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Colligative Properties. Ch 12. What is a colligative property?. Only apply to solutions (solvent and a solute) Colligative Property: A property that depends only upon the solute concentration and not on the solute’s identity. Four Important Colligative Properties of Solutions.

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Colligative Properties

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Colligative Properties

Ch 12

What is a colligative property?

  • Only apply to solutions (solvent and a solute)

  • Colligative Property: A property that depends only upon the solute concentration and not on the solute’s identity.

Four Important Colligative Properties of Solutions

  • Vapor Pressure Lowering — explains the value of putting antifreeze in your radiator to keep a car from overheating

  • Boiling Point Raising — explains how you can cook spaghetti faster in salt water

  • Freezing Point Lowering — explains why salt is placed on roads to keep ice from forming

  • Osmotic Pressure — explains why your little brother killed the family fish when he placed them in pure water

Vapor Pressure Lowering

On the surface of the pure solvent (shown on the left) there are more solvent molecules at the surface than in the right-hand solution flask. Therefore, it is more likely that solvent molecules escape into the gas phase on the left than on the right. Therefore, the solution should have a lower vapor pressure than the pure solvent.

Boiling Point Elevation

Adding solutes to boiling liquids (like salt to water) raises the boiling point.


For the Boiling Point Elevation: think of the solute molecules interfering with the liquid’s ability to become a gas. They get in the way… so need more energy to overcome = higher temperature

Freezing Point Depression

Adding solutes to freezing liquids (like salt to water) lowers the freezing point.


For Freezing Point Depression: solute molecules get in the way of the formation of crystals. In essence, the connections that need to be made are harder to create, so lower temperature is required.

Osmotic pressure

Osmosis refers to the flow of solvent molecules past a semipermeable membrane that stops the flow of solute molecules only. When a solution and the pure solvent are placed on either side of a semipermeable membrane, it is found that more solvent molecules flow out of the pure solvent side of the membrane than solvent flows into the pure solvent from the solution side. This is because more solvent molecules are at the membrane interface on the solvent side of the membrane than on the solution side. Therefore, it is more likely that a solvent molecule will pass from the solvent side to the solution side than vice versa. That difference in flow rate causes the osmotic pressure of the solution to rise.

Concentration and Colligative Properties

If colligative properties depend on the amount of the solute in the solvent, then the equations defining them must include a concentration term, and sure enough, they do.

Three colligative properties set equal to three different concentrations terms times a solvent constant

ΔP vapor pressure ΔT temperature raising or lowering π osmotic pressure

ΔP = χP0

ΔT f = mK f

π = MRT

χ (mole fraction)

ΔT b= mK b

M (molarity)

m (molality)

Expressing Concentration

Let’s start by imagining that we are placing 50.0 g (0.146 mole) of the solute, sugar, in 117 g (6.5 mole) of the solvent, H2O. Density of the solution is 1.34 g/ml.

What is χ (mole fraction) of 50g of sugar in 117 g of water?

χ = molesA

molesA + molesB

χ = moles sugar

moles sugar + moles H2O

χ = 0.146

0.146 + 6.5

= 0.0220 mole fraction sugar

What is m (molality) of 50.0 g of sugar in 117 g of water?

m = moles A

kg solvent

m = 0.146 moles sugar (solute)

0.117 kg H2O (solvent)

m = 1.25 molal

What is M (molarity) of 50.0 g of sugar in 117 g of water? First we need to find the volume of solution from a density calculation.

V solution = (mass)/(density) = (50 g + 117 g)/(1.34 g/mL) = 125 mL

M = moles sugar (solute)

Volume solution (L)

M = 0.146 moles sugar (solute)

0.125 L (solution)

M = 1.17 Molar

So we have 3 ways to describe 50.0 g of sugar in 117 g of water, each of which is used in a colligative property calculation.

0.022 mole fraction ≡ 1.25 m ≡ 1.17 M

So how big of an effect does a solute concentration have on a colligative property? Let’s use our calculated concentrations to find out…

Colligative property 1: Vapor pressure depression

ΔP = P0χ mole fraction which is the amount of solute added

Vapor pressure depression constant which is the vapor pressure of pure solvent at a given T. of pure solvent

Ex: For H2O at 25oC the pure vapor pressure is 23.76 mm Hg (see table 5.3 p. 200 for vapor pressure constants)

So the vapor pressure depression in

ΔP = 23.76 mm Hg (0.0220) = 0.523 mm Hg

And the new vapor pressure is now about 23.2 mm Hg.

By the way, this equation is referred to as Raoult’s Law which says simply that the vapor pressure above a solution is proportional to the mole fraction of the solute.

Colligative Property 2: boiling point elevation

ΔTb= K bm

molality which is the amount of solute added

Temperature increase

because of solute (b.p. of

solution minus b.p. of solvent)

temperature constant which is the boiling point elevation constant (see table on p. 532)

For water, Kbis 0.512 oC/molal

So the boiling point elevation for our previous example is ΔTb= Kbm = (0.512 oC/molal)(1.25 molal) = 0.64 oC

And the new b.p. of water with a heck of a lot of sugar in it is 100.64 oC

Colligative Property 3: freezing point depression

ΔTf = Kf m

molality which is the amount of solute added

Temp. decrease

because of solute (f.p. of

solvent minus f.p. of solution)

constant which is the freezing point depression constant (see table p. 532)

For water, Kb is 1.86 oC/molal

So the freezing point depression for our previous example is

ΔT f= Kfm = (1.86 oC/molal)(1.25 molal) = 2.32 oC

And the new f.p. of water with a heck of a lot of sugar in it is

-2.32 oC

Colligative Property 4: osmotic pressure

π = MRT

system temperature (K)

Osmotic pressure

increase because of solute

Gas constant, R

molarity which is the amount of solute added

R is 0.0821 L·atm/K·mol

So the osmotic pressure change for our previous example is

π = MRT = (1.17 M)(0.0821 L·atm/K·mol)(298 K) = 28.6 atm

Comparison of magnitudes of colligative property changes

  • ΔP lowers water vapor pressure from 23.76 to 23.2 mm Hg

  • ΔTbraises boiling point of water from 100 oC to 100.64 oC

  • ΔTflowers freezing point of water from 0 oC to -2.32 oC

  • π raises osmotic pressure from 1 atm to 28.6 atm

Time out for the Van’t Hoff equation

Very simply, Van’t Hoff corrects for the fact that the number of particles you throw into solution is not always the number of particles that determine the magnitude of the property. For example, think about what happens when you put the following one mole quantities into a liter of water. Which one raises the boiling point the most?

1. 1 mole NaCl 2. 1 mole Na2S 3. 1 mole CaS 4. 1 mole sugar

1 mole Na+

and 1 mole

Cl −

2 moles Na+

1 mole S2−

0 moles



1 mole


  • • 1mole of NaCl is 2 moles of particles in the solution

  • • 1 mole of Na2S is 3 moles of particles in the solution

  • • 1 mole of insoluble CaS is 0 moles of particles in solution

  • • 1 mole of sugar is 1 mole of particles in solution

  • What we need is a correction factor for each compound, i, the Van’t Hoff factor, which is i = 2, 3, 0 and 1, respectively for the four solutions. i is simply inserted into every colligative property equation to make the correction. The van't Hoff factor is 1 for all non-electrolyte solutes and equals the total number of ions released for electrolytes.

  • i = Number of Ions from each formula unit

The Van’t Hoff factor

∆Tf = i KfmΔP = i P0χ

∆Tb = i Kbm π = i MRT


What is the freezing point of a solution of 15.0 g of NaCl in 250 g of water? The molal freezing point constant, Kf, for water is 1.86 oC / molal.

For NaCl, i = 2 moles = 15.0 g = 0.257 mole

Kf =1.86 oC / molal. 58.4g/mol

molal = 0.257 mole/ 0.250 kg = 1.03 m in NaCl

∆Tf = i Kfm

∆Tf = (2)(1.86 oC/molal)(1.03 molal) = 3.8 oC. The freezing point of the solution is, therefore, 0 oC - 3.8 oC = -3.8 oC

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