Colligative Properties: Freezing, Boiling & Osmosis

Choose what to dissolve, how much, and in how much water. The page counts the particles and works out all four effects they cause: the freezing point, the boiling point, the osmotic pressure, and the fall in vapour pressure.

Worked example — Colligative Properties: Freezing, Boiling & Osmosis

Molality	1.000 mol/kg
Particles per kilogram of water	1.000 osmol/kg
Freezing-point depression	1.860 K
Boiling-point elevation	0.5131 K
Osmotic pressure at 25 °C	24.72 bar
Height of water that would hold it back	252.8 m
Fall in the vapour pressure of the water	1.770%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Colligative Properties: Freezing, Boiling & Osmosis

A colligative property depends on how many solute particles there are in a given amount of solvent, not on what they are. Four of them move together: the freezing point falls by ΔT_f = K_f · i · m, the boiling point rises by ΔT_b = K_b · i · m, the vapour pressure falls by the mole fraction of solute (Raoult's law), and an osmotic pressure Π = i · m · ρ · R · T appears across any membrane that passes the solvent and stops the solute. Here m is the molality, moles per kilogram of solvent, and i counts the particles each formula unit becomes.

What each symbol means

m: molality, in mol per kg of water. Per mass of solvent, so it does not change with temperature the way a concentration per litre does
i: the van 't Hoff factor: particles per formula unit. 1 for sugar, 2 for NaCl, 3 for CaCl₂, in the ideal count this page uses
K_f: the freezing-point constant, 1.8595 K·kg/mol for water. R T_f² M / ΔH_fus, so it is set by the latent heat
Π: osmotic pressure, in bar. The pressure that would have to be applied to the solution to stop pure water flowing into it

Assumes: An ideal dilute solution: the particles do not interact, the solute does not evaporate, and none of it goes into the ice. The van 't Hoff factor is the ideal count. A real salt solution behaves as if i were a little smaller, because ions of opposite charge stay close to each other, and a real protein carries counter-ions that add to its count. The osmotic pressure uses the density of water at the chosen temperature to turn molality into a concentration per volume, which is the dilute-solution form of van 't Hoff's law.
Breaks when: At high concentration, and for salt the page shows where. At 23.3 wt% NaCl the ideal line says −19.33 °C and the measured eutectic is −21.1 °C, 1.77 K lower; past that point the brine is saturated, and salt water can go no colder. The two bottom rows of the table are there to show the law being extrapolated past sense: 10 osmol/kg would claim 247.2 bar of osmotic pressure, a figure no real solution reaches in the form written here.

One kelvin of frost is thirteen bar of pressure 🖖

Press Sugar water. It freezes 1.860 K low and holds back 24.72 bar, and the ratio of the two is 13.29 bar per kelvin. Change the sugar for salt, urea or a protein and the ratio does not move, because both effects count particles and ignore what the particles are. That is why the chart can draw them as one line on two axes. Seawater freezes at about −1.9 °C, so the same arithmetic says it holds back about 25 bar at 25 °C, which is the pressure a reverse-osmosis plant has to beat before it gets a single drop of fresh water. Jacobus van 't Hoff wrote the law down in 1887 and won the first Nobel Prize in Chemistry partly for it; the equation for how K moves with temperature is his other law.

K_f is the latent heat in disguise 🖖

The freezing-point constant is not a separate fact to memorise. It is R T_f² M / ΔH_fus, and from water's 6.01 kJ/mol that is 1.8595 K·kg/mol; the boiling constant from its 40.65 kJ/mol of vaporisation is 0.5131. So a freezing point is 3.62 times as sensitive as a boiling point, and the reason is the latent heats: melting ice takes almost seven times less energy than boiling water away, which makes the solid–liquid balance far easier to push. A solvent that is easy to melt is easy to depress. Camphor, with a small heat of fusion, has a K_f near 40, which is why it was the classic solvent for weighing organic molecules by their freezing point.

A ruler can weigh a protein that a thermometer cannot see 🖖

Press A protein. One gram of a 100 kDa protein in 100 g of water moves the freezing point by 1.86×10⁻⁴ K, a fifth of a thousandth of a degree. The same solution, behind a membrane that lets water through and stops the protein, holds up a column of water 25.28 mm high. The osmotic pressure is 13.29 bar per kelvin at every concentration, but a millimetre of water is easy to read and a millikelvin is not. That is why molar masses of large molecules were measured by osmometry, and why the worked problem below runs the page backwards.

Problem solved in full

Weighing a protein with a water column 4 steps
You dissolve 1.00 g of an unknown protein in 100 g of water at 25 °C. Behind a membrane that passes water and stops protein, the solution rises until the column stands 25.3 mm above the pure water. Water's density at 25 °C is 997.05 kg/m³. Find the molar mass, then say how far the same solution's freezing point would move.

Open this problem in the tool
1. The column's weight is the osmotic pressure: 997.05 kg/m³ × 9.80665 m/s² × 0.0253 m, which is 247.4 Pa. About a four-hundredth of an atmosphere.
  
  $Π = ρ g h = 997.05 \times 9.80665 \times 0.0253 = 247.4 Pa$
2. Divide by RT to get the concentration: 0.09979 mol per cubic metre of solution.
  
  $c = \frac{Π}{R T} = \frac{247 . 4}{8 . 3145 \times 298 . 15} = 0.09979 mol/ m^{3}$
3. Per kilogram of water, divide by the density. That is 1.0009×10⁻⁴ mol/kg, and in 100 g of water, 1.0009×10⁻⁵ mol of protein.
  
  $m = \frac{0 . 09979}{997 . 05} = 1.0009 \times 1 0^{- 4} mol/kg \Rightarrow n = 1.0009 \times 1 0^{- 5} mol$
4. 1.00 g over 1.0009×10⁻⁵ mol gives the molar mass.
  
  $M = \frac{1 . 00}{1 . 0009 \times 1 0 ^{- 5}} = 99, 914 g/mol$
Answer

About 99,900 g/mol, a 100 kDa protein. The freezing point of the same solution would fall by 1.86×10⁻⁴ K. Both measurements carry the same information, since each is the other multiplied by 13.29 bar per kelvin, but one is a column you can read with a ruler and the other is a fifth of a millikelvin. A protein's counter-ions add particles and make the column read high, which is why osmometry is done in a salt buffer, so that small ions are present on both sides of the membrane and cancel.

$M \approx 99, 900 g/mol Δ T_{f} = 1.8595 \times 1.0009 \times 1 0^{- 4} = 1.86 \times 1 0^{- 4} K$

Learning path

Mixtures and solutions

References (3)

The law the page is built on, and the paper that gave osmotic pressure its gas-law form: J. H. van 't Hoff, "Die Rolle des osmotischen Druckes in der Analogie zwischen Lösungen und Gasen." Zeitschrift für Physikalische Chemie 1, 1887, pp. 481-508. DOI 10.1515/zpch-1887-0151
Water's density at each temperature, and the latent heats of fusion and vaporisation that K_f and K_b are computed from on this page: W. Wagner and A. Pruß, "The IAPWS Formulation 1995 for the Thermodynamic Properties of Ordinary Water Substance for General and Scientific Use." Journal of Physical and Chemical Reference Data 31(2), 2002, pp. 387-535. DOI 10.1063/1.1461829
The salt-water eutectic the lesson's breaks measures the ideal line against, −21.1 °C at 23.3 wt% NaCl: V. A. Drebushchak, A. G. Ogienko and A. S. Yunoshev, "Metastable eutectic melting in the NaCl-H2O system." Thermochimica Acta 647, 2017, pp. 94-100. DOI 10.1016/j.tca.2016.12.004

By the 42Math team · Reviewed against published sources. Corrections welcome.

Colligative Properties: Freezing, Boiling & Osmosis

What each symbol means

One kelvin of frost is thirteen bar of pressure 🖖

Kf is the latent heat in disguise 🖖

A ruler can weigh a protein that a thermometer cannot see 🖖

K_f is the latent heat in disguise 🖖