1 / 36

Solutions

Solutions. Chapter 16 in textbook. What is a solution?.

adila
Télécharger la présentation

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. Solutions Chapter 16 in textbook

  2. What is a solution? • Many people like to watch hummingbirds and put up hummingbird feeders in their yards. They fill the feeders with a red liquid made of water, sugar, and red food coloring. The sweet, colored liquid attracts hummingbirds. To make this food, you add sugar to water and stir. When you stir, the sugar crystals disappear. Next, you add a few drops of red food coloring and stir again. The red color spreads evenly. Why does this happen? • The red sugar-water is a solution. • A solutionis a mixture that has the same ingredients, color, density, and even taste mixed evenly throughout.

  3. Solutes and Solvents • The solute is the substance that dissolves. • Typically a solid but can sometimes be a gas. • Example: In carbonated soda, carbon dioxide gas is dissolved in water. The gas is the solute and the liquid is the solvent. • The solventis the substance that is doing the dissolving. • Typically a liquid

  4. Are there solutions that do not contain a liquid? • Solutions can also be mixtures of gases or even mixtures of solids. • The air you breath is a solution • Air is a solution of 78% Nitrogen, 20% Oxygen, and small amounts of other gases.

  5. How substances dissolve • Water is the universal solvent. **Water molecules, like all particles, are always moving. **Water molecules are polar. • Polar means that they have a positive end and a negative end.

  6. How does a solid dissolve in a liquid? • a) Water molecules move toward a sugar crystal. The positive ends of sugar molecules attract the negative ends of water molecules. • b) Water molecules pull sugar molecules into solution. • c) Process continues until the crystal is dissolved.

  7. Rate of Dissolving • Things that can speed up dissolving a solute in a solvent: • Stirring (agitation) • Temperature (increase heat) • The surface area of the dissolving particles (crushing the solute, or breaking it into smaller pieces)

  8. Solubility and Concentration

  9. How much can dissolve? • Suppose you like your lemonade very sweet. You add a tsp of sugar to it and stir. The sugar dissolves. If you keep adding sugar to the lemonade, you will reach a point when no more sugar will dissolve in it. The excess sugar crystals do not go into the solution. • Solubility is the greatest amount of solute that can dissolve in a specific amount of solvent at a given temperature.

  10. Concentration • You are making lemonade. You add 1 tsp of lemon juice to a glass of water and your friend adds 4 tsp lemon juice to the same amount of water. Which lemonade has more lemon flavor? • Your friend’s lemonade is concentrated, because it has large amounts of solute dissolved in the solvent. Your lemonade is dilute, because it has small amounts of solute dissolved in the solvent.

  11. Types of Solutions • A saturated solution is a solution that contains all the solute it can hold at a given temperature. • If you increase the temperature of a mixture, the more solute can dissolve

  12. Types of Solutions • An unsaturated solution is a solution that can dissolve more solute at a given temperature. • Supersaturated solution is a solution that has more solute than a saturated solution at the same temperature.

  13. Factors Affecting Solubility • Temperature: The solubility of most solid substances increases as the temperature increases • Pressure: Changes in pressure have little effect on the solubility of solids and liquids but pressure strongly influence the solubility of gases.

  14. Concentrations of Solutions • How to express the actual extent of dissolving, that is, the concentration of a solution: • Molarity (M) (AKA molar concentration) is the number of moles of solute dissolved in one liter of solution moles of solute Molarity (M) = liters of solution

  15. Sample Problem 16.2 • Calculating the Molarity of a Solution Intravenous (IV) saline solutions are often administered to patients in the hospital. One saline solution contains 0.90 g NaCl in exactly 100mL of solution. What is the molarity of the solution?

  16. Intravenous (IV) saline solutions are often administered to patients in the hospital. One saline solution contains 0.90 g NaCl in exactly 100mL of solution. What is the molarity of the solution? • Identify the Knowns and the Unknowns • Solution concentration = 0.90 g NaCl/100mL • Molar mass NaCl = 58.5g/mol • Solution concentration = ??M Solutions concentration = 0.90g NaCl1 mol NaCl1000 mL X X 100 mL 58.5 g NaCl 1 L = 0.15 mol/L = 0.15 M

  17. Making Dilutions • Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change. • Recall: Molarity (M) = • Rearranging the equation gives an expression for moles of solute. • Moles of solute before dilution is equal to the moles of solute after dilution Moles of solute = molarity (M) x liters of solution (V) Moles of solute = M1 x V1 = M2 x V2 moles of solute liters of solution

  18. Sample Problem 16.4 • How many milliliters of aqueous 2.00M MgSO4 solutions must be diluted with water to prepare 100.0 mL of aqueous 0.400M MgSO4? • Find the Knowns / Unknowns: • M1 = 2.00M MgSO4 • M2 = 0.400M MgSO4 • V1 = ?? • V2 = 100.0 mL of 0.400M MgSO4

  19. Sample Problem 16.4How many milliliters of aqueous 2.00M MgSO4 solutions must be diluted with water to prepare 100.0 mL of aqueous 0.400M MgSO4? • Find the Knowns / Unknowns: • M1 = 2.00M MgSO4 • M2 = 0.400M MgSO4 • V1 = ?? • V2 = 100.0 mL of 0.400M MgSO4 • Formula: M1 x V1 = M2 x V2 V1 = M2 x V2 M1

  20. Sample Problem 16.4How many milliliters of aqueous 2.00M MgSO4 solutions must be diluted with water to prepare 100.0 mL of aqueous 0.400M MgSO4? • Formula: V1 = (0.400M MgSO4) x (100.0 mL MgSO4) 2.00M MgSO4 V1 = 20.0 mL V1 = M2 x V2 M1

  21. Classwork Practice ProblemsThe following problems MUST be turned in at the end of class today! • A solution has a volume of 2.0 L and contains 36.0 g of glucose (C6H12O6). If the molar mass of glucose is 180 g/mL, what is the molarity of the solution? • A solution has a volume of 250 mL and contains 0.70 molNaCl. What is its molarity? • How many moles of ammonium nitrate are in 335 mL of 0.425M NH4NO3? • How many moles of solute are in 250 mL of 2.0M CaCl2? How many grams of CaCl2 is this? ANSWERS: 1) 0.1M; 2) 2.8 M; 3) 0.142 molNH4NO3 4) .5 mole CaCl2; 0.56g CaCl2

  22. Colligative Properties of Solutions

  23. Did you know? • The wood frog is a remarkable creature because it can survive being frozen. Scientists believe that a substance in the cells of this frog act as a natural antifreeze, which prevents the cells from freezing. Although fluids surrounding the frog’s cells may freeze, the cells themselves do not. • A solute can change the freezing point of a solution

  24. Colligative property • The physical properties of a solution differ from those of the pure solvent used to make the solution. • A colligative property is a property that depends only upon the number of solute particles in the solution.

  25. 3 Important Colligative Properties • Vapor-pressure lowering • Boiling-point elevation • Freezing-point depression

  26. 3 Important Colligative Properties • Vapor-pressure lowering • Pressure exerted by a vapor • The decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution. • Boiling-point elevation • Freezing-point depression

  27. 3 Important Colligative Properties • Vapor-pressure lowering • Boiling-point elevation • The temperature at which the vapor pressure of the liquid phase equals atmospheric pressure • The magnitude of the boiling-point elevation is proportional to the number of solute particles dissolved in the solvent. • Freezing-point depression

  28. 3 Important Colligative Properties • Vapor-pressure lowering • Boiling-point elevation • Freezing-point depression • Freezing-point of a solution is lower than the freezing point of the pure solvent • The magnitude of the freezing-point depression is proportional to the number of solute particles dissolved in the solvent and does not depend upon their identity.

  29. Calculating Colligative Properties • Cooking instructions for a wide variety of foods, from dried pasta to packaged beans to frozen fruits to fresh vegetables, often call for the addition of a small amount of salt to the cooking water. Most people like the flavor of food cooked with salt. But adding salt can have another effect on the cooking process. • Recall that dissolved salt elevates the boiling point of water. Suppose you added a tsp of salt to 2L of water. A tsp of salt has a mass of about 20g. Would the resulting boiling point increase be enough to shorten the time required for cooking?

  30. Calculating Colligative Properties Molality and Mole Fractions • The unit molality and mole fractions are 2 additional ways in which chemists express the concentration of a solution. • The unit molality (m) is the number of moles of solute dissolved in 1 kilogram (1000 g) of solvent. moles of solute Molality (m) = molal kilograms of solvent Please note that MOLALITY is NOT the same as MOLES.

  31. Sample Problem 16.6 • How many grams of potassium iodide must be dissolved in 500.0 g of water to produce a 0.060 molal KI solution? • Knowns & Unknowns: • Mass of water = 500.0 g = 0.5000 kg • Solution concentration = 0.060m • Molar mass KI = 166.0 g/mol • According to the definition of molal, the final solution must contain 0.060 mol KI per 1000g H2O. Use the molality as a conversion factor to convert from mass of water to moles of solute (KI). Then use the molar mass of KI to convert from mol KI to g KI. The steps are: mass of H2O mol KI  g KI

  32. Sample Problem 16.6How many grams of potassium iodide must be dissolved in 500.0 g of water to produce a 0.060 molal KI solution? • Knowns & Unknowns: • Mass of water = 500.0 g = 0.5000 kg • Solution concentration = 0.060m • Molar mass KI = 166.0 g/mol mass of H2O mol KI  g KI 0.060 mol KI 0.5000 kg H2O 166.0 g KI X X 1.000 kg H2O 1 mol KI = 5.0g KI

  33. Mole Fractions • A mole fraction is the ratio of the number of moles of one substance to the total number of moles of all substances in the solution. • Ethylene glycol (EG) is added to water as antifreeze in the proportions shown.

  34. Mole Fractions • A mole fraction is the ratio of the number of moles of one substance to the total number of moles of all substances in the solution. Total moles = Mole fraction EG = ____________ = 1.25 5.25 Mole fraction H2O= ____________ = 4.00 5.25 + B A A + A B B + B A

  35. Classwork Practice ProblemsThe following problems MUST be turned in at the end of class today! • How many milliliters of a solution of 4.00M KI are needed to prepare 250.0 mL of 0.760M KI? • How many grams of sodium fluoride are needed to prepare a 0.400mNaF solution that contains 750g of water? • Calculate the molality of a solution prepared by dissolving 10.0g NaCl in 600 g of water. • What is the mole fraction of each component in a solution made by mixing 300g of ethanol (C2H5OH) and 500g of water? ANSWERS: 5) 0.475 mL; 6) 12.6g NaF; 7) 0.285mNaCl; 8) C2H5OH = 0.190; H2O = 0.810

More Related