Transcript Powerpoint for Colligative Properties

Colligative Properties
Colligative properties

Properties that depend on the TOTAL number
of dissolved particles
Colligative properties

Properties that depend on the TOTAL number
of dissolved particles

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Vapor pressure lowering
Boiling point elevation
Freezing point depression
Osmotic pressure
Vapor pressure lowering
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Adding solute leads to
more intermolecular
attractions
It becomes harder for
solvent molecules to
escape into the gas
phase
Vapor pressure lowering

More solute  lower
vapor pressure than
pure solvent

Raoult’s Law
PA  X P
o
A A
Effects of non-volatile solutes
Boiling Point Elevation
Tbp  m  kb
Freezing Point Depression
Tfp  m  k f
Electrolytes vs. nonelectrolytes

Colligative properties depend on total number
of dissolved particles
Electrolytes vs. nonelectrolytes

Colligative properties depend on total number
of dissolved particles

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Non-electrolytes don’t dissociate
Electrolytes DO dissociate into ions in solution
Van’t Hoff factor, i
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For non-electrolyte, i = 1
For electolytes, i = # of ions in formula
(theoretical maximum)
Predict the van’t Hoff factor:
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Glucose, C6H12O6
NaCl
AlCl3
Methanol, CH3OH
Ion pairing

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One mole of NaCl does
not yield two moles of
ions
Some ions will
reassociate for a short
time
The actual van’t Hoff
factor will be slightly
lower than predicted
van’t hoff factors

More ion pairing occurs at higher
concentrations
van’t hoff factors

More ion pairing occurs at higher
concentrations
Incorporating the van’t Hoff
factor

Boiling point elevation:
Tbp  m  i  kb
Incorporating the van’t Hoff
factor

Boiling point elevation:
Tbp  m  i  kb

Freezing point
depression:
Tfp  m  i  k f
Problem

33.5 g of potassium chloride are dissolved in 459 g of water.
Calculate the boiling point and freezing point of the resulting
solution.
Problem

17.8 g of an unknown solute are dissolved in 276 g of water. If the
new freezing point is -1.67oC, calculate the molar mass of the
unknown substance.