FREEZING POINT DEPRESSION DISCUSSION The vapor pressure of a solution containing a non-volatile solute is lower than that of the pure solvent. Consequently, the boiling point of a solution is higher. The freezing point of a solvent is also affected by the solute; the freezing point of a solution is lower than that of the pure solvent. The freezing point depression, ΔTf, is defined as ΔTf = freezing point of solvent – freezing point of solution The freezing point depression, like the boiling point elevation, is termed a colligative property. Such properties depend only on the concentration of the solute particles, and not on their nature. For dilute solutions, ΔTf = mKf where m = molality (moles of soute particles per kg of solvent) Kf = molal freezing point depression constant, a property of the solvent. For solutions of electrolytes, such as NaCl, the solute ionizes in solution. Thus there are more solute particles than for the same number of moles of a non-electrolyte. For NaCl, Na+(aq) and Cl-(aq) exist in solution. As a result, the molality of the solute particles (ions in this case) is twice that for a non-electrolyte, and ΔTf is twice as large. However, if the solute particles associate in solution to form aggregates, then the number of solute particles (aggregates in this case) is reduced, causing ΔTf to be smaller than if the solute did not associate. The degree of dissociation or association of the solute particles is given by the van’t Hoff factor, i. This factor is the ratio of the number of moles of solute particles in solution to the number of moles of solute dissolved. If the solute dissociates in solution, i > 1 If the solute associates in solution, i < 1 An alternative form of the equation above that explicitly takes into account dissociation or association of the solute is shown below. ΔTf = imKf (m=molality of solute without taking dissociation or association into consideration) In this experiment, you are going to investigate the nature of acetic acid dissolved in a non-polar solvent, cyclohexane.OBJECTIVES 1. To understand colligative properties. 2. To find the freezing point depression of a solution. 3. To determine the van’t Hoff factor. 4. To deduce the nature of acetic acid dissolved in cyclohexane. PROCEDURE PART I: FREEZING POINT OF SOLVENT 1. Weigh a dry test tube supported in a 100-mL beaker. 2. Put about 15mL cyclohexane in the test tube and weigh again. 3. Put the test tube in a 600-mL beaker of water. Add a long stirring rod to the test tube. Clamp a thermometer in the water bath. Slowly cool the water bath to ice temperature by adding crushed ice. Stir gently with the glass rod. Continue the cooling and stirring until the first appearance of crystals. Note and record this temperature. 4. Allow the solid to melt. Change the water in the water bath. Repeat this determination at least one more time. PART II: FREEZING POINT OF SOLUTION 1. Drain ~0.2mL (read to within 0.005mL) of glacial acetic acid from the microburet directly into the cyclohexane. 2. Repeat the procedure as in Part I to determine the freezing point of the solution. CALCULATIONS 1. Calculate the freezing point depression from your data. 2. Calculate the van’t Hoff factor for acetic acid, CH3COOH (l), in cyclohexane, C6H12 (l). Given: Kf for cyclohexane is 20.0 degree per molal density of acetic acid at 20oC = 1.049 g/mL 3. Calculate the molecular mass of acetic acid in cyclohexane. 4. What acetic acid species exist in cyclohexane? Suggest a reason for their formation in cyclohexane.PROBLEMS 1. Calculate the freezing point of the following aqueous solutions. Kf for water = 1.86 degree per molal a. 5.00 g of NaCl in 50.0 g water. b. 5.00 g of AlCl3 in 50.0 g water. c. 5.00 g of glucose, C6H12O6, in 50.0 g water. 2. 5.50 g of a substance is dissolved in 50.0 g water. The solution was found to boil at 100.31oC. Assuming the solute does not dissociate nor associate, what is the molecular mass of the substance? (Kb for water = 0.51 degree per molal) O ‖ 3. Benzoic acid has the structure C6H5C-OH. A solution of 6.10 grams of benzoic acid in 250 grams of benzene, C6H6, had a boiling point of 80.253oC. a. Calculate the molecular mass of benzoic acid in benzene. Given: Boiling point of benzene is 80.0oC Kb for benzene is 2.53 degree per molal b. Suggest a reason for the difference between the molecular mass of benzoic acid determined in problem 3a above and its formula