1 / 40

Chapter 16 Solutions

Chapter 16 Solutions . Killarney High School. Section 16.1 Properties of Solutions. OBJECTIVES: Identify the factors that determine the rate at which a solute dissolves. Section 16.1 Properties of Solutions. OBJECTIVES:

Lucy
Download Presentation

Chapter 16 Solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 16Solutions Killarney High School

  2. Section 16.1Properties of Solutions • OBJECTIVES: • Identify the factors that determine the rate at which a solute dissolves.

  3. Section 16.1Properties of Solutions • OBJECTIVES: • Calculate the solubility of a gas in a liquid under various pressure conditions.

  4. Solution formation • Nature of the solute and the solvent • Whether a substance will dissolve • How much will dissolve • Factors determining rate of solution... • stirred or shaken (agitation) • particles are made smaller • temperature is increased • Why?

  5. Making solutions • In order to dissolve, the solvent molecules must come in contact with the solute. • Stirring moves fresh solvent next to the solute. • The solvent touches the surface of the solute. • Smaller pieces increase the amount of surface area of the solute. Solution Formation Flash

  6. Temperature and Solutions • Higher temperature makes the molecules of the solvent move around faster and contact the solute harder and more often. • Speeds up dissolving. • Usually increases the amount that will dissolve (exception is gases)

  7. How Much? • Solubility- The maximum amount of substance that will dissolve at a specific temperature (g solute/100 g solvent) • Saturated solution- Contains the maximum amount of solute dissolved • Unsaturated solution- Can still dissolve more solute • Supersaturated- solution that is holding more than it theoretically can; seed crystal will make it come out

  8. Liquids • Miscible means that two liquids can dissolve in each other • water and antifreeze, water and ethanol • Partially miscible- slightly • water and ether • Immiscible means they can’t • oil and vinegar

  9. Solubility? • For solids in liquids, as the temperature goes up-the solubility usually goes up (Fig. 16.5, p.373) • For gases in a liquid, as the temperature goes up-the solubility goes down (Fig. 16.6, p.374) • For gases in a liquid, as the pressure goes up-the solubility goes up (Fig. 16.6, p.373)

  10. Gases in liquids... • Henry’s Law - says the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid • think of a bottle of soda pop, removing the lid releases pres. • Equation: S1 S2 P1 P2 =

  11. Cloud seeding • Ever heard of seeding the clouds to make them produce rain? • Clouds- mass of air supersaturated with water vapor • Silver Iodide (AgI) crystals are dusted into the cloud • The AgI attracts the water, forming droplets to attract others

  12. Section 16.2Concentration of Solutions • OBJECTIVES: • Solve problems involving the molarity of a solution.

  13. Section 16.2Concentration of Solutions • OBJECTIVES: • Describe how to prepare dilute solutions from more concentrated solutions of known molarity.

  14. Section 16.2Concentration of Solutions • OBJECTIVES: • Explain what is meant by percent by volume [ % (v/v) ], and percent by mass [ % (m/v) ] solutions.

  15. Concentration is... • a measure of the amount of solute dissolved in a given quantity of solvent • Aconcentrated solution has a large amount of solute • Adilute solution has a small amount of solute • thus, only qualitative descriptions • But, there are ways to express solution concentration quantitatively

  16. Molarity - most important • The number of moles of solute in 1 Liter of the solution. • M = moles/Liter; such as 6.0 molar • What is the molarity of a solution with 2.0 moles of NaCl in 250 mL of solution? • Sample 16-2, page 378

  17. Making solutions • Pour in a small amount of solvent • Then add the solute (to dissolve it) • Carefully fill to final volume. • Fig. 18-10, page 509 • Also: M x L = moles • How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution?

  18. Making solutions • 10.3 g of NaCl are dissolved in a small amount of water, then diluted to 250 mL. What is the concentration? • How many grams of sugar are needed to make 125 mL of a 0.50 M C6H12O6 solution?

  19. Dilution Adding water to a solution

  20. Dilution • The number of moles of solute doesn’t change if you add more solvent! • The # moles before = the # moles after • M1 x V1 = M2 x V2 • M1 and V1 are the starting concentration and volume. • M2 and V2 are the final concentration and volume. • Stock solutions are pre-made to known Molarity

  21. Practice • 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? • You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make? • Need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. How do you make the required solution?

  22. Percent solutions... • Percent means parts per 100, so • Percent by volume: = Volume of solute x 100% Volume of solution • indicated %(v/v) • What is the percent solution if 25 mL of CH3OH is diluted to 150 mL with water?

  23. Percent solutions • Percent by mass: = Mass of solute(g) x 100% Volume of solution(mL) • Indicated %(m/v) • More commonly used • 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution? • How many grams of salt are there in 52 mL of a 6.3 % solution?

  24. Section 16.3Colligative Properties of Solutions • OBJECTIVES: • Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution.

  25. Section 16.3Colligative Properties of Solutions • OBJECTIVES: • Explain on a particle basis why a solution has an elevated boiling point, and a depressed freezing point compared with the pure solvent.

  26. Colligative Properties Depend only on the number of dissolved particles Not on what kind of particle

  27. Vapor Pressure decreased • The bonds between molecules keep molecules from escaping. • In a solution, some of the solvent is busy keeping the solute dissolved. • Lowers the vapor pressure • Electrolytes form ions when they are dissolved = more pieces. • NaCl ® Na+ + Cl- (= 2 pieces) • More pieces = bigger effect

  28. Boiling Point Elevation • The vapor pressure determines the boiling point. • Lower vapor pressure = higher boiling point. • Salt water boils above 100ºC • The number of dissolved particles determines how much, as well as the solvent itself.

  29. Freezing Point Depression • Solids form when molecules make an orderly pattern. • The solute molecules break up the orderly pattern. • Makes the freezing point lower. • Salt water freezes below 0ºC • How much depends on the number of solute particles dissolved. Simulation

  30. Section 16.4Calculations Involving Colligative Properties • OBJECTIVES: • Calculate the molality and mole fraction of a solution.

  31. Section 16.4Calculations Involving Colligative Properties • OBJECTIVES: • Calculate the molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound.

  32. Molality • a new unit for concentration • m = Moles of solute kilogram of solvent • m = Moles of solute 1000 g of solvent • What is the molality of a solution with 9.3 mole of NaCl in 450 g of water?

  33. Why molality? • The size of the change in boiling point is determined by the molality. • DTb= Kbx m x n • DTb is the change in the boiling point • Kb is a constant determined by the solvent (Table 16.2, page 387). • m is the molality of the solution. • n is the number of pieces it falls into when it dissolves.

  34. What about Freezing? • The size of the change in freezing point is also determined by molality. • DTf= -Kfx m x n • DTf is the change in freezing point • Kf is a constant determined by the solvent (Table 16.3, page 388). • m is the molality of the solution. • n is the number of pieces it falls into when it dissolves.

  35. Problems • What is the boiling point of a solution made by dissolving 1.20 moles of NaCl in 750 g of water? • What is the freezing point? • What is the boiling point of a solution made by dissolving 1.20 moles of CaCl2 in 750 g of water? • What is the freezing point?

  36. Mole fraction • This is another way to express concentration • It is the ratio of moles of solute to total number of moles of solute + solvent (Fig. 16-19, p.386) na na + nb Sample 18-8, page 521 X =

  37. Molar Mass • We can use changes in boiling and freezing to calculate the molar mass of a substance • Find: 1) molality 2) moles, and then 3) molar mass • Sample 16-10, page 388

  38. Tb = Kb x m x n m = tb thus Kb Kb = 0.512 C x kg of water Mol of solute 0.78 C m = = 1.5 m 0.512 C / m A solution of 7.50g of a nonvolatile compound in 22.60 g of water boils at 100.78 degrees celsius at 1 atm. What is the molar mass of the solute? 1. Use the BP elevation to calculate the molality of the solution.

  39. 0.0226 Kg 0.034 mol solute 1.5 mol solute Kg of water Molar mass = mass of solute Molar mass = 7.50 g Moles of solute 0.0344 mol 2. Now calculate the moles of solute I the solution 3. Use the # of moles of solute and its mass to get the molar mass Molar mass = 0.022 g/mol

  40. Key Equations • Note the key equations to solve problems in this chapter.

More Related