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Solutions

Learn about the different types of mixtures, including heterogeneous mixtures such as suspensions and colloids. Explore the properties and solubility of solutions. Understand the factors affecting the rate of dissolving and the concentration of solutions.

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Solutions

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  1. Solutions

  2. Types of Mixtures • Heterogeneous mixtures-mixtures that do not blend smoothly-not uniform throughout-individual substances remain distinct. • Two Types of Heterogeneous Mixtures are: Suspensions Colloids

  3. Suspensions • Suspensions-heterogeneous mixture that settles upon standing. Can be separated by filter paper.

  4. Colloids • Colloids-heterogeneous mixtures that appear cloudy, cannot be separated by filtration, and demonstrate the Tyndall effect.

  5. Tyndall Effect

  6. Types of Mixtures (continued) • Homogeneous mixtures-mixtures that do blend smoothly-uniform throughout-one set of properties. • Homogeneous mixtures are called solutions.

  7. Solutions • Solutions consist of: • Solute-part of the solution that gets dissolved • Solvent-part of the solutions that does the dissolving • Water is called the universal solvent.

  8. Types of Solutions • Enter answer text...

  9. Muddy water is an example of a 10 • colloid • solution • suspension

  10. Fog is an example of a 10 • colloid • solution • suspension

  11. Italian salad dressing is an example of a 10 • colloid • solution • suspension

  12. Kool-aid is an example of a 10 • colloid • solution • suspension

  13. Pure air is an example of a 10 • colloid • solution • suspension

  14. Salt is dissolved in water. Salt is the 10 • solute • solvent

  15. Sugar is dissolved in sweet tea. The tea is the 10 • solute • solvent

  16. Oxygen gas dissolved in lake water is an example of a 10 • Liquid dissolved in a gas • Gas dissolved in a liquid • Gas dissolved in a solid • Solid dissolved in a gas

  17. Carbon dissolved in iron to make steel is an example of a 10 • Solid dissolved in a solid • Solid dissolved in a liquid • Solid dissolved in a gas • Liquid dissolved in a solid • Gas dissolved in a solid

  18. Properties of Solutions • A substance that dissolves in a solvent is said to be soluble. • Two liquids that are soluble in each other are said to be miscible. • A substance that does not dissolve in a solvent is said to be insoluble. • Two liquids that are not soluble in each other are said to be immiscible.

  19. Insoluble Solids Immiscible Liquids

  20. “Like Dissolves Like” • Polar substances dissolve other polar substances (or ionic compounds). • Water is polar (due to asymmetrical arrangement of the molecule).

  21. Which of the following will dissolve in water? This one because it is polar!

  22. “Like Dissolves Like” (continued) • Nonpolar substances dissolve nonpolar substances. • Polar substances cannot dissolve nonpolar substances. Nonpolar iodine dissolves in nonpolar hexane Nonpolar iodine does NOT dissolve in polar water Nonpolar hexane and polar water do not mix.

  23. Corn oil does not dissolve in water. Corn oil is 10 • Polar • nonpolar

  24. Will calcium chloride (CaCl2) dissolve in water? 10 • Yes • No

  25. Which of the following is immiscible in water? 10 • Food coloring • Vinegar • Gasoline • Lemon-aid

  26. Factors Affecting the Rate of Dissolving • Agitation-(stirring)-stirring or shaking increases the rate at which solutes dissolve. • Surface area-breaking up a solid increases the surface area and increases the rate at which solids dissolve • Temperature-as temperature increases, the rate of dissolving of solid solutes increases.

  27. Solubility • Solubility is defined as the amount of solute that will dissolve in a given amount of solvent at a particular temperature

  28. Solubility (cont.) • Solutions that contain the maximum amount of dissolved soluteat a given temperature are said to be saturated. • Solutions that contain less than the maximum amount of solute are said to be unsaturated. • Solutions that contain more than the expected maximum amount of solute are said to be supersaturated (can be accomplished only through heating and careful cooling of the solvent).

  29. Temperature and Solubility • In warmer water, more solid will dissolve. • This is because a high temperature means water molecules are moving faster which keeps more solid molecules suspended. • Conversely a gas will be less soluble at a higher temperature. • This is because when a gas molecules are moving faster they are able to escape from the liquid surface.

  30. Solubility Curves • A solubility curve is a graph of the solubilities of various substances as a function of temperature. • When graphing the data for solubility, temperatureis the manipulated variable, and mass of solute dissolved is the responding variable.

  31. Interpreting Solubility Curves • 1. What is the maximum amount of NaNO3 that will dissolve in 100 g of water at 10oC? • 2. At what temperature will 70 g of NH4Cl dissolve in 100g of water? • 3. Which substance has the greatest solubility at 0oC? • 4. Which substance is the least soluble at 100oC? • 5. 15 g of KClO3 is dissolved in 100 g of water at 50oC. Is the solution saturated, unsaturated, or supersaturated?

  32. Concentration of Solutions • Qualitative descriptions: • Concentrated solutions-large amount of solute and small amount of solvent. • Dilute solutions-large amount of solvent and small amount of solute. • Quantitative descriptions: • Solubility-grams of solute/ml of solution • Molarity-moles of solute/L of solution • Molality-moles of solute/kg of solvent • Percent by Mass-massof solute/mass of solution x100% • Percent by Volume –volume of solute/volume of solution x 100%

  33. Molarity • A solution of NaCl has a molarity of 1 (1M). What does this mean? • 1 mole of NaCl is dissolved in enough water to make 1 L. • 1 mole = 58 g NaCl • A 1M solution of NaCl contains 58 g of NaCl dissolved in 1 liter of water. • How much salt is dissolved in a 2M solution? • 116 g • How much salt is dissolved in a 6 M solution? • 348 g • Which is more concentrated?

  34. Molality • A solution of NaCl contains 58 g of NaCl dissolved in 1 liter of water. The density of water is 1.00g/mL. What is the molality of the solution? • 1 liter = 1000 mL = 1000 g = 1 kg • 58 g = 1 mole • Molality = 1 mole/1 kg = 1 m • How much salt must be dissolved in 100 g of water in order to make a 2.0 molal (m) solution? • 100 g = .1 kg • 2.0 = x/.1 • X = .2 moles x 58 g = 11.6 g • Will molarity and molality for the same solution always be equal? Explain.

  35. Colligative Properties • Colligative propertiesare properties that depend on the number (not the type) of solute particles present in solution. • Colligative properties include: 1) freezing point 2) boiling point 3) vapor pressure

  36. Freezing Point Depression • The freezing point of a solution is lower than the freezing point of the pure solvent. • Example: Saltwater freezes at a lower temperature than pure water (below zero degrees Celsius). • The greater the number of ions in the solution, the lower the freezing point. • Example: NaCl consists of two ions; Na+ and Cl- CaCl2 consists of three ions; Ca2+ , Cl- , and Cl- Which has the lower freezing point? CaCl2

  37. Boiling Point Elevation • The boiling point of a solution is higher than the boiling point of the pure solvent. • Example: Saltwater boils at a higher temperature than pure water (at a temperature above 100 degrees Celsius). • The greater the number of ions in the solution, the higher the boiling point. • Which boils at a higher temperature; a solution of NaCl or a solution of CaCl2 • CaCl2

  38. Vapor Pressure Lowering • Vapor pressure is the pressure exerted by the vapor particles on the surface of a liquid. • The vapor pressure of a solution is lower than the vapor pressure of the pure solvent. • Due to the presence of solute particles, fewer solvent particles are able to escape from the surface of the liquid resulting in a lower pressure. • If the vapor pressure is lower, the boiling point will be higher.

  39. Which of the following is NOT a colligative property? 10 • Boiling point 2. Density • Freezing point • Vapor pressure

  40. The freezing point of a solvent will ____ when a solute is added. 10 • go up • go down • remain the same

  41. The boiling point of a solvent will ____ when a solute is added. 10 • go up • go down • remain the same

  42. Which of the following solutes will result in a solution having the highest boiling point? 10 • NaCl • CaCl2 • AlCl3 • C12H22O11

  43. If cost was not an issue, which of the following salts would be the most effective road deicer? 10 • NaCl • CaCl2 • AlCl3 • All would be equally effective

  44. Which solution would have the lowest freezing point? 10 • 1 M AlCl3 • 2 M NaCl • 3 M CaCl2 • 4 M C6H12O6

  45. Problems Involving Colligative Properties • The equation used to determine the Freezing Point Depression and Boiling Point Elevation is: • ∆T =i Kfm where: ∆T represents temperature change Kf is the freezing point depression constant (this value is specific to each solvent) m represents molality i represents the number of ions making up the solute. Note: molality = moles of solute/kg of solvent The same equation is used to determine the boiling point elevation however, Kb is substituted for Kf.

  46. Practice Problem • Sodium chloride is often used to prevent icy roads and to freeze ice cream. What is the freezing point of a 0.029 m aqueous solution of sodium chloride? • Molality = 0.029 • Kf for water = 1.86 • i = 2 • ∆T =i Kfm • ∆T = 2 (1.86) 0.029 • ∆T = .11 • 0 -.11 = -.11oC

  47. Practice Problem #2 • A lab technician determines that the boiling point of an aqueous solution of a calcium chloride solution (CaCl2) is 101.12oC. What is the solution’s molality? • ∆T = 101.12 – 100 = 1.12 • Kb for water = .512 • i = 3 for CaCl2 • ∆T =i Kbm • 1.12 = 3 (0.512) x • X=0.729 moles/kg of solvent

  48. Diluting Molar Solutions • You can prepare a less concentrated solution from a more concentrated solution by diluting the solution (increase the solvent particles) • The following equation can be used: • M1V1 = M2V2 where: • M1 and V1 represent the molarity and volume of the concentrated solutions • M2 and V2 represent the molarity and volume of the diluted solutions.

  49. Practice Problem • How many milliliters of a 5.0 M H2SO4 solution would you need to prepare 100 mL of a 0.25 M H2SO4 solution? • M1V1 = M2V2 • M1 = 5.0 M • V1 = x • M2 = 0.25 M • V2=100 mL • M1V1 = M2V2 • 5(x) = 0.25 (100) • X = 5 mL

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