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Chapter 7

Chapter 7. Properties of Water. Structure of Water. Sources of Water on Earth. Properties of Water I – Freezing Point. Both liquid and solid can be present together. H 2 O(s) → H 2 O(l) occurs at the same temperature as H 2 O(l) → H 2 O(s) 0 o C at normal atmospheric pressure.

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Chapter 7

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  1. Chapter 7 Properties of Water

  2. Structure of Water

  3. Sources of Water on Earth

  4. Properties of Water I – Freezing Point • Both liquid and solid can be present together. H2O(s) → H2O(l) occurs at the same temperature as H2O(l) → H2O(s) • 0oC at normal atmospheric pressure. • Heat of fusion = 334 J/g = 6.01 kJ/mol

  5. Properties of Water II – Boiling Point • The temperature at which the vapor pressure becomes equal to surrounding atmospheric pressure. • 100oC at 1 atmosphere. • Highly pressure-dependent. • Heat of vaporization = 2257 J/g = 40.67 kJ/mol

  6. Properties of Water III – Specific Heat • The amount of heat necessary to raise the temperature of one gram of a substance by 1oC. • S.H. of water = 4.184 J/gK • Higher for water than nearly any other substance. • Heat capacity = heat needed to raise a given mass of substance 1oC.

  7. Example: Heat Capacity Calculate the heat necessary to raise the temperature of 250 mL of water from 23oC to 75oC. S.H. of water = 4.184 J/gK

  8. Solution: Heat Capacity Calculate the heat necessary to raise the temperature of 250 mL of water from 23oC to 75oC. = 54,400 J = 54.4 kJ

  9. Water – The Universal Solvent

  10. Solubility of Selected Alcohols in Water AlcoholSolubility (g/L)Name CH3OH Infinite methanol C2H5OH Infinite ethanol C3H7OH Infinite 1-propanol C4H9OH 79 1-butanol C5H11OH 27 1-pentanol C6H13OH 5.9 1-hexanol C7H15OH 0.9 1-heptanol

  11. Colligative Properties • Properties of a solution that are changed by the number of solute particles, rather than their nature. • Vapor Pressure • Boiling Point Elevation • Freezing Point Depression • Osmotic Pressure

  12. Vapor Pressure of Pure Water and Water with a Solute

  13. Boiling Point ElevationFreezing Point Depression Tb = Kb x m Kb for water = 0.512 oC/m Tf = Kf x m Kf for water = 1.86 oC/m

  14. Example:Boiling Point Elevation Calculate the boiling point of a solution of 1.9 mol of sugar (C12H22O11) dissolved in 400 g of water. Tb = Kb x m Kb for water = 0.512 oC/m

  15. Solution:Boiling Point Elevation Calculate the boiling point of a solution of 1.9 mol of sugar (C12H22O11) dissolved in 400 g of water. molality = = 4.75m B.P elevation = 0.512 oC/m x 4.75m = 2.4 oC Boiling Point = 100 + 2.4 = 102.4oC

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