Colligative Properties

Properties of pure liquids change when a solution is formed. Some of these properties are referred to as follows:

• Vapor Pressure Lowering
• Freezing Point Depression
• Boiling Point Elevation
• Osmotic Pressure

The greater the concentration of the solute(s) in the solution, the larger the change in these properties. These properties are known as colligative properties meaning the change in property of the solution does not depend upon the identity of the solute, it only depends upon the number of solute particles.

Vapor Pressure Lowering

The vapor pressure of a solvent is lower than that of the pure liquid. The higher the concentration of the solution, the more the vapor pressure of the solvent is lowered. This relationship can be given as

$P_{\mathrm{A}} = \chi_{\mathrm{A}} P^{\circ}_{\mathrm{A}}$

where χA is the mole fraction of the solute, P°A is the vapor pressure of the pure liquid, and PA is the vapor pressure of the solvent.

Freezing Point Depression

The freezing point of a solution is lower than that of the pure liquid. The higher the concentration of the solution, the more the freezing point is lowered. This relationship can be given as

$\Delta T_{\mathrm{f}} = i m K_{\mathrm{f}}$

where i is the van’t Hoff factor of the solute, m is the concentration of the solution, Kf is the molal freezing point depression constant for the solvent (or cryoscopic constant), and ΔTf is the change in the freezing point from the pure solvent.

Boiling Point Elevation

The boiling point of a solution is higher than that of the pure liquid. The higher the concentration of the solution, the more the boiling point is raised This relationship can be given as

$\Delta T_{\mathrm{b}} = i m K_{\mathrm{b}}$

where i is the van’t Hoff factor of the solute, m is the concentration of the solution, Kb is the molal boiling point elevation constant (or ebullioscopic constant) for the solvent, and ΔTb is the change in the boiling point from the pure solvent.

Osmotic Pressure

A solution and pure solvent are initially separated by an osmotic membrane. Solvent molecules move to higher concentration (osmosis). The osmotic pressure is defined as the external pressure required to give zero net movement of solvent across the membrane.

The osmotic pressure of the solution is higher than that of the pure liquid. The higher the concentration of the solution, the higher the osmotic pressure it exhibits. The relationship can be given as

$\Pi = iMRT$

where i is the van’t Hoff factor of the solute, M is the concentration of the solution, R is the gas constant, T is the temperature of the solution, and Π is the osmotic pressure of the solution.

Determination of constants for solvents

The molal freezing (Kf) and boiling (Kb) point constants for a solvent can be determined experimentally and are reported as positive values.

Kf for benzene

The normal freezing point of benzene is 5.49 °C.1 Below is a table2 of solutes dissolved in benzene (1 g solute / 100 g solvent concentration). All listed solutes are neutral and do not dissociate in solution (i.e. they are non-electrolytes; i = 1). The change in the freezing point of benzene is recorded. Using the freezing-point depression equation, one can calculate Kf (given in °C m–1).

Solute Formula m.w. (g/mol) ΔT (°C) m Kf
methyl iodide CH3I 142 –0.335 0.070 4.76
chloroform CHCl3 119.5 –0.428 0.084 5.11
carbon tetrachloride CCl4 154 –0.333 0.065 5.13
carbon disulfide CS2 76 –0.654 0.132 4.97
ethyl iodide C2H5I 156 –0.331 0.064 5.16
ethyl bromide C2H5Br 109 –0.461 0.092 5.02
hexane C6H14 86 –0.597 0.116 5.13
ethylene chloride C2H5Cl 99 –0.491 0.101 4.86
turpentine (a-pinene) C10H16 136 –0.366 0.074 4.98
nitrobenzene C6H5NO2 123 –0.390 0.081 4.80
naphthalene C10H8 128 –0.391 0.078 5.00
anthracene C14H10 178 –0.287 0.056 5.11
methyl nitrate CH3NO3 77 –0.640 0.130 4.93
dimethyl oxalate C4H6O4 118 –0.417 0.085 4.92
methyl salicylate C8H8O3 152 –0.339 0.066 5.15
diethyl ether (C2H5)2O 74 –0.671 0.135 4.97
diethyl sulfide C4H10S 90 –0.576 0.111 5.18
ethyl cyanide C3H5N 55 –0.938 0.182 5.16
ethyl formate C3H6O2 74 –0.666 0.135 4.93
ethyl valerate C7H14O2 130 –0.384 0.077 4.99
allyl thiocyanate C4H5NS 99 –0.519 0.101 5.14
nitroglycerine C3H5N3O9 227 –0.220 0.044 4.99
tributyrin C15H26O6 302 –0.161 0.033 4.86
triolein C57H104O6 884 –0.056 0.011 4.95
acetaldehyde C2H4O 44 –1.107 0.227 4.87
chloral C2HCl3O 147.5 –0.342 0.068 5.04
benzaldehyde C7H6O 106 –0.473 0.094 5.01
camphor C10H16O 152 –0.338 0.066 5.14
acetone C3H6O 58 –0.850 0.172 4.93
5-Nonanone C9H18O 142 –0.359 0.070 5.10
Mean 5.01
Min 4.76
Max 5.18
Std. Dev 0.12

Calculating Kf

Since

$\Delta T = imK_{\mathrm{f}}$

$K_{\mathrm{f}} = \dfrac{\Delta T}{im}$

For the species below, i = 1. The change in a freezing point (ΔT ) can be measured. Therefore, to obtain Kf, simply punch in the concentration (m) of the solution and its change in freezing point.

Example: Determine the Kf for methyl iodide using the experimental data given above

\begin{align*} K_{\mathrm{f}} &= \dfrac{\Delta T}{im}\\[1.5ex] &= \dfrac{0.335~\mathrm{^{\circ}C}}{1\times 0.070~m}\\[1.5ex] &= 4.76~\mathrm{^{\circ}C~}m^{-1} \end{align*}

1 g / 100 g and molality?

The freezing point temperature changes in the table above were determined with solutions created with general solubility concentrations

$\dfrac{1~\mathrm{g~solute}}{100~\mathrm{g~solvent}}$

This can be expressed as

$\dfrac{n_{\mathrm{solute}}}{\mathrm{100~g~solvent}}$ One gram of methyl iodide, for example, is

$n_{\mathrm{methyl~iodide}} = 1.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{142~\mathrm{g}} \right ) = 0.007042$

Therefore, a solution of methyl iodide with a 1 g / 100 g benzene concentration would be

$\dfrac{0.007042}{100~\mathrm{g~benzene}} = 0.070~m$

However, molality concentration unit is defined as

$\dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}}$ If we scale up our solution from 100 g to 1 kg, we get

$\dfrac{0.007042}{100~\mathrm{g~benzene}} \left ( \dfrac{1000~\mathrm{g}}{1~\mathrm{kg}} \right) = 0.070~m$ which is the same concentration determined with our 1 g / 100 g solution.

It is interesting to note that Kf shows little variation across a large range of solutes. There is excellent agreement between the concentration of the solution and its change in freezing point. Note that each “point” on the plot represents a unique solute (at 1 g solute / 100 g benzene concentration).

The experimental Kf for benzene is 5.07 °C m–1.

François-Marie Raoult’s findings support that the freezing point depression of benzene had nothing to do with the identity of the solute, only the number of solute particles dissolved in solution. Raoult concludes:

“It is therefore permissible to say, from now on: In a multitude of cases, the lowering of the freezing point of a dissolving agent depends only on the ratio between the number of molecules of the dissolved body and of the solvent; it is independent of the nature, the number and the arrangement of the atoms which make up the dissolved molecules.”

Kf for water

Below is a table of freezing point temperatures (in °C) for 0.1 m aqueous solutions as reported by Linus Pauling.3

Solute Formula ΔT (°C)
Hydrogen peroxide H2O2 –0.186
Methanol CH3OH –0.181
Ethanol C2H5OH –0.183
Dextrose C6H12O6 –0.186
Sucrose C12H22O11 –0.188

Similar with Raoult’s findings, the change in the freezing point (ΔT ) shows little variation across a range of solutes. The molal freezing point depression constant for water is 1.86 °C m–1.

Constants

Molal freezing point depression (Kf) and boiling point elevation (Kb) constants (in oC m–1) for some solvents. Many can be found in the CRC Handbook.4

 Solvent Kf Kb Water, H2O 1.86 0.513 Ethanol, CH3CH2OH 1.99 1.22 Chloroform, CHCl3 4.68 3.63 Benzene, C6H6 5.07 2.64 Carbon tetrachloride, CCl4 29.8 5.02

CAUTION: Some of the constants given in the textbook are slightly different than those reported in the CRC Handbook.

When comparing the values of Kf and Kb, Kf is generally larger. This means that the presence of solutes tends to effect the freezing point of a solution more than its boiling point.

Practice Problems

Problem 1

122 g of sugar, sucrose (C12H22O11; m.w. = 342.3 g mol–1) is added to 350 g of water at 25 °C. What are the freezing and boiling points (in °C) of the solution?

• Kf = 1.86 °C m–1
• Kb = 0.513 °C m–1
Solution

\begin{align*} n_{\mathrm{solute}} &= 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356 \\[2ex] m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} = \dfrac{0.356~\mathrm{mol}}{0.350~\mathrm{kg}} = 1.017 \\[2ex] \Delta T_{\mathrm{f}} &= iK_{\mathrm{f}}m = \left ( 1 \right ) \left ( 1.86~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 1.017~m \right ) = 1.89~^{\circ}\mathrm{C} \\[2ex] \Delta T_{\mathrm{b}} &= iK_{\mathrm{b}}m = \left ( 1 \right ) \left ( 0.513~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 1.017~m \right ) = 0.522~^{\circ}\mathrm{C} \\[2ex] T_{\mathrm{f}} &= 0~^{\circ}\mathrm{C} - 1.89~^{\circ}\mathrm{C} = -1.89~^{\circ}\mathrm{C} \\ T_{\mathrm{b}} &= 100~^{\circ}\mathrm{C} + 0.522~^{\circ}\mathrm{C} = 100.5~^{\circ}\mathrm{C} \end{align*}

Problem 2

122 g of NaCl (m.w. = 58.44 g mol–1) is added to 350 g of water at 25 °C. What are the freezing and boiling points (in °C) of the solution? Kf = 1.86 °C m–1; Kb = 0.513 °C m–1

Solution

\begin{align*} n_{\mathrm{solute}} &= 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{58.44~\mathrm{g}} \right ) = 2.088 \\[2ex] m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} = \dfrac{2.088~\mathrm{mol}}{0.350~\mathrm{kg}} = 5.966 \\[2ex] \Delta T_{\mathrm{f}} &= iK_{\mathrm{f}}m = \left ( 2 \right ) \left ( 1.86~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 5.966~m \right ) = 22.2~^{\circ}\mathrm{C} \\[2ex] \Delta T_{\mathrm{b}} &= iK_{\mathrm{b}}m = \left ( 2 \right ) \left ( 0.513~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 5.966~m \right ) = 6.12~^{\circ}\mathrm{C} \\[2ex] T_{\mathrm{f}} &= 0~^{\circ}\mathrm{C} - 22.2~^{\circ}\mathrm{C} = -22.2~^{\circ}\mathrm{C} \\ T_{\mathrm{b}} &= 100~^{\circ}\mathrm{C} + 6.11~^{\circ}\mathrm{C} = 106.1~^{\circ}\mathrm{C} \end{align*}

Problem 3

122 g of sugar, sucrose (C12H22O11; m.w. = 342.3 g mol–1) is added to 350. g of water at 25 °C. What is the osmotic pressure (in atm) of this solution? dsucrose = 1.59 g cm–3; dwater = 1.0 g cm–3

Solution

\begin{align*} n_{\mathrm{solute}} &= 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356~\mathrm{mol} \\[1ex] V_{\mathrm{solute}} &= \dfrac{m_{\mathrm{solute}}}{d_{\mathrm{solute}}} = \dfrac{122~\mathrm{g}}{1.59~\mathrm{g~cm^{-3}}} = 76.73~\mathrm{mL} = 0.0767~\mathrm{L} \\[1ex] V_{\mathrm{solvent}} &= \dfrac{m_{\mathrm{solvent}}}{d_{\mathrm{solvent}}} = \dfrac{350~\mathrm{g}}{1.0~\mathrm{g~cm^{-3}}} = 350.0~\mathrm{mL} = 0.350~\mathrm{L} \\[1ex] V_{\mathrm{tot}} &= 0.0767~\mathrm{L} + 0.350~\mathrm{L} = 0.4267~\mathrm{L} \\[2ex] M &= \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}} = \dfrac{0.356~\mathrm{mol}}{0.4267~\mathrm{L}} = 0.834 \\[2ex] \pi &= iMRT \\ &= \left ( 1 \right ) \left ( 0.834~M \right ) \left ( 0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}} \right ) \left ( 298.15~\mathrm{K} \right ) \\ &= 20.6~\mathrm{atm} \end{align*}

Problem 4

122 g of NaCl (m.w. = 58.44 g mol–1) is added to 350.0 g of water at 25 °C. What is the osmotic pressure (in atm) of this solution? dNaCl = 2.16 g cm–3; dwater = 1.0 g cm–3

Solution

\begin{align*} n_{\mathrm{solute}} &= 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{58.4~\mathrm{g}} \right ) = 2.089~\mathrm{mol} \\[1ex] V_{\mathrm{solute}} &= \dfrac{m_{\mathrm{solute}}}{d_{\mathrm{solute}}} = \dfrac{122~\mathrm{g}}{2.16~\mathrm{g~cm^{-3}}} = 56.48~\mathrm{mL} = 0.0565~\mathrm{L} \\[1ex] V_{\mathrm{solvent}} &= \dfrac{m_{\mathrm{solvent}}}{d_{\mathrm{solvent}}} = \dfrac{350~\mathrm{g}}{1.0~\mathrm{g~cm^{-3}}} = 350.0~\mathrm{mL} = 0.350~\mathrm{L} \\[1ex] V_{\mathrm{tot}} &= 0.0565~\mathrm{L} + 0.350~\mathrm{L} = 0.4065~\mathrm{L} \\[2ex] M &= \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}} = \dfrac{2.089~\mathrm{mol}}{0.4065~\mathrm{L}} = 5.14 \\[2ex] \pi &= iMRT \\ &= \left ( 2 \right ) \left ( 5.14~M \right ) \left ( 0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}} \right ) \left ( 298.15~\mathrm{K} \right ) \\ &= 251.5~\mathrm{atm} \end{align*}

Problem 5

A 350 g sample of a nonelectrolyte is dissolved in 1.5. kg of water and the solution was found to boil at 104 °C. What is the molar mass of the compound? Kb = 0.513 °C m–1

Solution

\begin{align*} \Delta T_{\mathrm{b}} &= iK_{\mathrm{b}}m \longrightarrow \\[1ex] m &= \dfrac{\Delta T_{\mathrm{b}}}{iK_{\mathrm{b}}} = \dfrac{4~^{\circ}\mathrm{C}}{(1)\left (0.513~^{\circ}\mathrm{C}~m^{-1}\right )} = 7.7973~m \\[2ex] m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} \longrightarrow \\ n_{\mathrm{solute}} &= m_{\mathrm{solution}} \times \mathrm{kg~solvent} = \left ( 7.7973~m \right ) \left ( 1.50~\mathrm{kg~solvent} \right ) = 11.70~\mathrm{mol} \\[2ex] \mathrm{molar~mass} &= \dfrac{350~\mathrm{g}}{11.72~\mathrm{mol}} = 29.91~\mathrm{g~mol^{-1}} \end{align*}

Problem 6

Which aqueous solution is expected to have the highest boiling point?

1. 0.50 m C12H22O11
2. 0.50 m NaCl
3. 0.50 m MgCl2
4. cannot determine with the given information
Solution

$\Delta T_{\mathrm{b}} = iK_{\mathrm{b}}m$

m and Kb is the same for all solutions. Look for the largest i. Answer: 3

Problem 7

Which aqueous solution is expected to have the lowest freezing point?

1. 0.10 m NaCl
2. 0.50 m NaCl
3. 0.75 m NaCl
4. cannot determine with the given information
Solution

$\Delta T_{\mathrm{f}} = iK_{\mathrm{f}}m$

i and Kf same for all solutions. Look for the largest m. Answer: 3

Problem 8

A 5.00 L sample of water was found to contain 12 ppm of mercury. What is the mass of the mercury (in mg) in this sample of water? dsoln. = 1.00 g mL–1

Solution

\begin{align*} \mathrm{mass~solution} &= ( 5.00~\mathrm{L} ) \left ( \dfrac{10^3~\mathrm{mL}}{1~\mathrm{L}} \right ) \left ( 1.0~\mathrm{g~mL^{-1}} \right ) = 5000~\mathrm{g} \\ \mathrm{ppm} &= \dfrac{\mathrm{mass~solute}}{\mathrm{mass~solution}} \times 10^{6} \longrightarrow \\[2ex] \mathrm{mass~solute} &= \dfrac{\mathrm{ppm}\times\mathrm{mass~solution}}{\mathrm{10^6}}\\[2ex] &= \dfrac{12~\mathrm{ppm}\times 5000~\mathrm{g~solution}}{10^6} = 0.06~\mathrm{g} = 60~\mathrm{mg} \end{align*}

Problem 9

122 g of HF is added to 350 g of water at 25 °C. About 8% of the HF is dissociated. What is the freezing and boiling points (in °C) of the solution? Kf = 1.86 °C m–1; Kb = 0.513 °C m–1

Solution

\begin{align*} n_{\mathrm{solute}} &= 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{20.01~\mathrm{g}} \right ) = 6.097 \\[1ex] m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} = \dfrac{6.097~\mathrm{mol}}{0.350~\mathrm{kg}} = 17.42 \\[1ex] i &= 1.08 \\[2ex] \Delta T_{\mathrm{f}} &= iK_{\mathrm{f}}m = \left ( 1.08 \right ) \left ( 1.86~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 17.42~m \right ) = 34.99~^{\circ}\mathrm{C} \\[2ex] \Delta T_{\mathrm{b}} &= iK_{\mathrm{b}}m = \left ( 1.08 \right ) \left ( 0.513~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 17.42~m \right ) = 9.632~^{\circ}\mathrm{C} \\[2ex] T_{\mathrm{f}} &= 0~^{\circ}\mathrm{C} - 34.99~^{\circ}\mathrm{C} = -35.0~^{\circ}\mathrm{C} \\ T_{\mathrm{b}} &= 100~^{\circ}\mathrm{C} + 9.632~^{\circ}\mathrm{C} = 110.~^{\circ}\mathrm{C} \end{align*}

Timed Assessments

VIDEO

1. –1.89 °C
2. –15.91 °C
3. –2.49 °C

Boiling Point Elevation

VIDEO

NOTE: The boiling points were determined using Tb = 0.512 °C for water.

1. 100.5 °C
2. 104.4 °C
3. 101.4 °C

Osmotic Pressure

VIDEO

1. 20.4 atm
2. 187.6 atm
3. 63.6 atm

1. François-Marie Raoult, “Law of freezing of benzene solutions of neutral substances.”, Comptes Rendus 95, 187-189 (1882); https://gallica.bnf.fr/ark:/12148/bpt6k30518/f188.item↩︎

2. General Chemistry, 3rd ed., 459, (1970)↩︎

3. Lide et. al, CRC Handbook of Chemistry and Physics, 15-28, 15-27 (2008)↩︎

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