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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
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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
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Boiling Point Elevation: Liquid/Vapor Equilibrium
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Boiling Point Elevation: Addition of a Solute
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Boiling Point Elevation: Solution/Vapor Equilibrium
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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
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Freezing Point Depression: Solid/Liquid Equilibrium
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Freezing Point Depression: Addition of a Solute
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Freezing Point Depression: Solid/Solution Equilibrium
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Changes in Boiling Point and Freezing Point of Water
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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 °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 / 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
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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 °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 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
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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
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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
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Copyright © Cengage Learning. All rights reserved
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EXERCISE! When 33.4 mg of a compound is dissolved in 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
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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
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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
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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
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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
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Modified Equations Copyright © Cengage Learning. All rights reserved
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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
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Types of Colloids Copyright © Cengage Learning. All rights reserved
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Coagulation Destruction of a colloid.
Usually accomplished either by heating or by adding an electrolyte. Copyright © Cengage Learning. All rights reserved
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