purpose:
The purpose of this lab is to determine how antifreeze is able to lower the freezing point of water, in a car’s cooling system.
data table:
calculations:
conclusion:
Discussion of theory:
Antifreeze is a vital element in a car’s cooling system because it lowers the freezing point, which prevents the water inside the engine from becoming solid ice. Though the above commonly occurs in cold weathers, antifreeze can be used in warm weather because it also elevates the boiling point, which prevents the liquid water in an engine from changing into a gaseous state.
Freezing point depression describes how much a solvent’s, in the experiment water, freezing point lowers after it has been added a solute, in the experiment antifreeze. The previous occurred because when a solute is dissolved in a solvent, most likely water, the resulting solution does not freeze at 0.0°C due to a difference in vapors pressure: the solvent has acquired a lower vapor pressure than that of ice and requires lower temperatures to be able to be able to freeze. A lower vapor pressure occurs in a solution because the addition of a solute lowers the rate at which molecules change from liquid to solid. The new freezing point establishes the point at which the vapor pressure of the ice and that of the liquid are equal; this is also the point at which the rate of the liquid molecules leaving the solid is equal to the rate of formation for the solid. At both of these points equilibrium is reached. The freezing point depression equation for a non-electrolyte solution is ∆Tf = Kf ∙mSolute, and an electrolyte solution (solution with ions) is ∆Tf = iKf ∙mSolute : where ∆Tf is the freezing point depression , or change in temperature and it can be determined by subtracting the initial temperature from the final, i is the Van’t Hoff factor, Kf is the constant, known as the molal freezing-point depression constant, and mSolute is the molality of the solute in the solution.
Molality is the number of moles of solute per kilogram of solvent in a solution: because the freezing-point depression is dependent on the molality, the more moles of solute per kilogram of solvent in the solution, the higher the freezing point depression. This is why Solution 2 had a higher freezing point depression, because Solution 2 had twice as many grams as Solution 1, therefore it had a higher molality. Combined with the same constant Solution 2 provided a higher freezing point depression.The previous provides evidence that freezing-point depression is a colligative property. Colligative property: property dependent on the number of solute particles, and not on their actual identities. Another element that affects the freezing-point depression is the Van’t Hoff factor, i. The Van’t Hoff factor equals the moles of particles in solution divided by the moles of solute dissolved. In ionic compounds the Van’t Hoff factor is always a value greater than one, a covalent compound, or a compound that doesn’t split into ions is one, and an acid or base depends on whether the compound is a weak or strong base or acid.
Ionic compounds have a Van’t Hoff factor greater than one because when they are dissolved in water they are able to split about 99% into ions. For example, LiBr, when dissolved in water will dissolve into two ions Li+ and Br-, therefore its Van’t Hoff factor would be two. If the ionic compound were CaCl2, it would dissolve into three ions one Ca+ ion and two Cl- ions; therefore, its Van’t Hoff factor would be three. When dissolved in water, covalent molecules are only able to split 1%; therefore, they are unable to split into ions, and remain as one compound inside the solvent. For example; CO would no split almost completely into ions: it would remain as it is. Therefore, it has a Van’t Hoff factor of 1. Acid’s and Bases’ Van’t Hoff factors are dependent on whether they are weak or strong. A weak acid such as HF does not completely dissolve a water, so it’s Van’t Hoff factor is 1; but, since some the acid does dissolve its recognized Van’t Hoff factor is 1+, a little more than 1. A weak base follows this pattern. A strong acid, such as H2SO4 can dissolve into two ions H+ and HSO4- , but now HSO4- is a weak acid so some of its particles are able to dissolve; Therefore, H2SO4 has a Van’t Hoff factor of 2+.In both Solution 1 and 2 the Van’t Hoff factor was the same because the same compound was dissolve in both solutions, therefore the Van’t Hoff factor did not affect the freezing point depression in the experiment.
Freezing point depression describes how much a solvent’s, in the experiment water, freezing point lowers after it has been added a solute, in the experiment antifreeze. The previous occurred because when a solute is dissolved in a solvent, most likely water, the resulting solution does not freeze at 0.0°C due to a difference in vapors pressure: the solvent has acquired a lower vapor pressure than that of ice and requires lower temperatures to be able to be able to freeze. A lower vapor pressure occurs in a solution because the addition of a solute lowers the rate at which molecules change from liquid to solid. The new freezing point establishes the point at which the vapor pressure of the ice and that of the liquid are equal; this is also the point at which the rate of the liquid molecules leaving the solid is equal to the rate of formation for the solid. At both of these points equilibrium is reached. The freezing point depression equation for a non-electrolyte solution is ∆Tf = Kf ∙mSolute, and an electrolyte solution (solution with ions) is ∆Tf = iKf ∙mSolute : where ∆Tf is the freezing point depression , or change in temperature and it can be determined by subtracting the initial temperature from the final, i is the Van’t Hoff factor, Kf is the constant, known as the molal freezing-point depression constant, and mSolute is the molality of the solute in the solution.
Molality is the number of moles of solute per kilogram of solvent in a solution: because the freezing-point depression is dependent on the molality, the more moles of solute per kilogram of solvent in the solution, the higher the freezing point depression. This is why Solution 2 had a higher freezing point depression, because Solution 2 had twice as many grams as Solution 1, therefore it had a higher molality. Combined with the same constant Solution 2 provided a higher freezing point depression.The previous provides evidence that freezing-point depression is a colligative property. Colligative property: property dependent on the number of solute particles, and not on their actual identities. Another element that affects the freezing-point depression is the Van’t Hoff factor, i. The Van’t Hoff factor equals the moles of particles in solution divided by the moles of solute dissolved. In ionic compounds the Van’t Hoff factor is always a value greater than one, a covalent compound, or a compound that doesn’t split into ions is one, and an acid or base depends on whether the compound is a weak or strong base or acid.
Ionic compounds have a Van’t Hoff factor greater than one because when they are dissolved in water they are able to split about 99% into ions. For example, LiBr, when dissolved in water will dissolve into two ions Li+ and Br-, therefore its Van’t Hoff factor would be two. If the ionic compound were CaCl2, it would dissolve into three ions one Ca+ ion and two Cl- ions; therefore, its Van’t Hoff factor would be three. When dissolved in water, covalent molecules are only able to split 1%; therefore, they are unable to split into ions, and remain as one compound inside the solvent. For example; CO would no split almost completely into ions: it would remain as it is. Therefore, it has a Van’t Hoff factor of 1. Acid’s and Bases’ Van’t Hoff factors are dependent on whether they are weak or strong. A weak acid such as HF does not completely dissolve a water, so it’s Van’t Hoff factor is 1; but, since some the acid does dissolve its recognized Van’t Hoff factor is 1+, a little more than 1. A weak base follows this pattern. A strong acid, such as H2SO4 can dissolve into two ions H+ and HSO4- , but now HSO4- is a weak acid so some of its particles are able to dissolve; Therefore, H2SO4 has a Van’t Hoff factor of 2+.In both Solution 1 and 2 the Van’t Hoff factor was the same because the same compound was dissolve in both solutions, therefore the Van’t Hoff factor did not affect the freezing point depression in the experiment.
sources of error:
The first source of error in the experiment above was that tap water was used and not distilled water. The tap water may already have contained particles dissolved in the water which would affect the temperature of the solvent. In context, the fact that tap water was used and not distilled water would not affect any comparisons between the two because tap water was used for both; but, if compared to regular distilled water, the tap may have already had particles dissolved in it which might cause it to have a lower freezing point than regular distilled water. Another source of error was present when determining the freezing point of the tap water. The temperature should have been obtained when the water was first beginning to turn into ice (slush); however, the water froze at the bottom of the test tube, while the top remained a liquid. This might have caused the obtained value to be lower than the actual freezing point if it was measuring the ice, or higher if it was measuring the water in the test tube. It is concluded that the freezing temperature was too high, 0.5°C, because water’s regular freezing temperature was 0.0°C. The too high temperature could have caused the freezing-point depression value to be lower than it actually was. Also, Solution 2 did not freeze, so and estimate of an average of freezing-point depression value for Solution 2 was used. This value was higher than the true value of the freezing-point depression for Solution 2, because at -6.0°C Solution 2 was still a liquid. Consequently all the values and calculations using the above value were either to high if the variables were directly related or too low if they were inversely related. Also, for the second solution unknown quantities of salt and water were added to the beaker containing the second solution, this could have been the cause of the Second Solution not forming ice crystals.
critical thinking analysis:
3) Some of the major sources of error of the above investigation where that tap water was used instead of distilled water, causing it to have a lower freezing point than it should have originally had due to particles already being present inside the water, to reduce this source of error distilled water should have been used. Also, the calculated freezing point of water was 0.5°C when the freezing point of regular water was 0.0°C; this could have been eliminated by preventing the bottom of the test tube from being frozen and actually taking the temperature of the water when it was in the process of becoming a solid, or when it was slushy. Lastly, measuring the salt and water added to the beaker in the second Solution could have eliminated any uncertainties caused by adding unknown quantities of both salt and water.
critical thinking applications:
1) Freezing point depression can only be used for substances soluble in water. If a substance does not dissolve in water it is unable to lower the water’s vapor pressure and the rate at which molecules turn into a solid, therefore the freezing point of water would remain the same. If nothing is dissolved nothing changes. If a compound does not dissolve in water it cannot lower its freezing point. Also, in the freezing-point depression equation a value for molality is needed; Since the compound does not dissolve it is not recognized as a solute therefore its molality would be zero, because all the values are multiplied the equation would be invalid producing a result of 0=0.
2) Using a 1 molal solution of (NH4)3PO4 would still affect the freezing point depression. Even though (NH4)3PO4 has a molality of 1, it is an ionic compound and would dissolve into 4 ions: three NH4+ and one PO4-, and would have a Van’t Hoff factor of 4. Using the equation ∆Tf = iKf ∙mSolute, with -1.86 °C∙Kg/mol for the molal freezing-point depression constant of water , 1 molality, and a value of 4 for the Van’t Hoff factor, the freezing point depression value would be -7.44°C, which means the originals solvent’s freezing point was lowered by 7.44°C. The more ions a compound separates into the lower the freezing point depression because freezing point depression is a colligative property, depends on the amount and not the identity.
3) In the calculations taking place during the experiment it was assumed that the density of water was 1 g/ cm3, the density of water at 4°C. This value was used in the calculation for determining the grams of antifreeze per 1 kilogram, which in turn was used to determine the molar mass of antifreeze. If the value used was incorrect, the calculations determining the grams of antifreeze per kilogram of solvent and the molar mass of antifreeze are both incorrect.
4) Due to the fact that freezing point depression is a colligative property which means it depends on the amount and not the identity, other soluble substances can be used in place of antifreeze. Antifreeze is only one example of the various compounds that can be used. The freezing depression of a compound can be found by taking the freezing point temperature of distilled water, and then that of the compound dissolved in water. The temperature of the freezing point of the solvent is then subtracted from the temperature of the freezing point of the solution. Thus, providing the freezing point depression value. The resulting value can be used to determine the molality of a substance by using its Van’t Hoff factor, and the freezing-point depression constant unique to the solvent. Then the grams of the substance per kilograms of the solvent can be figured out by sing the density of the solvent. The resulting value would then be used in the process needed to discover the molar mass of the substance. Thus, providing an identical experiment with a different substance.
2) Using a 1 molal solution of (NH4)3PO4 would still affect the freezing point depression. Even though (NH4)3PO4 has a molality of 1, it is an ionic compound and would dissolve into 4 ions: three NH4+ and one PO4-, and would have a Van’t Hoff factor of 4. Using the equation ∆Tf = iKf ∙mSolute, with -1.86 °C∙Kg/mol for the molal freezing-point depression constant of water , 1 molality, and a value of 4 for the Van’t Hoff factor, the freezing point depression value would be -7.44°C, which means the originals solvent’s freezing point was lowered by 7.44°C. The more ions a compound separates into the lower the freezing point depression because freezing point depression is a colligative property, depends on the amount and not the identity.
3) In the calculations taking place during the experiment it was assumed that the density of water was 1 g/ cm3, the density of water at 4°C. This value was used in the calculation for determining the grams of antifreeze per 1 kilogram, which in turn was used to determine the molar mass of antifreeze. If the value used was incorrect, the calculations determining the grams of antifreeze per kilogram of solvent and the molar mass of antifreeze are both incorrect.
4) Due to the fact that freezing point depression is a colligative property which means it depends on the amount and not the identity, other soluble substances can be used in place of antifreeze. Antifreeze is only one example of the various compounds that can be used. The freezing depression of a compound can be found by taking the freezing point temperature of distilled water, and then that of the compound dissolved in water. The temperature of the freezing point of the solvent is then subtracted from the temperature of the freezing point of the solution. Thus, providing the freezing point depression value. The resulting value can be used to determine the molality of a substance by using its Van’t Hoff factor, and the freezing-point depression constant unique to the solvent. Then the grams of the substance per kilograms of the solvent can be figured out by sing the density of the solvent. The resulting value would then be used in the process needed to discover the molar mass of the substance. Thus, providing an identical experiment with a different substance.