Colligative Properties Depend only on the number, not on the identity, of the solute particles in an ideal solution: Boiling-point elevation Freezing-point depression Osmotic pressure Copyright © Cengage Learning. All rights reserved

Boiling-Point Elevation Nonvolatile solute elevates the boiling point of the solvent. ΔT = Kbmsolute ΔT = boiling-point elevation Kb = molal boiling-point elevation constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved

Boiling Point Elevation: Liquid/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Boiling Point Elevation: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Boiling Point Elevation: Solution/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Freezing-Point Depression When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = Kfmsolute ΔT = freezing-point depression Kf = molal freezing-point depression constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved

Freezing Point Depression: Solid/Liquid Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Freezing Point Depression: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Freezing Point Depression: Solid/Solution Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Changes in Boiling Point and Freezing Point of Water Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution. 100.35 °C The change in temperature is ΔT = Kbmsolute. Kb is 0.51 °C·kg/mol. To solve formsolute, use the equation m = moles of solute/kg of solvent. Moles of solute = (25.00 g glucose)(1 mol / 180.16 g glucose) = 0.1388 mol glucose Kg of solvent = (200.0 g)(1 kg / 1000 g) = 0.2000 kg water msolute = (0.1388 mol glucose) / (0.2000 kg water) = 0.6938 mol/kg ΔT = (0.51 °C·kg/mol)(0.6938 mol/kg) = 0.35 °C. The boiling point of the resulting solution is 100.00 °C + 0.35 °C = 100.35 °C. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

EXERCISE! You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be -0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture. 72.8% sucrose and 27.2% sodium chloride; mole fraction of the sucrose is 0.313 The solution is 72.8% sucrose and 27.2% sodium chloride. The mole fraction of the sucrose is 0.313. To solve this problem, the students must assume that i = 2 for NaCl. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

EXERCISE! A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of –0.246°C. Will the cell explode, shrivel, or do nothing? The cell will explode (or at least expand). The concentration of the solution is 0.186 m. Thus, the cell has a higher concentration, and water will enter the cell. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

M = molarity of the solution R = gas law constant Osmosis – flow of solvent into the solution through a semipermeable membrane. = MRT = osmotic pressure (atm) M = molarity of the solution R = gas law constant T = temperature (Kelvin) Copyright © Cengage Learning. All rights reserved

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Osmosis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

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EXERCISE! When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound. 111 g/mol The molar mass is 111 g/mol. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

van’t Hoff Factor, i The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as: Copyright © Cengage Learning. All rights reserved

Ion Pairing At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle. Copyright © Cengage Learning. All rights reserved

Examples The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl i = 2 KNO3 i = 2 Na3PO4 i = 4 Copyright © Cengage Learning. All rights reserved

Ion Pairing Ion pairing is most important in concentrated solutions. As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs. Ion pairing occurs to some extent in all electrolyte solutions. Ion pairing is most important for highly charged ions. Copyright © Cengage Learning. All rights reserved

Modified Equations Copyright © Cengage Learning. All rights reserved

A suspension of tiny particles in some medium. Tyndall effect – scattering of light by particles. Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm. Copyright © Cengage Learning. All rights reserved

Types of Colloids Copyright © Cengage Learning. All rights reserved

Coagulation Destruction of a colloid. Usually accomplished either by heating or by adding an electrolyte. Copyright © Cengage Learning. All rights reserved