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CHAPTER 14 Solutions. The Dissolution Process  Solutions are homogeneous mixtures of two or more substances solvent  Dissolving medium is called the.

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Presentation on theme: "CHAPTER 14 Solutions. The Dissolution Process  Solutions are homogeneous mixtures of two or more substances solvent  Dissolving medium is called the."— Presentation transcript:

1 CHAPTER 14 Solutions

2 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

3 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

4 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

5 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

6 Mole Fraction  For the mixture of components A, B, C, …

7 Mole Fraction: Example  What are the mole fractions of glucose (C 6 H 12 O 6 ) and water in a 10.0% glucose solution?

8 Mole Fraction  If a solution contains only one type of solute A

9 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

10 P solvent = (1–X A )  P 0 solvent Raoult’s Law

11 Raoult’s Law: Example  Determine the vapor pressure lowering for the 10.0% glucose solution?

12 Molality  In Chapter 3 we introduced two important concentration units: molality  Now we introduce another unit - molality

13 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

14  Calculate the molality of a 10.0% aqueous solution of glucose, C 6 H 12 O 6 Molality: Example 1

15  10 g of NaOH was dissolved in 250 mL of water. What is the molality of the obtained solution? Molality: Example 2

16 Boiling Point Elevation  According to Raoult’s law, addition of a solute lowers the vapor pressure of the solvent:

17 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

18 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

19  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

20  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

21 Example 2 (continued)

22 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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