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Colligative Properties

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Presentation on theme: "Colligative Properties"— Presentation transcript:

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2 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

3 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

4 Boiling Point Elevation: Liquid/Vapor Equilibrium
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5 Boiling Point Elevation: Addition of a Solute
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6 Boiling Point Elevation: Solution/Vapor Equilibrium
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7 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

8 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

9 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

10 Freezing Point Depression: Solid/Solution Equilibrium
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11 Changes in Boiling Point and Freezing Point of Water
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12 EXERCISE! A solution was prepared by dissolving g of glucose in g water. The molar mass of glucose is 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. 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 / g glucose) = mol glucose Kg of solvent = (200.0 g)(1 kg / 1000 g) = kg water msolute = ( mol glucose) / ( kg water) = mol/kg ΔT = (0.51 °C·kg/mol)( mol/kg) = 0.35 °C. The boiling point of the resulting solution is °C °C = °C. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

13 EXERCISE! A plant cell has a natural concentration of 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 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

14 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

15 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

16 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

17 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

18 Modified Equations Copyright © Cengage Learning. All rights reserved


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