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Solutions

Solutions. Chapter 16. S What is a Solution? p124. Warm up: give an example of a solution. Solutions: a homogeneous mixture of two or more substances in same phase. video. P 124. The dissolving medium is the solvent . The substance being dissolved is the solute (smaller amount).

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Solutions

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  1. Solutions • Chapter 16

  2. SWhat is a Solution? p124 • Warm up: give an example of a solution. • Solutions: a homogeneous mixture of two or more substances in same phase. • video

  3. P 124 • The dissolving medium is the solvent. • The substance being dissolved is the solute (smaller amount). • 10g of sugar is poured into 100g tea, which is the solute? Solvent?

  4. P 124 Will 2 substances mix together? 1. Polar (partial + or – charges) -ex. Water, ammonia 2. Nonpolar (equal sharing of e-) -ex. Benzene, methane • Remember the rule: Like Dissolves Like • Left side p123: CREW statement will water dissolve benzene? • make an acrostic for: SOLUTIONS

  5. The Dissolving Process p126 • Warm up: what happens to a substance when it is dissolved? • Water molecules surround and isolate the surface ions. They move away from each other. (dissociation)

  6. The Dissolving Process Draw and explain what happens when a piece of salt gets dissolved in water on top half of your left page.

  7. STypwsspensTypes ions Types of homogeneous mixtures p.126 • Solution: Particles in a solvent are small (less than 1nm in diameter) • If the particles in a solvent are large and settle out unless the mixture is stirred, its called a suspension. • Particles over 1000nm in diameter form suspensions.

  8. Suspensions Types of Solutions p126 • Particles that are intermediate in size(1-1000nm) between those in solutions and suspensions form colloids.

  9. Alloys p126 • Alloys are solid solutions that are made by melting metals together and letting them cool.

  10. Draw on bottom half your left page and fill in the box below with: 1. label each particle 2. list the size of each particle 3. give an example of each type. Solutions Colloids Suspensions

  11. Solubilityp128 Solubility • The tyndall effect occurs when light is reflected from larger particles in solutions, but goes right through smaller particles. • . Warm Up: How you know which of the above is a colloid and which is a solution? Use CREW.

  12. Solubility p 128 • Solubility :the amount of a substance that can be dissolved in a quantity of solvent. • less than 0.01 mol/L is insoluble.

  13. (don’t write) • The graph represents the solubility of NaCl, NaNO3, and KNO3 at different temps. • Notice that when temp increases, solubility increases

  14. Factors Affecting Solubility p128 • like dissolves like • most solids dissolve easier in warm water and as particle size decreases. • For gases, solubility goes down as temp goes up and increases as pressure goes up. • Stirring increases solubility

  15. Left page questions: p127 • How do you make a solid more soluble in water? • How do you make a gas more soluble in water? • Do you think fish prefer cold or warm water to live in? Why?

  16. Saturated Solutions p128 • Saturated- cannot dissolve any more solute • Unsaturated- can still dissolve more solute. • Supersaturated solution are made by dissolving the solute under high temps. • On your left page draw cartoons or pictures representing saturated, unsaturated, and super saturated solutions.

  17. Solution Concentration Molarity p130 • Warm up: what does concentration mean? • Molarity: the concentration of solute in a given solvent, * (M) = moles solute Liters solution Review: Change 300ml to L: 10.0 g NaCl = _______ moles?

  18. MolarP ity Calculations Example, p 130 • Calculate the molarity of: *35.2 grams of CO2 in 500. mL of solution Step 1: convert 35.2 g of CO2 into moles 35.2 g (1 mole ) = 0.800 mol (44.0 g) Step 2: divide moles by volume in liters 0.800 mol CO2 = 1.60 M 0.500 L

  19. Molarity practice p 129 • How many moles of NaCl are needed to make 750 ml of a 2.4 M solution? • How many grams of KBr are needed to make 2.0 L of a 0.5 M solution?

  20. Making Diluted Solutions Dilution of Solutions p132 • Warm up: what is a diluted solution? • Use the following equation: M1V1 = M2V2 • M1 and M2 are the initial and final molar solutions. • V1 and V2 are the initial and final volumes of solutions. • Ex. You need to make 200. mL of a 0.10 M solution, from 1.0 M HC2H3O2

  21. Dilution practice p131 • How many ml of 18.0 M HCl are needed to make 550 mL of a 0.50 M solution? How much water do you need? • How many ml of 5.0 M H2SO4 are needed to make 50. mL of a 1.50 M solution? ? How much water do you need?

  22. Molality p134 • Warm up: what are the units for Molarity? Where do the units come from? • Molality (m) is the measure of the number of moles of a solute per Kg of solvent. (remember: g to Kg, divide by 1000) m=moles solute (kg) solvent

  23. Examples p134 • What’s the molality when 3.4 g NaCl is dissolved in 500.0 g water? • How many grams NaOH are needed to make 3.5 kg of a 1.90 m solution?

  24. Practice p 133 • What’s the molality when 29.5 g KBr is dissolved in 500.0 g water? • How many grams NaCl are needed to make 1.5 kg of a 3.0 m solution?

  25. Boiling Point and Freezing Point p 134 • Warm up: Why are you supposed to put antifreeze in your car? • The addition of a solute will require a higher temp for solvent to boil, thus: Boiling point elevation • The addition of a solute will require a lower temp for solvent to freeze, thus: Freezing point depression

  26. Draw and label the diagram on top half of p 135

  27. Write the formulas: p134 • Boiling point elevation is: ΔTb = (Kb)(m)(# particles) (molality)(change in bp) (bp constant) • Freezing point depression is: ΔTf = (Kf)(m)(# particles) (molality) (change in fp) (fp constant)

  28. Determining # of particles: p 134 • Molecular: two nonmetals! Will not split in solution, #of particles=1 • Ex. H2O NO2 CO2 C6H2O6 • Ionic compound: First element is a metal. Will split in solution. • NaNO3 makes 2 (1 Na & 1 NO3) • NaCl makes 2 (1 Na & 1 Cl) • Ca(NO3)2 makes 3 (1 Ca and 2 NO3)

  29. Calculating Freezing and Boiling Points p134 • Write down Kb & Kf for water & Benzene

  30. Example: p136 • Calculate both the boiling point and freezing point of the following solution: • 27.6 g Mg(ClO4)2 dissolved in 100.g of water

  31. Practice p 135 • Compute both boiling and freezing points of these solutions: • 100.0 g of C10H8 (naphthalene) in 250. g of C6H6 (benzene). • 55.6 g of Al2(SO4)3 in 500. g of water.

  32. If 55.0 g of glucose (C6H12O6) are dissolved in 525 g of water, what will be the change in bp&fp? Step 1: Convert g of glucose to moles 55.0 g (1 mol) (180.18 g) = 0.305 mol Step 2: Convert g of water to kg 525g  0.525kg Step 3: Calculate m 0.305/0.525 = 0.581 m

  33. Step 4: Obtain molal Kb and Kf from table. • Step 5: Place values into equation ΔTb = Kbm# ΔTb = (0.52°C/m)(0.581m)(1) = 0.302 °C • This means that the boiling point will be elevated by 0.302 °C. This solution will reach boiling point at 100.302 °C. • Now calculate the change in freezing.

  34. Calculate the change in freezing point of 24.5g nickel(II) bromide dissolved in 445 g of water. (assume 100% dissociation) Step 1: Convert g of NiBr2 into moles 24.5g ( 1 mol) (218.5 g) = 0.112mol Step 2: Convert solvent to kg 445 g  0.445kg Step 3: Calculate m 0.112 mol/0.445kg = 0.252m

  35. We now have to take 0.252 and multiply by 3 because the dissociation of the ionic compound makes 3 moles of ions (solute) per kg of solvent: NiBr2 Ni+2 + 2Br – 0.252 x 3 = 0.756m • Step 4: Obtain molal Kf from table. • Step 5: Place values into equation ΔTf = (1.86°C/m)(0.756m) = 1.41 °C *Freezing point has been depressed to -1.41 °C.

  36. The data for the table was found by doing experiments. • It has been found that 1 mole of solute will raise the boiling pt. of 1 kg of water by 0.52 C°. • The same concentration of solute will lower the freezing point of 1 kg of water by 1.86 C°. • These two figures are the molal boiling point constant (Kb) and the molal freezing point constant (Kf).

  37. Coolant is used because it takes higher temperatures to reach boiling point. • Antifreeze needs lower temperatures in order to freeze. • This also why salt is used on frozen roads and walkways. The salt dissolves in the water and lowers the freezing point of water. It now takes colder temps to turn the water into ice. • A 10-percent salt solution freezes at 20 F (-6 C), and a 20-percent solution freezes at 2 F (-16 C).

  38. The salt industryHighway deicing accounts for the majority of sodium chloride use in the United States. Nationally, a total of 31.5 million tons of salt was used in 1994 in the following major end-user percentages (according to The Salt Institute): • 59.8% Highway deicing • 8.7% Industrial processes • 8.2% Water conditioning • 6.3% Agriculture • 4.2% Human consumption Deicing Company Slogan

  39. In order to obtain a 200. mL volume of the diluted solution you would: *Pipet 20 mL of the acidinto a flask that contains 180. mL of water. *Always add acid to water.

  40. What Does This Represent? NaCl(aq) NaCl(aq) CCCCCCC Saline Saline Over The 7 Seas

  41. Colligative Properties • Colligative comes from the Greek word kolligativ meaning glue together. • We use this term for the properties of substances (solutes and solvents) together.

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