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Chapter 12 Solutions

Ethylene glycol is added to a radiator to form a aqueous solution that has a lower freezing point than water. Chapter 12 Solutions. 12.7 Properties of Solutions. Solutions. Solutions contain small particles (ions or molecules) that are uniformly dispersed in the solvent

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Chapter 12 Solutions

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  1. Ethylene glycol is added to a radiator to form a aqueous solution that has a lower freezing point than water. Chapter 12 Solutions 12.7Properties of Solutions

  2. Solutions Solutions • contain small particles (ions or molecules) that are uniformly dispersed in the solvent • are not separated by filters or a semipermeable membrane

  3. Colloids Colloids • contain large solute particles that are uniformly dispersed in the solvent • cannot be separated by filtration • can be separated by semipermeable membranes

  4. Examples of Colloids

  5. Suspensions Suspensions • have very large particles that settle out • are separated by filters • must be stirred to stay suspended Examples: blood platelets, muddy water, and calamine lotion

  6. Properties of Different Types of Mixtures

  7. Solutions Colloids, and Suspensions

  8. Moles of Particles The number of moles of particles in water depends on the type of solute • nonelectrolytes dissolve as molecules 1 mol of nonelectrolyte = 1 mol of particles • strong electrolytes dissolve as ions 1 mol of a strong electrolyte = 2 to 4 mol of particles Examples: Nonelectrolyte: 1 mol of C2H6O2(l) = 1 mol of C2H6O2(aq) Strong electrolyte: 1 mol of KCl (s) = 1 mol of K+(aq) + 1 mol of Cl(aq) =2 mol of particles(aq)

  9. Molality The molality, abbreviation m,is a concentration unit that gives • number of moles of solute particles • 1 kg (1000 g) of solvent Molality (m) = moles of solute particles kilograms of solution

  10. Example of Calculating Molality What is the molality of a solution containing 80.0 g of the nonelectrolyte glucose (C6H12O6, molar mass 180.16) in 0.250 kg of water? STEP 1 Given 80.0 g of glucose; 0.250 kg of water Need molality (m) of glucose solution STEP 2 Plan grams of glucose moles of glucose moles of glucose particles molality

  11. Example of Calculating Molality (continued) STEP 3 Equalities/Conversion Factors 1 mol of glucose = 180.16 g of glucose 1 mol glucose and 180.16 g glucose 180.16 g glucose 1 mol glucose 1 mol of glucose = 1 mol of particles 1 mol glucose and 1 mol particles 1 mol particles 1 mol glucose

  12. Example of Calculating Molality (continued) STEP 4 Set Up Problem Moles of glucose particles = 80.0 g glucose x 1 mol glucose x 1 mol particles 180.16 g glucose 1 mol glucose = 0.444 mol of glucose particles Molality (m) = 0.444 mol particles = 0.178 m 0.250 kg water

  13. Learning Check What is the molality of particles in a solution of 125 g of MgCl2, a strong electrolyte, dissolved in 0.500 kg of water?

  14. Solution STEP 1 Given 125 g of MgCl; 0.500 kg of water Need molality (m) of particles in water STEP 2 Plan grams of MgCl2 moles of MgCl2 moles of particles molality STEP 3 Equalities/Conversion Factors 1 mol of MgCl2 = 95.21 g of MgCl2 1 mol MgCl2 and 95.21 g MgCl2 95.21 g MgCl2 1 mol MgCl2

  15. Solution (continued) STEP 3 Equalities/Conversion Factors (continued) 1 mol of MgCl2(s) = 1 mol of Mg2+(aq) + 2 mol of Cl(aq) = 3 mol of particles(aq) 1 mol of MgCl2 = 1 mol of particles 1 mol MgCl2 and 3 mol particles 3 mol particles 1 mol MgCl2

  16. Solution (continued) STEP 4 Set Up Problem Moles of glucose particles = 125 g MgCl2 x 1 mol MgCl2 x 3 mol particles 95.21 g MgCl2 1 mol MgCl2 = 3.94 mol of particles Molality (m) = 0.394 mol particles = 7.88 m 0.500 kg water

  17. Freezing Point and Boiling Point Change When a solute is added to water, • the number of particles produced in water (molality) changes the physical properties of water • freezing point is lower than pure water For a 1 molal solution, the freezing point depression (Tf ) is 1.86 °C • boiling point is higher than pure water For a 1 molal solution, the boiling point elevation (Tb ) is 0.52 °C

  18. Effect of Solute Particles

  19. Learning Check A 1.00-kg sample of water contains 0.500 mol of Na3PO4. What is the freezing point (FP) of the solution? A) FP = 1.86 °C B) FP = 3.72 °C C) FP = 7.44 °C

  20. Solution STEP 1 Given 0.500 mol of Na3PO4; 1.00 kg of water Need freezing point of the solution STEP 2 Plan grams of Na3PO4 moles of Na3PO4 moles of particles molality Tf FP STEP 3 Equalities/Conversion Factors 1 mol of Na3PO4(s) = 3 mol of Na+(aq) + 1 mol of PO43(aq) =4 mol of particles(aq) 1 mol of Na3PO4 = 4 mol of particles 1 mol Na3PO4 and 4 mol particles 4 mol particles 1 mol Na3PO4

  21. Solution (continued) STEP 4 Set Up Problem Moles of Na3PO4 particles = 0.500 mol Na3PO4 x 4 mol particles 1 mol Na3PO4 = 2.00 mol of particles Molality (m) = 2.00 mol particles = 2.00 m 1.00 kg water (Tf ) = 2.00 m x 1.86 °C = 3.72 °C 1 m Freezing point = 0.00 – 3.72 °C = – 3.72 °C B)

  22. 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

  23. 4% starch 10% starch H2O Example of Osmosis A semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens? semipermeable membrane

  24. Example of Osmosis (continued) • The 10% starch solution is diluted by the flow of water out of the 4% solution, and its volume increases. • The 4% starch solution loses water, and its volume decreases. • Eventually, the water flow between the two becomes equal. 7% starch H2O 7% starch H2O

  25. 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

  26. 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.

  27. 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. 1. Solution (A) has the greater osmotic pressure. 2. Water initially flows from (B) into (A). 3. The level of solution (B) will be lower.

  28. Osmotic Pressure of the Blood Red blood cells • have cell walls that are semipermeable membranes • maintain an osmotic pressure that cannot change without damage occurring • must maintain an equal flow of water between the red blood cell and its surrounding environment

  29. Isotonic Solutions An isotonic solution • exerts the same osmotic pressure as red blood cells • is a physiological solution measured as a mass/volume percent (m/v) • of 5.0% (m/v) glucose or 0.9% (m/v) NaCl gives the same osmotic pressure as in the red blood cells (RBCs) • retains the shape of the RBCs

  30. Hypotonic Solutions A hypotonic solution • has a lower osmotic pressure than in RBCs • has a lower concentration than physiological solutions of 5.0% (m/v) glucose or 0.9% (m/v) NaCl • causes water to flow into RBCs • causes hemolysis: RBCs swell and may burst

  31. Hypertonic Solutions A hypertonic solution • has a higher osmotic pressure than RBCs • has a higher concentration than physiological solutions of 5.0% (m/v) glucose or 0.9% (m/v) NaCl • causes water to flow out of RBCs • causes crenation: RBCs shrink in size

  32. Learning Check Indicate if each of the following solutions is 1) isotonic 2) hypotonic 3) hypertonic A. ____ 2% (m/v) NaCl solution B. ____ 1% (m/v) glucose solution C. ____ 0.5% (m/v) NaCl solution D. ____ 5% (m/v) glucose solution

  33. Solution Indicate if each of the following solutions is 1) isotonic, 2) hypotonic, or 3) hypertonic. A. _3)_ 2% (m/v) NaCl solution B. _2)_ 1% (m/v) glucose solution C. _2)_ 0.5% (m/v) NaCl solution D. _1)_ 5% (m/v) glucose solution

  34. Learning Check When placed in each of the following, indicate if a red blood cell will 1) not change 2) hemolyze 3) crenate A. ____ 5% (m/v) glucose solution B. ____ 1% (m/v) glucose solution C. ____ 0.5% (m/v) NaCl solution D. ____ 2% (m/v) NaCl solution

  35. Solution When placed in each of the following, indicate if a red blood cell will 1) not change 2) hemolyze 3) crenate A. _1_ 5%(m/v) glucose solution B. _2_ 1%(m/v) glucose solution C. _2_ 0.5%(m/v) NaCl solution D. _3_ 2%(m/v) NaCl solution

  36. Dialysis In dialysis, • solvent and small solute particles pass through an artificial membrane • large particles are retained inside • waste particles, such as urea from blood, are removed using hemodialysis (artificial kidney)

  37. Hemodialysis • When the kidneys fail, an artificial kidney uses hemodialysis to remove waste particles such as urea from blood. [[page 398]]

  38. Concept Map

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