Lab 9. Colligative Properties an Online Lab Activity

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1 Prelab Assignment Before coming to lab: Lab 9. Colligative Properties an Online Lab Activity Chemistry K. Marr Revised Winter 2014 This lab exercise does not require a report in your lab notebook. The report for this exercise consists of completing the attached Report pages that start on page 3 as you carry out the procedure on page 2. Answer questions 1 and 2 on page 3. Answer the pre-lab questions on the last 2 pages, to be turned in at the start of the lab period. Purpose In this online lab activity you will use freezing point depression and boiling point elevation data to calculate the K f and K b of water and explore the van t Hoff factor for electrolytes. Introduction The physical properties of solutions that depend solely on the number of dissolved solute particles (molecules or ions) and not their chemical identities are known as colligative properties. Some colligative properties are vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure. When a nonvolatile solute is added to a solvent, the vapor pressure of the solvent above the solution is less than the vapor pressure above the pure solvent as demonstrated in figure 1. The boiling point of the solution will be greater than the boiling point of the pure solvent because the solution will need to be heated to a higher temperature in order for the vapor pressure to become equal to the external pressure of the atmosphere. This boiling point elevation depends only on the concentration of the solute particles. For a nonelectrolyte solute the relationship is: where: T b = K b m T is the change in the boiling point i.e. by how much the boiling point is elevated. K b is the molal boiling point elevation constant, different for each solvent, with units of o C/m. m is the molality of the solution, a unit of concentration that does not vary with temperature and thus a good unit to use in colligative property calculations. molality moles solute m kg solvent Molarity M moles solute Liter solution When a non-volatile solute is added to a pure solvent, the freezing point of the solution is always lower than that of the pure solvent. This freezing point depression depends only on the concentration of the solute and not on its chemical identity. The amount by which the freezing point is lowered by a non-electrolyte is given by the following relationship: T f = K f m T f is the change in the freezing point i.e. by how much the freezing point of a solvent is lowered due to the presence of dissolved solute(s) K f is the freezing point depression constant (units, o C/m). K f is different for each solvent. m is the molality of the solution. Page 1 of 10

2 When a pure liquid freezes, the temperature remains constant until all of the liquid has frozen. The plateau in the curve for liquid A in figure 2 represents this. When a solution freezes, only the pure solvent actually freezes. The solute remains in solution. Thus, as the pure solvent begins to freeze out, the remaining solution becomes increasingly concentrated, causing the freezing point to drop even lower. This represented by the sloped line in the graph for liquid B in figure 2. When the solute is an electrolyte, it is necessary to include the van t Hoff factor, i, in these equations. T b = i K b m T f = i K f m This factor represents the number of moles of ions that would be produced after complete dissociation per one mole of electrolyte in a very dilute solution (ideal conditions). Experimentally, i is normally less than the ideal value due to the formation of ion pairs in the solution. Ion pairing in solution serves to decrease the concentration of solute in solution. Consider this What Figure 2. Cooling curves for a pure solvent and a solution. The freezing point depression, T f, is the difference in the freezing point of the pure liquid solvent (A) and that of the solution (B). effect do you think ion pairing has on T f and T b? Why? What effect do you think solute concentration has on the formation of ion pairs? Why? In today s online experiment, you will measure both the freezing point depression, T f, and the boiling point elevation, T f, when known amounts of solutes are added to known amount of solvents. From the data collected you will be able to calculate the K f and K b of your solvent, and calculate the van t Hoff factor for two electrolytes. Procedure 1. Go to: 2. For each experiment in Tables 1 and 2 on page 3: a. Click on the switch on the water bath to choose the correct bath setting, cold (for freezing point measurements) or hot (for boiling point measurements). b. Select the appropriate amount of solvent and solute for experiment A in Table 1 on page 3. c. Click Start and record the freezing point in Table 1 once the temperature becomes relatively constant. d. Click Reset and then perform each of the experiment in Tables 1 and 2 on page 3. Acknowledgement: This lab is adapted from a similar lab produced by my chemistry colleague at GRCC, Dr. Sue Critchlow Figure 3. Online simulation of boiling point elevation and freezing point depression. Pictured is a study of the freezing point for of a solution of sulfur in carbon tetrachloride. Page 2 of 10

3 Name Lab 9 Report Sheet Colligative Properties Team No. Date Section 1. In your own words, briefly state the purpose of the lab. 2. List the freezing point depression and boiling point elevation equations (there are total of 4!). Table 1. Freezing Point Data (Use a pen to record all results!) Experiment Freezing Point ( ) T f ( ) Moles of Solute Added Molality ( ) i (from expmtl Data) A g water, no solute B g water, 4.28 g C 12 H 22 O 11 C g water, 8.56 g C 12 H 22 O 11 D g water, 1.46 g NaCl E g water, 2.92 g NaCl F g water, 5.55 g CaCl 2 G g water, 8.32 g CaCl 2 Table 2. Boiling Point Data (Use a pen to record all results!) Experiment Boiling Point ( ) T b ( ) Moles of Solute Added Molality ( ) i (from expmtl Data) H g water, no solute I g water, 4.28 g C 12 H 22 O 11 J g water, 8.56 g C 12 H 22 O 11 K g water, 1.46 g NaCl L g water, 2.92 g NaCl M g water, 5.55 g CaCl 2 N g water, 8.32 g CaCl 2 Page 3 of 10

4 Analysis of the Results for the Freezing Point Data (It s O.K. to use a pencil for this section.) 1. Use the freezing point of the pure solvent (H 2 O) measured in Expmts A and H and the freezing point of each solution measured in Expmts B G and I N to calculate the freezing depression, T f, for each solution. Record the T f results in table 1. Show below a sample calculation for experiment B and circle your answer. 2. Calculate the moles of solute added in each experiment in tables 1 and 2. Show below a sample calculation for experiments B, D and F. Circle your answers. Solution B Solution D Solution F 3. Use the moles of solute in step 2, above, to calculate the molality for each solution and record the results in tables 1 and 2. Show below a sample calculation for experiments B, D and F. Circle your answers Solution B Solution D Solution F 4. a) When you doubled the mass of sucrose in experiments B and C, what effect did this have on the change in freezing point? Use specific numerical data to support your response. b) Explain why doubling the mass of sucrose had the effect that it did on the freezing point. Page 4 of 10

5 5. Compare the molalities of the solutions in experiments E and F. Why does the T f differ? Use specific numerical data to support your response. 6. a) Calculate the K f for water for solutions B and C, find the average value, and compare it to the accepted value, 1.86 o C/m. Comment on the accuracy and the likely source of error. Circle your answers. Calculation of K f of for Expmt B Calculation of K f for Expmt C: Average K f for Experiments B and C Comment on the likely source or Error 7. a) Use the accepted value for the K f of water and the data from experiments D, E, F and G to calculate the van t Hoff factor for NaCl and CaCl 2. Circle your answers. Calculation i for NaCl (using Expmt D data) Calculation i for NaCl (using Expmt E data) Calculation i for CaCl 2 (using Expmt F data) Calculation i for CaCl 2 (using Expmt G data) Page 5 of 10

6 b) Are the van t Hoff factor values calculated above in part a what you would have predicted? Do they change with concentration of the electrolyte? Explain using specific numerical data to support your response. Analysis of the Results for the Boiling Point Data (It s O.K. to use a pencil for this section.) 8. Use the experimentally determined boiling points to find the boiling point elevation, T b, for each solution in Table 2. Record the results in table 2. Show below a sample calculation for experiment I. Circle your Answer. 9. When you doubled the amount of solute in experiments I and J, what effect did this have on the change in boiling point? Explain why there is a difference. Use specific numerical data to support your response. 10. Compare solutions J and K. What do they have in common? Why does the T b differ? Use specific numerical data to support your response. Page 6 of 10

7 11. a) Calculate the K b for water for solutions I and J. Calculate the average value and compare it to the accepted value, o C/m. Comment on the accuracy and the likely source of error. Circle your answers. Calculation of K b of for Expmt I Calculation of K b for Expmt J Average K b for Experiments I and J Comment on the likely source or Error 12. a) Use the accepted value for the K b of water and the data from experiments K, L, M and N to calculate the van t Hoff factor for NaCl and CaCl 2. Record your results in Table 2. Circle your answer below. Calculation i for NaCl (using Expmt K data) Calculation i for NaCl (using Expmt L data) Calculation i for CaCl 2 (using Expmt M data) Calculation i for CaCl 2 (using Expmt N data) b) How do the van t Hoff factor values calculated above compare to the values calculated from the freezing point data? Explain any differences. Use specific numerical data to support your response. Page 7 of 10

8 13. Use the data in figure 3 on page 2 to calculate the molecular formula of sulfur (S x ) that is, determine the whole number value of x. The freezing point of the solution in figure 3 is o C. Circle your answer. Error Analysis Questions (It s O.K. to use a pencil for this section.) How would the calculated value of the K f be affected (i.e. high, low or unaffected) by each of the errors below? In your explanations discuss the experimentally measured value(s) that will be impacted by the error, how the measured value(s) will be affected (i.e. high, low or unaffected) and how this affects the experimentally determined K f. 14. If some of the solvent (i.e. water) evaporated during the experiment, then the calculated value for the K f would be... a.) too high b.) too low c.) unaffected Circle your choice and explain below. 15. If some of the solute did not dissolve, then the calculated value for the K f would be... a.) too high b.) too low c.) unaffected Circle your choice and explain below. 16. If the freezing point of the pure solvent was determined accurately, but the freezing point of the solution was poorly estimated and was 1.0 o C too high, then the calculated value for the K f would be... a.) too high b.) too low c.) unaffected Circle your choice and explain below. 17. If the temperature probe was not properly calibrated and was reading high by 5.0 o C, then the calculated value for the K f would be... a.) too high b.) too low c.) unaffected Circle your choice and explain below. Page 8 of 10

9 Lab 9. Colligative Properties Name Prelab Questions Team Date Section Instructions: Complete the following 6 questions and hand in at the start of your lab period. Show your work with units and correct significant figures for all questions that involve a calculation. Circle all numerical answers. 1. Distinguish between molarity and molality and then explain why molality and not molarity is used as the unit of concentration in colligative property calculations such as freezing point depression and boiling point elevation. 2. What is the freezing point of a solution of 6.90 g or pure methanol, CH 3 OH, in grams of water? K f for water is 1.86 o C/m and the freezing point of pure water is defined as 0 o C. Circle your answer. 3. The freezing point of a solution made by dissolving g of a certain sugar in 225 g of water is 4.00 o C. What is the molar mass of the sugar? K f for water is 1.86 o C/m. Circle your answer. Page 9 of 10

10 4. Recall that an electrolyte is a compound that dissociates into ions in aqueous solution (hence the solution is a conductor of electricity), whereas a nonelectrolyte dissociates into neutral molecules when dissolved in water (hence the solution is a nonconductor). As stated in the introduction of this lab, for a solution involving an electrolyte the change in the freezing temperature can be expressed as T f = i K f m where i is the van t Hoff factor and represents the number of moles of ions that would be produced after complete dissociation of one mole of the electrolyte in a very dilute solution (i.e. ideal conditions). Assuming 100% dissociation of ions, what is the van t Hoff factor for each substance below in aqueous solution? a.) Nonelectrolyte: i = d.) CH 3 CH 2 OH (ethanol): i = b.) NaCl: i = c.) Na 3 PO 4 : i = e.) CaCl 2 : i = f.) HNO 3 (nitric acid, a strong acid): i = 5. Without doing a calculation, how does the freezing point of m KCl compare to that of m C 12 H 22 O 11 (sucrose i.e. table sugar, a nonelectrolyte)? The freezing point of m KCl is a.) lower than the FP of m C 12 H 22 O 11 b.) greater than the FP of m C 12 H 22 O 11 c.) the same as the FP of m C 12 H 22 O 11 Circle your choice and explain your reasoning below: 6. Assuming 100% dissociation in aqueous solution, calculate the freezing point of a solution made by dissolving 5.00 g NaCl in ml DI water. Assume the density of DI water = g/ml. K f for water is 1.86 o C/m. Circle your answer. Page 10 of 10

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