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CHAPTER 14 Solutions
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The Dissolution Process Solutions are homogeneous mixtures of two or more substances solvent Dissolving medium is called the solvent solute Dissolved species are called the solute Upon dissolution, solute molecules are surrounded (solvated) by solvent molecules Solvent molecules have to rearrange around solute molecules NaCl(s) Na + (aq) + Cl – (aq) H2OH2O
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Dissolution of Ionic Salts Forces to overcome (energy loss): Cation-anion electrostatic attraction Attractive forces between solvent molecules Energy gain: Solvent-cation and solvent-anion electrostatic attractions Increased disorder If energy gain exceeds energy loss, the compound is soluble
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Colligative Properties of Solutions When solute is introduced into the volume of solvent, the solvent properties change Colligative properties are properties of solutions that depend solely on the number of particles dissolved in the solution We will look at three types of colligative properties of solutions: Vapor pressure lowering Freezing point depression Boiling point elevation
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Raoult’s Law Rearrangement of solvent molecules and new solvent-solute interactions affect the vapor pressure of the solvent – it becomes lower Raoult’s Law: The vapor pressure (P ) of a solvent in an ideal solution is directly proportional to the mole fraction (X ) of the solvent in the solution P solvent = X solvent P 0 solvent (P 0 – vapor pressure of the pure solvent) This is true only for non-volatile non-ionizing solutes at low solute concentrations
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Mole Fraction For the mixture of components A, B, C, …
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Mole Fraction: Example What are the mole fractions of glucose (C 6 H 12 O 6 ) and water in a 10.0% glucose solution?
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Mole Fraction If a solution contains only one type of solute A
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Raoult’s Law P 0 – the vapor pressure of the pure solvent P solvent – the vapor pressure of the solvent over the solution of compound A: P solvent = X solvent P 0 solvent P solvent = (1–X A ) P 0 solvent The vapor pressure lowering: P solvent = P 0 solvent – P solvent P solvent = X A P 0 solvent
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P solvent = (1–X A ) P 0 solvent Raoult’s Law
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Raoult’s Law: Example Determine the vapor pressure lowering for the 10.0% glucose solution?
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Molality In Chapter 3 we introduced two important concentration units: molality Now we introduce another unit - molality
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Molality m Molality (m) – the number of moles of solute per kilogram of solvent Boiling point elevation and freezing point depression are calculated based on this unit of concentration
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Calculate the molality of a 10.0% aqueous solution of glucose, C 6 H 12 O 6 Molality: Example 1
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10 g of NaOH was dissolved in 250 mL of water. What is the molality of the obtained solution? Molality: Example 2
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Boiling Point Elevation According to Raoult’s law, addition of a solute lowers the vapor pressure of the solvent:
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Boiling Point Elevation The boiling temperature elevation is determined by the number of moles of solute dissolved in the solution: T b = K b m T b – the change in boiling point K b – boiling point elevation constant (depends only on the nature of solvent) m – molality of the solution
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Freezing Point Depression Addition of a nonvolatile solute to a solution lowers the freezing point of the solution relative to the pure solvent T f = K f m T f – the change in freezing point K f – freezing point depression constant (depends only on the nature of solvent) m – molality of the solution
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Calculate the freezing and boiling points of a solution that contains 8.50 g of benzoic acid (C 6 H 5 COOH) in 75.0 g of benzene, C 6 H 6 Example 1
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3.75 g of a nonvolatile compound was dissolved in 108.7 g of acetone. The solution boiled at 56.58°C. The boiling point of pure acetone is 55.95°C and K b = 1.71°C/m. Calculate the molecular weight of the compound. Example 2
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Example 2 (continued)
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Assignments & Reminders Read Sections 14-8, 14-9, 14-11, 14-12 & 14-13 Homework #8 is due by 12/07 @ 9:00 p.m. Review Sessions: Sunday – 5:15-8:00 p.m. in 100 Held Wednesday – 5:15-8:00 p.m. in 100 Held Final Exam: Friday – 3:00-5:00 p.m. in 100 Held
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