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Chapter 12 Solutions 12.6 Properties of Solutions

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1 Chapter 12 Solutions 12.6 Properties of Solutions
Learning Goal Identify a mixture as a solution, a colloid, or a suspension. Describe how the number of particles in a solution affects the freezing point, boiling point, and osmotic pressure. © 2014 Pearson Education, Inc.

2 Solution and Solute Particles
The solute particles in a solution play an important role in determining the properties of that solution are dispersed uniformly through the solution and cannot be visibly distinguished from the solvent are so small that they go through filters and semipermeable membranes © 2014 Pearson Education, Inc.

3 Colloids Colloids are homogeneous mixtures that do not settle out. The solute particles in a colloidal suspension or colloid are much larger than the particles in a solution large molecules such as proteins, or groups of molecules or ions small enough to pass through filters but too large to go through a semipermeable membrane

4 Colloids

5 Suspensions Suspensions are heterogeneous, nonuniform mixtures contain very large particles that are trapped by filters and do not pass through semipermeable membranes contain particles that settle out of solution and must be stirred to stay suspended Examples: muddy water, liquid penicillin

6 Solutions, Colloids, Suspensions

7 Solutions, Colloids, Suspensions

8 Colligative Properties
Colligative properties of solutions depend on the number of solute particles in the solution. Adding a solute to water lowers the freezing point of the solvent. Adding a solute to water elevates the boiling point of the solution.

9 Freezing Point Lowering
Ethylene glycol (antifreeze) HO—CH2—CH2—OH is an organic compound with two –OH groups, that make it very soluble in water is added to water in a car radiator to lower its freezing point and raise its boiling point prevents the water in a car radiator from freezing in cold weather and boiling in hot weather

10 Ethylene Glycol A 50-50% by mass mixture of ethylene glycol and water freezes at about −30 oF and does not boil until it reaches 225 oF. Ethylene glycol dissolves as molecules.

11 Particles in Solution Solutes that are nonelectrolytes dissolve as molecules 1 mole of C2H6O2(l) = 1 mole of C2H6O2(aq) strong electrolytes dissolve as ions 1 mole of NaCl(s) = 2 moles of particles, Na+(aq) and Cl–(aq)

12 Molality (m) Molality is a unit of concentration: Molality is used to calculate freezing point lowering or boiling point elevation.

13 Guide to Calculating Molality

14 Learning Check What is the molality of particles in a solution of 125 g of MgCl2, a strong electrolyte, dissolved in kg of water?

15 Solution What is the molality of particles in a solution of 125 g of MgCl2, a strong electrolyte, dissolved in kg of water? Step 1 State the given and needed quantities.

16 Solution What is the molality of particles in a solution of 125 g of MgCl2, a strong electrolyte, dissolved in kg of water? Step 2 Write the molality expression.

17 Solution What is the molality of particles in a solution of 125 g of MgCl2, a strong electrolyte, dissolved in kg of water? Step 3 Substitute the solute and solvent quantities into the expression and calculate.

18 Solution What is the molality of particles in a solution of 125 g of MgCl2, a strong electrolyte, dissolved in kg of water? Step 3 Substitute the solute and solvent quantities into the expression and calculate.

19 Solution What is the molality of particles in a solution of 125 g of MgCl2, a strong electrolyte, dissolved in kg of water? Step 3 Substitute the solute and solvent quantities into the expression and calculate.

20 Molality, Colligative Properties
Freezing point lowering (ΔTf) is determined from the molality (m) of the particles in the solution and the freezing point constant, Kf. To determine the new solution freezing point The freezing point constant (Kf) for water is 1.86o C/m.

21 Molality, Colligative Properties
For nonelectrolyte solutions such as glucose with 1 mole of particles per one mole of solute, For electrolyte solutions, such as NaCl with 2 moles of particles per mole of solute, NaCl(s)  Na+(aq) + Cl−(aq)

22 Molality, Colligative Properties
Boiling point elevation (ΔTb) is determined from the molality (m) of the particles in the solution and the boiling point constant, Kb. To determine the new boiling point The boiling point constant (Kb) for water is oC/m.

23 Molality, Colligative Properties
For nonelectrolyte solutions such as glucose with 1 mole of particles per one mole of solute, For electrolyte solutions, such as NaCl with 2 moles of particles per mole of solute, NaCl(s)  Na+(aq) + Cl−(aq)

24 Solute Concentration and Freezing, Boiling Points

25 Guide to Using Molality

26 Learning Check In the northeastern United States during freezing temperatures, CaCl2 is spread on icy highways to melt the ice. Calculate the freezing point lowering and freezing point of a solution containing 225 g of CaCl2 in 500. g of water.

27 Solution Calculate the freezing point lowering and freezing point of a solution containing 225 g of CaCl2 in 500. g of water. Step 1 State the given and needed quantities.

28 Solution Calculate the freezing point lowering and freezing point of a solution containing 225 g of CaCl2 in 500. g of water. Step 2 Write the expression for the change in freezing or boiling point.

29 Solution Calculate the freezing point lowering and freezing point of a solution containing 225 g of CaCl2 in 500. g of water. Step 3 Substitute the molality into the expression and calculate.

30 Step 3 Substitute the molality into the expression and calculate.
Solution Calculate the freezing point lowering and freezing point of a solution containing 225 g of CaCl2 in 500. g of water. Step 3 Substitute the molality into the expression and calculate. Calculating the molality of the CaCl2 solution: moles of particles

31 Solution Calculate the freezing point lowering and freezing point of a solution containing 225 g of CaCl2 in 500. g of water. Step 3 Substitute the molality into the expression and calculate.

32 Solution Calculate the freezing point lowering and freezing point of a solution containing 225 g of CaCl2 in 500. g of water. Step 3 Substitute the molality into the expression and calculate.

33 Osmosis In osmosis, water (solvent flows from the lower solute concentration into the higher solute concentration the level of the solution with the higher solute concentration rises the concentrations of the two solutions become equal with time

34 Osmotic Pressure Osmotic pressure is produced by the solute particles dissolved in a solution is the pressure that prevents the flow of additional water into the more concentrated solution increases as the number of dissolved particles in the solution increases

35 Osmosis and Osmotic Pressure
Water flows into the solution with a higher solute concentration until the flow of water becomes equal in both directions.

36 Learning Check A semipermeable membrane separates a 10% starch solution (A) from a 5% starch solution (B). If starch is a colloid, fill in the blanks in the statements below. 1. Solution ____ has the greater osmotic pressure. 2. Water initially flows from ___ into ___. 3. The level of solution ____ will be lower.

37 Solution A semipermeable membrane separates a 10% starch solution (A) from a 5% starch solution (B). If starch is a colloid, fill in the blanks in the statements below. Solution (A) has the greater osmotic pressure. Water initially flows from (B) into (A). 3. The level of solution (B) will be lower.

38 pressure as body fluids such as blood.
Isotonic Solutions Solutes in body solutions such as blood, tissue fluids, lymph, and plasma exert osmotic pressure on semipermeable cell membranes. Most intravenous solutions used in hospitals are isotonic solutions, which exert the same osmotic pressure as body fluids such as blood. A 0.9% NaCl solution is isotonic with the solute concentration of the blood cells of the body.

39 Hypotonic Solutions When a red blood cell is placed in a hypotonic solution, which has a lower solute concentration, water flows into the cell by osmosis. The increase in fluid causes the cell to swell and possibly burst—a process called hemolysis.

40 Hypotonic Solutions When a red blood cell is placed in a hypertonic solution, which has a higher solute concentration, water flows out of the cell by osmosis. The decrease in fluid causes the cell to shrink and possibly burst—a process called crenation.

41 Isotonic, Hypotonic, Hypertonic Solutions
(a) In an isotonic solution, a red blood cell retains its normal volume. (b) Hemolysis: In a hypotonic solution, water flows into a red blood cell, causing it to swell and burst. (c) Crenation: In a hypertonic solution, water leaves the red blood cell, causing it to shrink.

42 Isotonic, Hypotonic, Hypertonic Solutions


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