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Colligative Properties Colligative Properties Colligative Properties Are physical properties that depend upon the quantity or concentration of the solution They do not depend upon the identity of the solute Colligative Properties 1. Vapor pressure lowering (Raoult’s law) 2. Boiling point elevation 3. Freezing point depression 4. Osmosis Colligative properties -Vapor Volatile liquids are easily evaporated and nonvolatile liquids do not evaporate easily. Examples of volatile liquids are acetone and gasoline Examples of nonvolatile liquids oils and glycerin Vapor Pressure lowering A volatile liquid is mixed with a non- volatile substance The nonvolatile liquid or particles will get in the way of the evaporation of the liquid This action lowers the vapor pressure since less gas is formed Raoult’s law is used to predict the vapor pressure of a system PA = XA P°A PA = the vapor pressure of the solution XA = the mole fraction of the solvent P°A = the vapor pressure of pure solvent at a particular temperature Problem Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/mL. Calculate the vapor pressure of 25°C of a solution made by adding 50.0 mL glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr. Problem 1st determine the mole fraction 1.26 g/ml *50ml = 63 g glycerin 63g glycerin* (1mol/92.09g) = 0.684mol glycerin 500.0 g H2O * (1mol/18.01 g) = 0.684 glycerin/ (0.684 +27.76) = 0.024 Xglycerin 2nd plug in numbers and solve 23.8 torr *.024 = 0.57 torr Problem The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? Problem PA = Xglycol P°A 1.0 atm = 760 torr 760 torr = Xglycol * 1070 torr 760 torr/1070 torr = Xglycol Xglycol = 0.71 Xglycol Boiling point elevation By adding a solute to a liquid, the boiling temperature is increased Tb = Kb * m Tb = the change in boiling point Kb = the molal-boiling-point elevation constant (depends upon the solvent) m = the molality of the solution * if you work with ions you must multiply the molality times the number of ions formed. 4.7 m solution of NaCl How many ions formed? And how does that change the molality? 2 4.7 * 2 9.4 m is plugged into the formula 0.7 m of CaCl2 How many ions formed? And how does that change the molality? 3 0.7m * 3 = 2.1 m Plug in the higher molalities into the boiling point elevation when ionic compounds are the solute. Your text may refer to ionic solutions as electrolyte solutions (electrolyte solutions will conduct electricity because the dissociate in water forming ions) Freezing point depression The lowering of the freezing point by adding solute Tf = kf * m Tf = change in freezing point kf = molar freezing-point depression constant (solvent dependent) m = molality of solution * multiply the molality of an ionic solution by the number of ions formed in solution problem Calculate the boiling point and freezing point of a 25.0 mass percent solution of ethylene glycol, C2H6O2, in water. Kf water = 1.86 °C/m Kb water = 0.52 °C/m Solution Step 1: determine the molality of solution Assume 100g, we have 25g of ethylene glycol and 75 g of solvent (water) Convert 25 g of ethylene glycol to mol 25g*(1 mol/ 62.07g)= 0.403mol ethylene glycol The density of water is about 1g/mL with means we have 75ml or 0.075L 0.403 mol / 0.075 L = 5.37m Solution Step 2: Plug into freezing point formula then subtract from normal freezing point of water 5.37m * 1.86°C/m = 9.99°C 0°C-9.99°C = -9.99°C Step 3: plug into boiling point formula then add to normal boiling point of water 5.37m*0.52°C/m = 2.79°C 100°C + 2.79°C = 102.8°C Problem Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42 g of C10H18O. Solution Step 1: solve for molality 42 g C10H18O * (1 mol/154.25 g) =0.272 mol 0.272 mol /.600kg = 0.453 m Step 2: kf = 4.68 C°/m 4.68 C°/m * 0.453 m = 2.1C° Step 3: normal freezing point is -63.5°C -62.5 – 2.1 = -64.6°C .
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