Colligative Properties

Nathaniel P. DugosAdamson UniversityManila, Philippines

Colligative properties

Properties determined by the number of particles in solution rather than the type of particles.

Vapor pressure loweringFreezing point depressionBoiling point elevationOsmotic pressure

How Vapor Pressure Lowering Occurs

Solute particles take up space in a solution.

Solute particles on surface decrease number of solvent particles on the surface.

Less solvent particles can evaporate which lowers the vapor pressure of a liquid.

Vapor Pressures of Pure Water and a Water Solution

The vapor pressure of water over pure water is greater than the vapor pressure of water over an aqueous solution containing a nonvolatile solute.

Solute particles take up surface area and lower the vapor pressure

Vapor Pressure Lowering

*AAA PXP

AP

AX

Let component A be the solvent and B the solute.solute B is nonvolatile

Applying Raoult’s Law:

where:

= vapor pressure of the solvent in solution= vapor pressure of the solution= vapor pressure of the pure solvent*

AP= mole fraction of the solvent

The lowering in vapor pressure,

AA PPP *

P

AAA XPP **

*)1( AA PX*ABPXP

where:

BX = mole fraction of solute

1. What mass of urea, CON2H4, must be added to 450 g water to get a solution with a vapor pressure

of 29.3 mmHg? The vapor pressure of pure water is 31.8 mmHg at this temperature.

Sample Problems (Vapor Pressure Lowering)

2. The vapor pressure of 2-propanol is 50.00 kPa at 338.8oC, but fell to 49.62 kPa when 8.69 g of an involatile organic compound was dissolved in 250-g of 2-propanol. Calculate the molar mass of the compound.

Sample Problems (Vapor Pressure Lowering

Boiling Point ElevationBoiling Point ElevationWhen a non volatile solute is added to solvent:• Vapor pressure of solvent is lowered

• solution formed must be heated to higher temperature than boiling point of pure solvent to reach a vapor pressure of 1 atm.

• This means that non volatile solute elevates the boiling point of the solvent which we call boiling point elevation

Boiling Point Elevation

Bvap

XH

RTT

2*

ABBA

BB Mm

mMmX

/1(for dilute solutions)

where AM is the molar mass of the solvent and

Bm the molality of the solute in mol/kg

Boiling Point Elevation

Bvap

Ab mHMRTT

2*

mKT bb

HMRT

vap

Ab2*

where bK

bK = boiling point constant or ebullioscopic constant of the solvent

for dilute solutions

Freezing Point DepressionFreezing Point Depression

Addition of a Addition of a nonvolatile nonvolatile solute to a solution solute to a solution lowers the freezing point of the solution lowers the freezing point of the solution relative to the pure solvent.relative to the pure solvent.

Freezing Point Depression

mKT ff

HMRT

K Aff

fus

2*

where

(for dilute solutions)

fK = molal freezing point depression constant or cryoscopic constant

What is the value of the freezing point constant for water? The enthalpy of fusion at 273.15 K is 6.00 kJ mol-1

The molal freezing point depression constant of benzene is 5.12. A 0.450% solution of monoclinic sulfur in benzene freezes 0.088 K below the freezing point of pure benzene. Find the molecular formula of the sulfur in benzene.

OsmosisThere are many times in nature when a solvent will pass spontaneously through a semipermeable membrane, which is a membrane permeable to solvent, but not solute

The osmotic pressure, Π, is the pressure that must be applied to stop the influx of solvent

Examples:

(a) the transport of fluids through living cell membranes (b) basis of osmometry, the determination of molecular mass by measurement of osmotic pressure

Osmosis

OsmosisOsmosis Eventually the pressure difference between the arms stops osmosis.

OsmosisTo treat osmosis thermochemically, we note that at equilibrium, the chemical potential on each side of the membrane must be equal

Equilibrium is established when the hydrostatic pressure of the solution in the column is equal to the osmotic pressure

tosoluteby lowered is so 1 :sideSolution

pressureat )( :sidesolvent Pure*

*

AAA

A

x

pp

mB

mBAA

p

pmA

p

pmAA

AAA

AAA

A

VRTxVxxx

dpVxRT

dpVpp

xRTppx

pxp

p

constant is and )1ln(by replaced ln :solutions Dilute

ln

have weequations, combiningAsolvent pure of memolar volu theisV where

)()(

account into pressure ofeffect thetakeln)(),(

solute of presence for theaccount ),()(

pressuregreater by increased now is However

m

**

*A

*

mB VRTx

solute theofion concentratmolar theis ][ where

tosimplifiesequation thesolvent, theof volumetotal

, Because . dilute, is solute When the

VnB

VVnnnx

B

mAA

BB

RTB][

A solution of polystyrene in benzene contains 10 g/L. The equilibrium height of the column of solution (density 0.88 g cm-3) in the osmometer corrected for capillary rise is 11.6 cm at 25oC. What is the molar mass of polystyrene, assuming the solution is ideal.

The osmotic pressure of an aqueous solution at 300 K is 120 kPa. Calculate the freezing point of the solution.

Osmometry

Osmotic pressure is easily measured, and is quite large. Osmometry can be applied for the determination of molecular weights of large molecules (proteins, synthetic polymers), which dissolve to produce less than ideal solutions. The Van’t Hoff equation can be rewritten in virial form:

where B is the empirically determined osmotic virial coefficient

Π= [B] RT {1 + B [B] + ...}

Consider the example of poly (vinyl chloride) PVC, in cyclohexanone at 298 KPressures are expressed in terms of heights of solution, ρ=0.980 g cm-3 in balance with the osmotic pressure

c (g L-1) 1.00 2.00 4.00 7.00 9.00h (cm) 0.28 0.71 2.01 5.10 8.00Use Π = [B] RT {1 + B [B] + …} with [B] = c/M,

where c is the mass concentration and M is the molar mass. The osmotic pressure is related to the hydrostatic pressure by Π = ρgh, where g = 9.81 m s-2. Then:

cgMRTB

gMRT

MBc

gMRT

ch

21

Plot h/ c vs. c to find M, expecting a straight line with intercept RT/ ρgM at c = 0.

Data set:c(g L-1) 1.00 2.00 4.00 7.00 9.00h/c(cm g-1 L) 0.28 0.36 0.53 0.729 0.889

The data give anintercept of 0.21 g/mL

The data give an intercept of 0.21 cm g-1 L, which is equal to RT/ ρgM

Thus:

LcmggRTM 1 21.0

1

12

14323

11

102.1

101.21

)81.9()980()298(314.8

kgmol

kgmmskgmKmolJK

where we have used 1 kg m2 s-2=1J