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Chapter 7Chapter 7

Properties of WaterProperties of Water

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Structure of Water

O

HH

....

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Sources of Water on Earth

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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

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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

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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.

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Example: 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

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Solution: 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= 54,400 J = 54.4 kJ

C52 x Cg

J 4.184 x

mL 1

g 1 x mL 250 o

o

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Water – The Universal Solvent

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Solubility of Selected Alcohols in WaterAlcohol Solubility (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

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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

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Vapor Pressure of Pure Water and Water with a Solute

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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

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Example: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

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Solution: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 = = 4.75 m

B.P elevation = 0.512 oC/m x 4.75 m = 2.4 oC

Boiling Point =Boiling Point = 100 + 2.4 = 102.4102.4ooCC

waterkg 0.400

sugar mol 91.

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