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Chemistry 205 Colligative Properties Freezing Point Depression Purpose • To measure the change in temperature of water as it freezes, and to determine its freezing point. • To measure the drop in the freezing temperature (freezing-point depression) caused by the dissolution of a solute in water. • To study the effect of different solutes (electrolytes and non-electrolytes) on the freezing point of water. Concentration Units (Review!) The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass mass of solute x 100% % by mass = mass of solute + mass of solvent mass of solute x 100% = mass of solution Mole Fraction (X) moles of A XA = sum of moles of all components Ref. Chang, Sect. 12.3 Concentration Units Continued Molarity (M) M = moles of solute liters of solution Molality (m) m = moles of solute mass of solvent (kg) Ref. Chang, Sect. 12.3 Background • A pure liquid such as water or ethanol has characteristic physical properties: melting point, boiling point, density, vapor pressure, etc. • Addition of a soluble solute to the liquid forms a homogeneous mixture called a solution. • The solvent of the solution assumes physical properties that are no longer definite, but dependent on the amount of solute added. These properties are called colligative properties Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles (whether atoms, molecules or ions) in solution and not on the nature of the solute particles. Interesting Colligative properties: Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Vapor-Pressure Lowering The vapor pressure of a solution (nonvolatile solute and a solvent) is less than that of a pure solvent. P1 = X1 P 10 Raoult’s law P1 = partial pressure of a solvent over a solution P 10 = vapor pressure of pure solvent X1 = mole fraction of the solvent If the solution contains only one solute, X1 = 1 – X2, X2 = mole fraction of solute P1 = (1 – X2) P 01 P 10 - P1 = ∆P = X2 P 10 Boiling-Point Elevation Boiling point of a liquid: The temperature at which the vapor pressure equals the external atmospheric pressure. (Normal boiling point: external pressure = 1atm.) ∆Tb = Tb – T b0 T b0 is the boiling point of the pure solvent T b is the boiling point of the solution Tb > T b0 ∆Tb > 0 ∆Tb = Kb m m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) Ref. Chang, Sect. 12.6 Freezing-Point Depression Freezing point of a liquid: is the temperature at which solid and liquid phases coexist in equilibrium. (Normal freezing point: external pressure = 1atm.) ∆Tf = T 0f – Tf T 0 f is the freezing point of the pure solvent T f is the freezing point of the solution 0 T f > Tf ∆Tf > 0 ∆Tf = Kf m m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) Ref. Chang, Sect. 12.6 Ref. Chang, Table 12.2 Colligative Properties of Electrolyte Solutions 0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions 0.1 m NaCl solution 0.2 m ions in solution van’t Hoff factor (i) = nonelectrolytes NaCl CaCl2 actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be 1 2 3 Boiling-Point Elevation ∆Tb = iKb m Freezing-Point Depression ∆Tf = iKf m Ref. Chang, Sect. 12.7 For electrolyte solutions: Measured colligative properties are smaller than those calculated. Explanation: Formation of ion pairs (cations and anions held together by electrostatic forces of attraction). Presence of ion pairs reduces the number of particles in solution, causing a Ref. Chang, Table 12.3 reduction in the colligative properties. What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? moles of solute moles of solute m = M = mass of solvent (kg) liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg moles of solute m = mass of solvent (kg) = 5.86 moles C2H5OH = 8.92 m 0.657 kg solvent Ref. Chang, Sect. 12.3 What is the freezing point of a solution containing 478 g of ethylene glycol,C2H6O2, (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g. ∆Tf = Kf m Kf water = 1.86 0C/m moles of solute m = mass of solvent (kg) 478 g x 1 mol 62.01 g = = 2.41 m 3.202 kg solvent ∆Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C ∆Tf = T 0f – Tf Tf = T 0f – ∆Tf = 0.00 0C – 4.48 0C = -4.48 0C Ref. Chang, Sect. 12.6 Procedure Refer to the hand-out. Calculations • In Part I, determine the freezing point of water. • In Part II, calculate the number of moles of solute and the molality of each solution. • Knowing that the molal freezing-point depression constant, Kf is equal to 1.86°C kg/mol, determine the expected freezing point of each solution and compare it with the observed value. ∆Tf = i Kf x m Practice Exercises 1. Which of the following aqueous solutions has the highest boiling point? Kb for water is 0.52°C/m. Which has the highest freezing point? 0.2 m KCl, 0.2 m Na2SO4, 0.3 m Ca(NO3)2 2. What is the molarity of a solution of 10% by mass cadmium sulfate, CdSO4 (molar mass = 208.46 g/mol)? The density of the solution is 1.10 g/mL. Answers: 1. Highest boiling point: 0.3 m Ca(NO3)2 Highest freezing point: 0.2 m KCl 2. 0.528 mol/L