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Objectives

Objectives. II. States of Matter A. Gases 1. Laws of ideal gases a. Equation of state for an ideal gas b. Partial pressures. Objectives (cont.). 2. Kinetic molecular theory a. Interpretation of ideal gas laws on the basis of this theory b. Avogadro’s hypothesis and the mole concept

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Objectives

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  1. Objectives • II. States of Matter • A. Gases • 1. Laws of ideal gases • a. Equation of state for an ideal gas • b. Partial pressures

  2. Objectives (cont.) • 2. Kinetic molecular theory • a. Interpretation of ideal gas laws on the basis of this theory • b. Avogadro’s hypothesis and the mole concept • c. Dependence of kinetic energy of molecules on temperature • d. Deviations from ideal gas laws

  3. Objectives (cont.) • C. Solutions • 1. Types of solutions and factors affecting solubility • 2. Methods of expressing concentration (use of normalities is not tested) • 3. Raoult’s law and colligative properties (nonvolatile solutes); osmosis • 4. Nonideal behavior (qualitative aspects)

  4. Gas Laws • PV/T = P’V’/T’ • P is pressure in atm, kPa, mmHg, torr • V is volume in cm3, mL, L, etc. • T is temperature and must be in Kelvins • K = oC+ 273 This formula is used to change the conditions of a gas only.

  5. How to use ideal gas law • Need moles? PV/(RT) • Need density? PM/(RT) • If you have density, then you have g and the volume is 1 L

  6. Gas Laws Examples • What is the density of CO2 at 25oCand 115 kPa? • What pressure will O2 exert on a flask at 50oC if the density if 1.435 g/L?

  7. Ideal Gas Law • PV = nRT • The number of moles is n • R is the gas constant and is dependent on how the pressure is measured • V in Liters only; T in Kelvin • PVM = gRT • Since n = g/M, then substitute into PV=nRT and get this version • M is molar mass and g is grams • V in Liters only; T in Kelvin

  8. Partial Pressures of Gas Mixtures • Pressure is defined as the collisions of gas molecules with the sides of a container • The identity of individual gas molecules has no effect • Dalton’s Law of Partial Pressures: P total = P1 + P2 + P3 +…. • In a given container, the pressure is related then the # of molecules of each gas

  9. Partial Pressure and Moles • The ideal gas law demonstrates that P1 = n1 (RT/V) P2 n2 (RT/V) Hence, the ratio n1/n2 is the mole fraction and can be used to calculate the P of each gas in a container.

  10. Example • A container is filled with 54% O2, 26% N2, 12% Ar, 8% H2O by mass. At 250K, the pressure in the tank is 811 torr. What is the partial pressure of each gas? • If these % were given by volume, instead of mass, it is a different problem. How so?

  11. Collecting Gases over Water • This is a version of Dalton’s Law and will be important in a stoichiometry problem. • If oxygen was collected over water at a certain temperature, the pressure reported would have to be adjusted to remove the water. • The water vapor chart is used.

  12. Effusion • The property of effusion has to do with the fact that gases will leak out of very tiny openings. The speed that occurs is related to the size of the gas molecule. • Graham’s Law: r1/r2 = sqrt(M2/M1) • This can be used to determine MM

  13. Example • An unknown gas effuses at a rate that is only 0.355 times that of oxygen at the same temperature. What is the molar mass of the gas? • Calculate the ratio of the rate of effusion of He to oxygen. rHe/rO2

  14. Deviations from Ideal Behavior • Gases behave ideally at higher temperatures and lower pressures. Why? • Johannes van der Waals gave an equation to correct for this behavior. • The equation corrects for the size of the molecule and molecular attractions

  15. Van der Waals equation • If 1.000 mol of an ideal gas was confined to 22.41 L at 273K, it would exert 1.000 atm of pressure. If the “a” value for chlorine gas is 6.49 and “b” is 0.0562, what pressure would 1.000 mol of chlorine exert?

  16. Principles of Solubility Nature of solute/solvent Temperature Pressure of a gaseous solute

  17. Temperature • Read the section on page 263 • Be able to read a solubility graph such as the one on p. 263

  18. Look at KI at 30C and 60C. • How many grams KI will dissolve in 25.6 g water at 30C? • How many grams of water will be needed to dissolve 300 g KI at 60C? • How many grams of KI will fall out of solution if it is cooled to 30C? • Saturated, unsaturated, supersaturated— • What is the difference?

  19. Nature of Solute/Solvent • Read section 10.2 on pgs. 261-263

  20. Density • Density = Mass/Volume • Mass = ? • Volume = ? • The density of mercury is 13.5 g/mL. How many grams are in 2.4 mL? • What is the mass of 2.76 L of CO2 if the density is 1.85 g/L?

  21. Concentration • Concentration is a measure of the solute dissolved in the particular solvent to make a solution. • A solution is defined as a homogeneous mixture; therefore, the solute must be soluble in the solvent at the amounts described. • There are several ways to describe concentration:

  22. Molarity: mol solute / L solution • Most used for chemical reactions. • Use the stoichiometry relationship for calculations: • g / (cF) = M * L / c • The g and F are of the solute and M and L are of the solution. • Most of the time, the coefficients are 1.

  23. M example • How many grams of NaOH (FM = 40.00) are needed to make 500 mL of a 0.75 M solution? • What is the M of a solution made by dissolving 14 g of NaOH in 250 mL of solution? • How many g of aluminum chloride are needed to mix 100 mL of a solution that is 0.2M in chloride ions? (c doesn’t = 1)

  24. molality:mol solute / kg solvent • The solute and solvent are not combined into the mass of solution. • Most used for colligative relationship of freezing and boiling point changes. • Can also use stoichiometry relationship • g / (cF) = m * kg / c • Again g & F are the solute and m & kg are the solvent; coefficients usually = 1

  25. m examples • How many g of sugar (MM = 342.3) are needed to add to 300 g of water to make a 0.55 m solution? • What is the molality of a solution that has 10 g of sugar in 150 g of water?

  26. % solution:g solute / 100 g solution • Easiest to mix; often on household labels; practically useless as a chemistry measurement. • Solve using ratios: • % solute = g solute 100 g solution g solution g solution – g solute = g solvent

  27. ppm and ppb • Used for very small amounts of solute. • If you think of % as parts per hundred, these are the same, except parts per million and parts per billion. • ppm solute = g solute 106 g solution g solution g solution – g solute = g solvent

  28. pph, ppm, ppb • What is the % NaCl in a solution made by adding 43 g to 157 g of water? • How many g of AgCl are needed to mix 500.0 g of a 150 ppm solution? • What is the concentration of Ag+1 ions in the previous solution?

  29. Mole Fraction (X):molsolute / (molsolute + molsolvent) • Percent solutions are called mass fraction, so mole fraction is similar. The solute and solvent are both changed to moles and it is likewise part / total. • Mole fraction can be used for solutes or solvents. Simply change numerator to mol of solvent. • Mole fractions are always < 1

  30. X examples • What is the mole fraction of glucose (MM = 180.18) in a solution prepared by adding 25 g to 200 g of water? • Describe how to mix a 0.30 mole fraction glucose solution in 500 g water.

  31. Dilutions • Many solutions are mixed by diluting a more concentrated solution. It does not matter which units of concentration are used, as long as they are the same. • CA = NT or (M*V)Concentrated = (M*V)Dilute • C-concentration of original, A-aliquot • N-new concentration of dilute, T-total volume

  32. Dilution example • How many mL of 18 M sulfuric acid solution is needed to make 500 mL of 0.75 M? • What is the concentration of a solution made by diluting 200 g of a 20% solution up to 500 g?

  33. Comparisons of Concentration • A solute has a M = 200. Taking 20 g of this solute = 0.1 moles. • 20 g in a 500 mL volumetric flask, filled with water = 0.2 M • 20 g + 500 g water = 0.2 m • 20 g + 80 g water = 20% (0.2 mass fraction) • 20 g + 7.2 g water = .20 mole fraction (Xsolute)

  34. Solution Conversions • To make conversions among the 4 concentrations, the following will be needed: the density and concentration. • You provide: • g solute, mol solute (FM conversion) • g solvent, mol solvent (FM conversion) • g solution, mL solution (density conversion)

  35. Solutions Isopropanol in water • Isopropanol (I) • M = 60.11 • ?g I • ?mol I • ?g water • ?mol water • ?g solution • ?mL solution

  36. Pressure of a gaseous solute • Henry’s Law: At low to moderate pressures, gas solubility is directly proportional to pressure • C (gas) = k * P (gas), where C = concentration and P = pressure; k is a constant characterized by the gas/liquid • Use the mole fraction (X) to find the partial pressure of a gas • See example p. 264

  37. Henry’s Law Example 2 • A certain soft drink is bottled so that a bottle at 25oC contains CO2 at a pressure of 5 atm. Assuming that the partial pressure of CO2 in the atmosphere is 0.00040 atm, calculate the concentrations of CO2 in the soda before and after the bottle is opened. The Henry’s law constant is 0.031 M atm at 25oC.

  38. Colligative Properties • Those properties of a solution that are independent of the type of solute dissolved. The # of moles of particles dissolved is all that matters. • Nonelectrolytes (molecular compounds) provide 1 mol particles/mol solute • Electrolytes (ionic) provide > 1 mole of particles/mol solute

  39. Freezing Point DepressionBoiling Point Elevation • The colligative property where the solute “interferes” with the solvent’s ability to freeze or boil and is directly proportional to the molality of the solution. • DT = k * molality • T = Normal + DT • k is the freezing/boiling point constants and depend on the solvent (on “Data”)

  40. Freezing/Boiling Point Examples • See p. 267 • What is the freezing point if 11.3 g sugar (M = 342.3) is dissolved in 50.0 g of camphor? • What is the boiling point if 26.3 g of the electrolyte aluminum chloride is dissolved in 125 g of water?

  41. Raoult’s Law • The vapor pressure of a solution is directly proportional to the mole fraction of solvent. Ps = Xsolvent Ppure solvent • This is a colligative property—the lowering of the vapor pressure of a solvent is independent of the solute and may be expressed: DP = Xsolute Ppure solvent

  42. Raoult’s Law Examples • What is the vapor pressure of a solution made by dissolving 158 g sugar (M = 342.3) in 641.6 g water at 25oC? • What is the vapor pressure of a solution prepared by mixing 35 g of the electrolyte Na2SO4 in 175 g water at 25oC? • How do you prepare 500 g of an aqueous glucose solution (M = 180.18) that will have a vapor pressure of 21.97 mm Hg if pure water at 24oC has a vapor pressure of 22.38 mm Hg?

  43. Osmotic Pressure • The colligative property of a solvent that the osmotic pressure is directly proportional to the concentration (M) of a solute. p = MRT • Osmotic pressure (p) is equal to the external pressure, P, sufficient to prevent osmosis (the flow of a solvent from a more concentrated to less concentrated region).

  44. Osmotic Pressure Examples • If the human eye has an osmotic pressure at 25oC of 8 atm, what concentration of solute in water will provide an isotonic eyedrop solution? (isotonic = equal osmotic pressure) • How would you prepare a 1 L aqueous saline solution with an osmotic pressure of 12 atm at 22oC? (saline = electrolyte NaCl)

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