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Chapter 12

Chapter 12. Properties of Solutions. Liquids. 2 Properties of Liquids A. Viscosity B. Surface Tension. Some Properties of Liquids. Viscosity Viscosity is the resistance of a liquid to flow. The stronger the intermolecular forces, the higher the viscosity. Some Properties of Liquids.

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Chapter 12

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  1. Chapter 12 Properties of Solutions

  2. Liquids • 2 Properties of Liquids • A. Viscosity • B. Surface Tension

  3. Some Properties of Liquids • Viscosity • Viscosity is the resistance of a liquid to flow. • The stronger the intermolecular forces, the higher the viscosity.

  4. Some Properties of Liquids Viscosity

  5. Surface Tension

  6. Some Properties of Liquids • Surface Tension • Surface tension is the amount of energy required to increase the surface area of a liquid. • Resistance to increased surface area or the tendency of a liquid to maintain minimum surface area. • Cohesive forces bind molecules to each other. • Adhesive forces bind molecules to a surface. • Ex. miniscus

  7. Some Properties of Liquids • Surface Tension • Meniscusis the shape of the liquid surface. • If adhesive forces are greater than cohesive forces, the liquid surface is attracted to its container more than the bulk molecules. Therefore, the meniscus is U-shaped (e.g. water in glass). • If cohesive forces are greater than adhesive forces, the meniscus is curved downwards.

  8. Definitions • Solvent – what does the dissolving • The one that is greater quantity • Solute – what gets dissolved • can be liquid, solid or gas • The one that is lesser in quantity

  9. A solution is always homogeneous!

  10. Factors Affecting Solubility • Temperature • Pressure • Solvent-Solvent Interactions

  11. Rule of Thumb • LIKE DISSOLVES LIKE! • Polar solutes dissolve in polar solvents. • Non-polar solutes dissolve in non-polar solvents. • Compounds with metals (Na, K, Li, etc…) are ionic and therefore polar.

  12. Solution • Concentration • Molarity (M) • % Mass, % Volume, %Mass/Vol. • Mole Fraction • Molality (m) • Normality (will be skipped)

  13. Molarity (M) • Molarity = moles of solute Liter of solution • Moles = Molarity x Liter of soln. • Moles = grams molar mass

  14. Sample Problem • Calculate the molarity of a solution containing 3.65 grams of HCl in enough water to make 500.00 mL of solution. • Ans: 0.200 M

  15. Molality (m) • Molality = moles of solute kg. solvent

  16. Mole Fraction (c) • To Solve for the Mole Fraction of A : cA = Mole A Mole A + Mole B

  17. Sample Problem • 1.00 gram of CH3CH2OH (ethanol) is mixed with 100.0 grams of water. Calculate the mole fraction of ethanol in this solution.

  18. 3 Types of % Concentrations • Percent by Mass • Percent by Volume • Percent by Mass-Volume

  19. Percent by Mass • % Mass = mass of solute x 100 total mass of solution • Sample Problem: Find the concentration of 20 grams of sugar in enough water to make 350 grams of solution. • Answer: 5.7%

  20. Things to Note: • Identity of the substance (molecular formula) is not taken into account since only the masses are needed in the equation.

  21. Percent by Volume • % Volume = volume of solute x 100 Total Vol. of solution

  22. Sample Problem • 10.0 mL of benzene is added to 40.0 mL of carbon tetrachloride. Find the concentration of the solution. • Ans: 10 mL / 50 mL) x 100 = 20%

  23. Percent by Mass-Volume • % Mass-Volume = mass of solute x 100 Total Vol. of Solution

  24. Answers to Problem Set • 1. 357 oC • 2. a. molar mass = 181 • b. Tantalum • c. 1.4315 x 10-8 • 3.  • 4. a. 0.0247 • b. 4.90 • c. 0.547

  25. Molality (m) • Molality = moles kg. solvent

  26. Usefulness of Molality • Chemists find that molality (m) is a useful concept when they must deal with the effect of a solute on boiling point and freezing point of a solution.

  27. Problem • A solution is made containing 25.5 g phenol (C6H5OH) in 495 g ethanol (CH3CH2OH) Calculate: • A. mole fraction of phenol • B. mass % of phenol • C. molality of phenol

  28. Colligative Properties • Depend upon the number of particles in solution rather than the identity of the compound • Therefore if solute does not ionize, the concentration of the solution does not change and there is no change in the calculation of the mole fraction (csoln)

  29. Colligative Properties If solute ionizes, then the number of particles the solute breaks down into should be factored in. • Example: NaCl Na+ + Cl- • 1 mole 1 mole 1 mole

  30. Colligative Properties • If solute is Na2SO4 • Then number of particles is 3 • Na2SO4 = 2Na+ + SO42-

  31. Colligative Properties • Boiling Point Elevation • Freezing Point Depression • Osmotic Pressure • Vapor Pressure lowering

  32. Freezing Point Depression • Presence of solute in solution decreases the freezing point of solution DT = kfmsolute where DT = change in Temperature in oC kf = molal freezing point depression constant (oC/molal) m = molality

  33. Effect of Ionizing substances • The presence of solutes that ionize have to be factored into the equation such that if : DT = kfmsolute , Now: DT = ikfmsolute where i = the # of ions produced.

  34. Boiling Point Elevation • Non-volatile solute elevates the boiling point of solvent • Presence of non-volatile solute decreases vapor pressure, thus solution has to be heated to a higher temperature to boil.

  35. Boiling Point Elevation DT = kbmsolute Where: DT = change in Temp. (in oC) kb = molal boiling point depression constant (oC/molal) m = molality

  36. Sample Problem • A solution is prepared by dissolving 4.9 grams sucrose (C12H22O11 = Mol. Mass: 342.295 g/mol) in 175 grams of water. Calculate the boiling point of this solution. Calculate the freezing point of this solution. Sucrose is a non-electrolyte. Kb of water is 0.51 oC/molal. Kf of water is 1.86 oC/molal.

  37. Sample Problem • Calculate the boiling point and freezing point of 0.50 m FeCl3 solution. Assume complete dissociation. Kb of water is 0.51 oC/molal. Kf of water is 1.86 oC/molal.

  38. Sample Problem • A solution of an unknown nonvolatile compound was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the resulting solution was 0.357 oC higher than that of the pure solvent. Calculate the molar mass of the solute. Kb for CCl4 is 5.02. Normal boiling point of CCl4 is 76.8 oC

  39. Lab Experiment • Determine molality (m) using the equation DT = kfmsolute • Determine molecular mass of sulfur using the equation molality (m) = gram mol. mass Kg. solvent

  40. Osmosis • Osmosis – direction of flow is from less concentrated to more concentrated - shrivelling • Reverse Osmosis - from more concentrated to less concentrated - bloating

  41. Equation Osmotic Pressure p = MRT where M = Molarity R = ideal gas constant (.08206 L-atm/K-mol) T = Temperature in Kelvin p = osmotic pressure in atm

  42. Effect of Ionizing substances • Again the presence of solutes that ionize has to be considered: p = MRT , Now: p = iMRT where i = the # of ions produced.

  43. Sample Problem • A 0.020 grams sample of a non-dissociating protein is dissolved in water to make 25.0 mL of solution. The osmotic pressure of the solution is 0.56 torr at 25 oC. What is the molar mass of the protein?

  44. Solution Formation • Energies are involved in solution formation to: • A. Break the solute apart (endothermic) • B. Break the intermolecular forces of the solvent (endothermic = DH is positive) • C. Allow solute and solvent molecules to interact (exothermic = DH is negative)

  45. Therefore the enthalpy of the solution DHsolution = DHA + DHB + DHC • If DHsoln is (+) = means DHC is small • If DHsoln is (-) = means DHC is large

  46. Factors Affecting Solubility • A. Structure Effects • Hydrophilic – loves water • Hydrophobic – hates water • B. Pressure Effects – more important for gases • C. Temperature effects

  47. IDEAL SOLUTION • When Raoult’s Law is closely followed • Very little interaction between molecules (i.e., bonding between solute and solvent) • If only LDF were present, Raoult’s Law closely followed.

  48. Temperature Effect • Increasing temperature ONLY increases the RATE of solubility and not solubility itself. In some cases, higher Temp. decreases solubility. Experimentation needed.

  49. Raoult’s Law • States that the partial pressure exerted by solvent vapor above the solution (PA), equals the product of the mole fraction of the solvent in the solution (CA) times the vapor pressure of the pure solvent (Po) • PA = (CA)( Po) • Partial pressure = vapor pressure above soln.

  50. Vapor Pressure • The total vapor pressure of the solution is the sum of the partial pressures of each volatile component. • Ptot (above soln) = Psolvent(above soln) + Psolute(above soln) • Partial Pressure of A(solv.) in soln: Psolvent = CsolventPo • Partial Pressure of B(solute) in soln: Psolute = CsolutePo

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