percent Error
Solution 1
(62.11 - 62.00) x 100% = .1774%
62.00
Solution 2
(53.19 - 62.00) x 100% = -14.21%
62.00
3. What do you think are the major sources of error in this investigation? How might some of them be reduced?
1. The water used was distilled water from the sink. The water contained other impurities that may have distorted the data. It may have been washed with flourine and other chemicals to be cleansed.
Solution: Pure water could be used to reduce the sources of error.
2. When measuring the freezing point of the antifreeze solution, the observer may have failed to record the exact temperature in which the first ice crystal has began to form. Pinpointing the exact point where the solution starts to freeze is very difficult for humans to do, and this may have affected some of the calculations.
3. The measuring equipments might not have been the most accurate. The triple-beam-balance and the thermometer might have defects and could have affected the data and the results.
Solution: Use of digital scales and trusted thermometer will help to minimize error.
Critical Thinking: Applications
1. Could freezing point depression be used for substances not soluble in water?
The freezing point depression cannot be used for insoluble substances. This has to do with its colligative property, which only applies to solubles. The colligative property allows the freezing point depression to be used for substances soluble in water only, because it allows the ions to separate into smaller particles. This ion is called Van't Hoff Factor, or i, and it allows for the freezing point depression to be used in soluble substances. Insoluble substances, however, will not break into particles with freezing point depression. Van't Hoff Factor will not increase and stay as 1, which is not enough for the freezing point depression to be used for insoluble substances.
2. What effect on the freezing point depression of water would a 1 m solution of the ionic substance (NH4)3PO4 have?
The freezing point depression of water would decrease because the ionic substance (NH4)3PO4 would break into ions, raising the i Van't Hoff Factor. According to colligative property, increasing ions in the solution would lower the freezing point of the solution.
3. What assumption is made about the density of distilled water in this investigation?
Distilled water is assumed to be pure, as 1 mL was converted to 1 g. The density of the water in the lab was assumed to be 1 g/mL.
4. Would this method of molar mass determination be practical for other substances soluble in water?
This method of molar mass determination can be practical for other soluble substances. The ions will break into particles and raise the Van't Hoff Factor, allowing freezing point depression to be used.
The freezing point depression cannot be used for insoluble substances. This has to do with its colligative property, which only applies to solubles. The colligative property allows the freezing point depression to be used for substances soluble in water only, because it allows the ions to separate into smaller particles. This ion is called Van't Hoff Factor, or i, and it allows for the freezing point depression to be used in soluble substances. Insoluble substances, however, will not break into particles with freezing point depression. Van't Hoff Factor will not increase and stay as 1, which is not enough for the freezing point depression to be used for insoluble substances.
2. What effect on the freezing point depression of water would a 1 m solution of the ionic substance (NH4)3PO4 have?
The freezing point depression of water would decrease because the ionic substance (NH4)3PO4 would break into ions, raising the i Van't Hoff Factor. According to colligative property, increasing ions in the solution would lower the freezing point of the solution.
3. What assumption is made about the density of distilled water in this investigation?
Distilled water is assumed to be pure, as 1 mL was converted to 1 g. The density of the water in the lab was assumed to be 1 g/mL.
4. Would this method of molar mass determination be practical for other substances soluble in water?
This method of molar mass determination can be practical for other soluble substances. The ions will break into particles and raise the Van't Hoff Factor, allowing freezing point depression to be used.