Raoult’sLaw Colligaveproperes A’property… · Lecture’notes’33’ Raoult’sLaw...

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Lecture notes 33

Raoult’s Law

Colliga4ve proper4es A property that depends on the ra4o of the number

of solute molecules to the number of solvent molecules

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Vapor pressure of solu.ons

Ideal solu.on: intermolecular forces between solvent and solute are comparable

Consider a solute red and solvent blue

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Ideal solu.on: intermolecular forces between solvent and solute are comparable

Consider a solute red and solvent blue

red -‐ red and

blue – blue

comaparble to

red -‐ blue

Ideal solu.on

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vapor

solu4on

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Vapor pressure above an ideal solu.on is equal to the sum of the vapor pressures of the the components

Raoult’s Law

P = PA + PBThe vapor pressure of a component is equal to

the product of its mole frac.on and its pure vapor pressure

PA = χAPA0 PB = χBPB

0

P = χAPA0 + χBPB

0

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χB →0 1

P

PA0

PB0

Pure A Pure B

PA + PB

PA

PB

Consider a solu4on of A and B

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CH 3CH2CH 3 & CH 3CH 3

Representa.ve Ideal solu.ons

C6H6 & C6H5CH 3

H2O & CH 3OH

CH 3CH2OH & CH 3OH

Calculate the vapor pressure of a solu.on prepared by mixing 250 ml of H2O (vapor pressure = 23.8 torr) and 250 ml of C2H5OH (vapor presure = 59.8 torr).

density of H2O = 0.997 g/ml & density of C2H5OH = 0.786 g/ml

Need mole frac4on of components

gH2O = 250ml × 0.997 gml

= 249.3g

moles H2O = 249.3g18.0g /mol

= 13.85

gC2H5OH = 250ml × 0.786 gml

= 196.5g

moles C2H5OH = 196.5g46.0g /mol

= 4.27mol

total moles = 18.12

χH2O= 13.8518.12

= 0.764

χC2H5OH= 4.2718.12

= 0.236

PH2O = 0.764 × 23.8torr = 18.2torr

PC2H5OH = 0.236 × 59.8torr = 14.1torr

Ptotal = 32.3torr

What is the mole frac.on of C2H5OH in the vapor phase?

χH2O(liquid) = 0.764

χC2H5OH(liquid) = 0.236

18.2 torr H2O14.1 torr C2H5OHPtotal = 32.3torr

χH2O(vapor) = 18.2torr

32.3torr= 0.56

χC2H5OH(vapor) = 14.1torr

32.3torr= 0.44

Liquid composi4on

pressures in vapor






32.3torr= 0.56


32.3torr= 0.44

Liquid composi4on

pressures in vapor

nH2O = PH2OVRT

⎛⎝⎜

⎞⎠⎟

nC2H5OH = PC2H5OHVRT

⎛⎝⎜

⎞⎠⎟

χC2H5OH=

nC2H5OHnC2H5OH + nH2O

=PC2H5OH

PC2H5OH + PH2O






32.3torr= 0.56


32.3torr= 0.44

Liquid composi4on

pressures in vapor



χH2O(vapor) = 0.56

χC2H5OH(vapor) = 0.44

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Freezing point lowering & boiling point eleva.on &

Entropy

Premise : addi.on of a non-‐vola.le solute increases the entropy of the solu.on

This results in an increase in the boiling point and a decrease

in the freezing point

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vapor vapor

Pure solvent solu.on

ΔGf = ΔH f −TmΔSf = 0

Tb =ΔHV

ΔSsolvent

SΔSsolution

ΔSsolvent

Tb increases

ΔSsolution < ΔSsolvent

Tb =ΔHV

ΔSsolution

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A similar argument holds for the freezing point depression

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Pure liquid

liquid solu.on

Pure solid Solid solu.on

ΔGf = ΔH f −TfΔSf = 0

Tf =ΔHV

ΔSsolvent

S ΔSsolutionΔSsolvent

Tf decreases

ΔSsolution > ΔSsolvent

Tf =ΔHV

ΔSsolution

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Extension of Raoult’s Law

Adding a non-‐vola.le solute to a solu.on always

lowers the vapor pressure of the solu.on

Ptotal = Psolvent + Psolute = χ solventPsolvent0

non-‐vola.le

Example of Raoult’s Law

What is the vapor pressure of a solu.on made by adding 50.0 mL of glycerin (non-‐vola.le) to 500.0 mL of

water at 25 0C? Density of glycerin (C3H8O3) 1.26g/mL

Vapor pressure of pure water at 25 0C 23.8 torr

Example of Raoult’s Law

What is the vapor pressure of a solu.on made by adding 50.0 mL of glycerin (non-‐vola.le) to 500.0 mL of

water at 25 0C? Density of glycerin (C3H8O3) 1.26g/mL

Vapor pressure of pure water at 25 0C 23.8 torr

gramsC3H8O3 = 50.0ml ×1.26gml

= 63.0g

grams H2O = 500.0g

moles C3H8O3 = 63.0g × 1mole112g

= 0.56

moles H2O = 500.0g × 1mole18.0g

= 27.78

total moles = 28.34

χH2O= 27.7828.34

= 0.980

PH2O = 0.980 × 23.8torr = 23.3torr

Ptotal = Psolvent + Psolute = χvolatilePvolatile0

Phase diagram of water

P

T

pure liquid ice

vapor

1 atm

00C 1000C

solu4on

ΔTb ΔT f

Consequences of Vapor Pressure lowering

ΔTb = kb *m* iΔTf = k f *m* i

m = molality

i =Van 't Hoff factor

k f = freezing point depression constant (solvent dependent)

kb = boiling point elevation constant (solvent dependent)

k f (H2O) = 1.86

kb (H2O) = 0.50

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Van’t Hoff Factor : measures the extent to which solutes break up in solu.on

In a moderately dilute solu.on it is simply the number of ion generated from one formula unit.

NaCl

Na2SO4

sugar

KNO3

Mg(NO3)2

Ca3(PO4)2

HCl

CH3CO2H

i=2

i=3

i=1

i=2

i=3

i=5

i=2

i=1

Which of the following aqueous solu.ons has the : lowest freezing point ?

1.5 m NaCl 1.2 m Na2SO4 3.0 m sugar 1.6 m KNO3

: highest vapor pressure?

: lowest vapor pressure?

LFP

HVP

HVP

LVP

I mole of Magnesium acetate is dissolved in water and the boiling point increased by 60C. How much water was in the solu.on?

kb(H2O) = 0.50 0C/m

Mg(CH3COO)2 ΔTb = kb *m* i

m = ΔTbkbi

= 60.5 × 3

= 4moles / kgH2Om = moles

kgH2O

kgH2O = molesm

= 14kgH2O

Raoult’sLaw Colligaveproperes A’property… · Lecture’notes’33’ Raoult’sLaw...

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