Chemistry · General chemistry II · Concept

Colligative properties of solutions

A dissolved solute lowers a solvent’s vapor pressure and freezing point, raises its boiling point and creates osmotic pressure, by amounts that depend on how many solute particles there are, not on what they are. Calculate each change, and run the equations backward to find a molar mass or a van ’t Hoff factor.

A concrete case

Three solutes, one freezing point

Dissolve $0.100 mol$ of table sugar, sucrose, in $1.00 kg$ of water. That is $34.2 g$ . The solution no longer freezes at $0.00 ° C$ . It freezes at $- 0.186 ° C$ .

Now dissolve $0.100 mol$ of glucose instead, only $18.0 g$ of a different molecule. The solution freezes at $- 0.186 ° C$ again. The kind of molecule and its mass made no difference.

Last, dissolve $0.100 mol$ of $NaCl$ , $5.84 g$ . Each formula unit comes apart into ${Na}^{+}$ and ${Cl}^{-}$ , so the water now holds $0.200 mol$ of particles. The freezing point falls about twice as far, to near $- 0.372 ° C$ . A careful measurement gives a little less, because a few ions cling together in pairs.

The water responded to the number of dissolved particles and nothing else. A property that depends only on that count is called colligative, from a Latin word for “bound together.” Four of them follow from it: vapor-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure.

$0.100 mol$ of solute in $1.00 kg$ of water (ideal values)
Solute	Mass dissolved	Particles	Freezing point
Sucrose, $C_{12} H_{22} O_{11}$	$34.2 g$	$0.100 mol$	$- 0.186 ° C$
Glucose, $C_{6} H_{12} O_{6}$	$18.0 g$	$0.100 mol$	$- 0.186 ° C$
Sodium chloride, $NaCl$	$5.84 g$	$0.200 mol$	$- 0.372 ° C$

The rule

Four equations, one count

Raoult’s law gives the vapor pressure of the solvent over a solution with a nonvolatile solute. $P$ is the solution’s vapor pressure, $χ_{solvent}$ the solvent’s mole fraction and $P °$ the pure solvent’s vapor pressure at the same temperature.

The freezing point falls by $Δ T_{f}$ and the boiling point rises by $Δ T_{b}$ . Here $m$ is the molality, in moles of solute per kilogram of solvent. $K_{f}$ and $K_{b}$ are constants of the solvent. For water, $K_{f} = 1.86 ° C \cdot kg / mol$ and $K_{b} = 0.512 ° C \cdot kg / mol$ .

The osmotic pressure $Π$ is the pressure that stops solvent flowing through a membrane into the solution. $M$ is the molarity, $R = 0.08206 L \cdot atm / (mol \cdot K)$ and $T$ is the temperature in kelvin.

The van ’t Hoff factor $i$ counts the particles each formula unit gives. A nonelectrolyte, such as glucose or sucrose, stays whole in water, so $i = 1$ . Ideally $i$ is $2$ for $NaCl$ and $3$ for ${CaCl}_{2}$ .

P = χ_{solvent} P^{\circ}, Δ T_{f} = i K_{f} m, Δ T_{b} = i K_{b} m, Π = i M R T

\begin{array}{c} P = χ_{solvent} P^{\circ} \\ Δ T_{f} = i K_{f} m \\ Δ T_{b} = i K_{b} m \\ Π = i M R T \end{array}

Why it works

Why the count is all that matters

At the surface of pure water, every molecule is water, and some escape into the vapor. Dissolve a solute and part of the liquid is no longer solvent. A smaller share of the molecules can escape, so the vapor pressure falls in proportion to the solvent’s mole fraction. That is Raoult’s law, and a mole fraction counts particles, not grams.

The other three effects follow from the lower vapor pressure. A liquid boils when its vapor pressure reaches the outside pressure, so the solution must be heated past $100 ° C$ to get there. A liquid freezes where its vapor pressure meets that of the solid. Pure ice is unchanged by the solute, so the meeting point moves to a lower temperature.

Solvent also moves on its own from where it makes up more of the liquid to where it makes up less. Across a membrane that passes only water, the water flows into the solution. The pressure needed to stop that flow is the osmotic pressure.

In a dilute solution the solvent’s moles barely change, so the solute’s mole fraction is close to its molality times the solvent’s molar mass. That is why the freezing and boiling equations can use molality.

Which concentration each property uses
Property	Equation	Concentration	Why
Vapor pressure	$P$ $= χ_{solvent} P °$	Mole fraction $χ$	Counts particles at the surface
Freezing point	$Δ T_{f}$ $= i K_{f} m$	Molality $m$ ( $mol / kg$ solvent)	Mass of solvent doesn’t change with temperature
Boiling point	$Δ T_{b}$ $= i K_{b} m$	Molality $m$	Same as freezing point
Osmotic pressure	$Π$ $= i M R T$	Molarity $M$ ( $mol / L$ solution)	Like a gas pressure, amount per volume

The method

Freezing and boiling points

1. Convert the solute to moles with its molar mass.

2. Divide by the mass of the solvent in kilograms to get the molality.

3. Decide $i$ . A molecular solute such as sugar or ethylene glycol has $i = 1$ . For an ionic solid, count the ions in one formula unit.

4. Multiply, $Δ T = i K m$ , with $K_{f}$ for freezing or $K_{b}$ for boiling.

5. Apply the change in the right direction. The freezing point is the pure solvent’s minus $Δ T_{f}$ . The boiling point is the pure solvent’s plus $Δ T_{b}$ .

Vapor pressure and osmotic pressure

For a vapor pressure, find the moles of solvent and of solute particles, form the solvent’s mole fraction, and multiply by $P °$ . If both liquids evaporate, give each one its own Raoult’s-law term, $P_{A} = χ_{A} P °_{A}$ , and add them. Dalton’s law says the total is their sum.

For an osmotic pressure, find the molarity, convert the temperature to kelvin, and multiply $i M R T$ . With $R = 0.08206 L \cdot atm / (mol \cdot K)$ , the answer is in atmospheres.

Working backward to a molar mass or $i$

Each equation can be run in reverse. From a measured $Δ T_{f}$ , find the molality, $m = Δ T_{f} \div (i K_{f})$ . Multiply by the kilograms of solvent to get the moles of solute. Then divide the solute’s mass by those moles to get its molar mass.

From a measured $Π$ , find the molarity, $M = Π \div (R T)$ , then the moles in the volume, then the molar mass the same way.

For a known ionic solute, divide the measured $Δ T$ by the change $i = 1$ would give. The quotient is the measured van ’t Hoff factor.

molar mass = \frac{mass of solute}{n_{solute}}, i_{measured} = \frac{Δ T_{measured}}{K m}

\begin{array}{c} molar mass = \frac{mass of solute}{n_{solute}} \\ i_{measured} = \frac{Δ T_{measured}}{K m} \end{array}

Where it breaks

The equations describe ideal, dilute solutions. Above about $1 mol / kg$ , measured changes drift away from them.

The ideal $i$ is an upper limit. Ions attract one another, and some travel in pairs, so measured values sit a little below $2$ for $NaCl$ and below $3$ for ${CaCl}_{2}$ . A solute can also have $i$ below $1$ . Acetic acid in benzene pairs up into dimers, so it gives fewer particles than formula units.

The boiling-point and Raoult equations assume the solute does not evaporate. Ethanol in water is volatile and adds its own vapor pressure, so the two-component form of Raoult’s law is needed. Real mixtures also stray from it. When the two kinds of molecules attract each other more strongly than themselves, the total pressure falls below the ideal line, and when less strongly, it rises above.

Molality is not molarity. $Δ T_{f} = i K_{f} m$ divides by kilograms of solvent, while $Π = i M R T$ divides by liters of solution. In a dilute water solution the two numbers are close, and in a concentrated one they are not.

Check your answer

Check the direction. A solute always lowers the vapor pressure and the freezing point and raises the boiling point. A freezing point above $0 ° C$ for a water solution means a sign slip.

Compare the two temperature shifts. For water, $Δ T_{f} \div Δ T_{b}$ $= 1.86 \div 0.512$ $= 3.63$ . The antifreeze example’s $7.49 ° C$ and $2.06 ° C$ give $3.64$ , the same within rounding.

Ask whether a molar mass is reasonable. A small molecule comes out at tens or hundreds of grams per mole, and a protein at tens of thousands.

In a Chemistry box, write the relation on its own row, each value on a row below it, then the substitution. Under $Δ T_{f} = K_{f} m$ , with $K_{f} = 1.86 ° C \cdot kg / mol$ and $m = 4.028 mol / kg$ above, the row $Δ T_{f}$ $= (1.86 ° C \cdot kg / mol)$ $(4.028 mol / kg)$ $= 7.49 ° C$ reads ✓. The Checker says “Substitution matches row 3 with the values defined above.”

For an ionic solute, write $i$ on its own row, such as $i = 3$ under $Δ T_{f} = i K_{f} m$ . Without it the substitution reads ? with “i isn’t defined above.”

Where

K_{f}

comes from

$K_{f}$ is not an arbitrary constant. A thermodynamic argument at the freezing point gives $K_{f} = R T_{f}^{2} M \div Δ H_{fus}$ , where $T_{f}$ is the solvent’s freezing point in kelvin, $M$ its molar mass in $kg / mol$ and $Δ H_{fus}$ its enthalpy of fusion. For water, $(8.314) {(273.15)}^{2} (0.01802) \div 6010$ $= 1.86 ° C \cdot kg / mol$ . A solvent that melts with little heat, such as camphor, has a much larger $K_{f}$ .

Osmotic pressure is the most sensitive of the four. A protein at $1.0 \times 10^{- 4} mol / L$ raises an easily measured column of water, while it lowers the freezing point by less than a thousandth of a degree. That is why osmometers weigh large molecules.

Pressing on the solution side with more than $Π$ drives water the other way, out of the solution. That is reverse osmosis, which turns seawater into drinking water. Cells face the same balance. A red blood cell swells in pure water and shrinks in strong brine, because water crosses its membrane toward the side with more dissolved particles.

Common mistakes

Dividing by the solution’s volume in $Δ T_{f} = i K_{f} m$ . The $m$ is a molality, moles per kilogram of solvent, so the antifreeze uses $0.4028 mol \div 0.1000 kg$ $= 4.028 mol / kg$ , not a molarity.
Forgetting $i$ for an ionic solute. $0.100 mol / kg {CaCl}_{2}$ holds $0.300 mol / kg$ of ions, so $Δ T_{f}$ is $0.558 ° C$ , not $0.186 ° C$ . Each formula unit releases three ions, and the water counts them all.
Moving the boiling point the wrong way. A solute raises it, so the antifreeze boils at $100.00 ° C + 2.06 ° C$ $= 102.06 ° C$ , not $97.94 ° C$ . Only the freezing point goes down.
Using $° C$ in $Π = i M R T$ . At $25.0 ° C$ the temperature is $298.2 K$ . Putting in $25.0$ makes the osmotic pressure about twelve times too small, because $R$ is defined per kelvin from absolute zero.
Using the mass fraction in Raoult’s law. $50.0 g$ of glucose in $250.0 g$ of water is $16.7 %$ glucose by mass, but χ(glucose) is only $0.0196$ . Raoult’s law counts molecules, so $P = 0.9804 \times 23.8$ $= 23.3 torr$ .
Assuming only one liquid evaporates. Benzene and toluene both have vapor pressures, so their partial pressures add to $50.6 torr$ . Counting toluene alone gives $18.9 torr$ , far too low.

Key terms

Colligative properties: Properties of a solution that depend on how many solute particles are dissolved, not what they are: vapor-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure. Real solutions deviate at higher concentrations.
Molality: Moles of solute per kilogram of solvent $(mol / kg)$ . Unlike molarity, it uses the solvent’s mass rather than the solution’s volume, so it doesn’t change with temperature.
Van’t Hoff factor: The number of particles, $i$ , that a solute gives per formula unit when it dissolves: $1$ for sugar, about $2$ for $NaCl$ and about $3$ for ${CaCl}_{2}$ . Measured values are often a little lower because ions pair up.
Raoult’s law: For an ideal solution, each volatile component’s vapor pressure equals its mole fraction in the liquid times the vapor pressure of the pure component at that temperature. In symbols, $P = χ P °$ , where $χ$ is the mole fraction in the liquid and $P °$ the pure component’s vapor pressure.
Vapor pressure: The pressure of a vapor in equilibrium with its liquid or solid at a given temperature. It rises with temperature. A gas collected over water includes water’s vapor pressure.
Freezing-point depression: The drop in a solvent’s freezing point caused by dissolved particles: $Δ T_{f} = i \cdot K_{f} \cdot m$ , where $i$ is the van’t Hoff factor, $K_{f}$ is the solvent’s freezing-point constant and $m$ is the molality.
Boiling-point elevation: The rise in a solvent’s boiling point caused by a dissolved nonvolatile solute: $Δ T_{b} = i \cdot K_{b} \cdot m$ , where $i$ is the van’t Hoff factor, $K_{b}$ the solvent’s boiling constant and $m$ the molality. Use the solvent’s boiling constant $K_{b}$ , not its freezing constant.
Osmotic pressure: The pressure needed to stop solvent from flowing through a semipermeable membrane into a solution. For a dilute solution, $Π = i M R T$ , where $i$ is the van’t Hoff factor, $M$ the molarity, $R$ the gas constant ( $0.08206 L \cdot atm / (mol \cdot K)$ for $Π$ in $atm$ ) and $T$ the kelvin temperature.
Mole fraction: $mole fraction of A$ $= (moles of A)$ $\div (total moles of everything in the mixture)$ . It has no units, and all the mole fractions in a mixture add up to $1$ .
Ideal solution: A solution in which attractions between unlike particles equal those between like ones. It follows Raoult’s law at every composition: a component with mole fraction $χ$ has vapor pressure $P = χ \cdot P °$ , where $P °$ is that component’s vapor pressure when pure. Mixing the components releases or absorbs no heat.

Work through an example

$25.0 g$ of ethylene glycol, $C_{2} H_{6} O_{2}$ , is dissolved in $100.0 g$ of water. Find the freezing and boiling points of the solution. For water, $K_{f} = 1.86 ° C \cdot kg / mol$ and $K_{b} = 0.512 ° C \cdot kg / mol$ .