Colligative Properties: Colligative property is an important concept in Chemistry. The word “colligative” has been adapted from the Latin word “colligatus” which translates to “bound together”. Colligative properties help us understand how the properties of the solution are linked to the concentration of solute in the solution.

In chemistry, colligative properties are those properties of solutions that depend on the ratio of the number of solute particles to the number of solvent particles in a solution and not on the chemical species. There are four colligative properties which are boiling point elevation,zing point depression, relative lowering of vapour pressure, and osmotic pressure. Let us see in detail what colligative properties are. 

What Are Colligative Properties of Solutions?

Colligative properties are mostly seen in dilute solutions. Dilute solutions which contain non-volatile solute, exhibit some properties. These properties which depend only on the number of solute particles present instead of the type of solute present are called colligative properties.

Colligative properties are obtained by the dissolution of a non-volatile solute in a volatile solvent. Generally, the solute changes the solvent properties where its particles remove some of the solvent molecules in the liquid phase. This also results in the reduction of the concentration of the solvent.

Colligative properties are inversely proportional to the solute molar mass. Dilute solution of non-volatile solutes exhibit a particular set of properties which are related to the number of solute and solvent particles present in the solution. These do not depend upon the nature of the solute. Thus, a colligative property may be defined in the following manner:

The properties of dilute solutions of non-volatile solutes that depends upon the concentration of solute particles in the solution but not on the chemical nature of solute are called colligative properties.

What Are the 4 Colligative Properties?

The various colligative properties are:

1. Relative lowering of vapour pressure
2. Elevation of boiling point
3. Depression inzing point
4. Osmotic pressure

Types of Colligative Properties

Let us now understand each type of the colligative property in detail:

Relative Lowering of Vapour Pressure

The vapour pressure of a volatile solvent gets lowered when a non-volatile solute is dissolved in it. If \({{\rm{p}}^{\rm{^\circ }}}\) represents then the vapour pressure of a pure solvent and \({\rm{p}}\) represents the vapour pressure of the solution, we have-

Lowering of pressure \({\rm{ = }}{{\rm{p}}^{\rm{^\circ }}}{\rm{ – p}}\)
Relative lowering of vapour pressure \({\rm{ = }}\frac{{{{\rm{p}}^{\rm{^\circ }}}{\rm{ – p}}}}{{\rm{p}}}\)
The relative lowering of pressure and lowering of pressure are colligative property.

A relation between the pressure of the solution, the vapour pressure of the pure solvent and the mole fraction of the solute was discovered by a French chemist Raoult. As per Raoult’s law, which states that the lowering in vapour pressure of a dilute solution is equal to the mole fraction of the solute in the solution.

\(\frac{{{{\rm{p}}^{\rm{^\circ }}}{\rm{ – p}}}}{{\rm{p}}}{\rm{ = }}\frac{{\rm{n}}}{{{\rm{n + N}}}}\)

Here, \({\rm{n}}\) is the moles of the solute dissolved in \({\rm{N}}\) moles of the solvent.

Measuring Relative lowering of pressure diagram-

Elevation in Boiling Point

The boiling point of a liquid may be defined as a temperature at which its vapour pressure becomes equal to its atmospheric pressure.

Whenever a non-volatile solute is added to a pure solvent, its vapour pressure decreases, which further elevates its boiling point. This elevation happens while making the vapour pressure equal to the atmospheric pressure. To achieve this, we need to increase the temperature of the solution. The elevation in boiling point can be shown graphically-

The difference in the boiling point of the solution and the boiling point of the pure solvent is the expression for the elevation in boiling point.

If  \({{\rm{T}}_{\rm{b}}}{\rm{\;}}\) and \({\rm{T}}_{\rm{b}}^{\rm{^\circ }}\) are the boiling point of the solution and pure solvent, then
\({\rm{\Delta Tb = \;T}}{\rm{b}}^{\rm{^\circ }}{\rm{ – }}{{\rm{T}}{\rm{b}}}\) 

The magnitude of \({\rm{\Delta Tb}}\) depends on \({\rm{\Delta p}}{\rm{.}}\) Now, according to Raoult’s law-

\({\rm{\Delta p\;\alpha \;m}}\)

Therefore, \({\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{\;\alpha \;m}}\)

\({\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\,{\rm{ = }}\,{{\rm{k}}_{\rm{b}}}{\rm{m}}\)

\({\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\,{\rm{ = }}\,\frac{{{\rm{100\;}}{{\rm{k}}_{\rm{b}}}{{\rm{w}}_{\rm{2}}}}}{{{{\rm{w}}_{\rm{1}}}{{\rm{M}}_{\rm{2}}}}}\)

 where, \({{\rm{k}}_{\rm{b}}}\) is the molal elevation constant.

Depression inzing Point

Thezing point of a substance is when the solid and liquid phase has the same vapour pressure. Thezing point of a pure liquid is fixed, but when a non-volatile solute is dissolved in the pure liquid to constitute a solution, there occurs a lowering in thezing point.

We know that when a non-volatile solute is added to the pure solvent, its vapour pressure decreases, and this lowered vapour pressure becomes equal to that of the solid solvent at a lower temperature.

The difference in thezing point of the solution and thezing point of the pure solvent is the expression for the depression in thezing point.

If \({{\rm{T}}_{\rm{f}}}\) and \({{\rm{T}}_{\rm{f}}}^\circ\) are thezing points of the solution and pure solvent, then \(\Delta {{\rm{T}}_{\rm{f}}}{\rm{ = }}\;{{\rm{T}}_{\rm{f}}}^\circ {\rm{ – }}{{\rm{T}}_{\rm{f}}}\)

The magnitude of \({\rm{\Delta Tf}}\) depends on \({\rm{\Delta p}}{\rm{.}}\) Now, according to Raoult’s law-

\({\rm{\Delta p\;\alpha \;m}}\)

Therefore, \({\rm{\Delta }}{{\rm{T}}_{\rm{f}}}{\rm{\;\alpha \;m}}\)

\(\Delta {{\rm{T}}_{\rm{f}}}\,{\rm{ = }}\,{{\rm{k}}_{\rm{f}}}{\rm{m}}\)

\(\Delta {{\rm{T}}_{\rm{f}}}\,{\rm{ = }}\,\frac{{{\rm{100}}\;{{\rm{k}}_{\rm{f}}}\;{{\rm{w}}_{\rm{2}}}}}{{{{\rm{w}}_{\rm{1}}}{{\rm{M}}_{\rm{2}}}}}\)

where, \({{\rm{k}}_{\rm{f}}}\) is the molal depression constant.

Osmotic Pressure

The passage of solvent from the pure solvent or from the solution of lower concentrations into the solution of higher concentrations through a semi-permeable membrane is called Osmosis.

The flow of solvent through the semi-permeable membrane can be stopped by applying extra pressure on the solution. This pressure is known as the osmotic pressure of the solution.

Thus, the osmotic pressure of the solution at a given temperature may be defined as the excess hydrostatic pressure that builds up when the solution is separated from the solvent by a semi-permeable membrane. It is denoted by \({\rm{\pi }}\)

OR

The osmotic pressure may also be defined as the excess pressure that must be applied to the solution side to prevent the passage of solvent through a semi-permeable membrane.

The expression for osmotic pressure is-

\({\rm{\pi }}\,{\rm{ = }}\,{\rm{CRT}}\) where \({\rm{R}}\) is the gas constant.

\({\rm{\pi = }}\frac{{{{\rm{n}}_{\rm{2}}}}}{{\rm{V}}}{\rm{RT}}\)

Here, \({\rm{V}}\) is the volume of solution in litres and \({{\rm{n}}_{\rm{2}}}\) are moles of solute

If \({{\rm{m}}_{\rm{2}}}{\rm{\;}}\) is the weight of solute and \({{\rm{M}}_{\rm{2}}}{\rm{\;}}\) molar mass of solute, then \({{\rm{n}}_2}{\rm{ = }}\frac{{{{\rm{m}}_2}}}{{{{\rm{M}}_{\rm{2}}}}}\)

Then, \({\rm{\pi }}\,{\rm{ = }}\,\frac{{{{\rm{m}}_2}}}{{{{\rm{M}}_2}{\rm{V}}}}{\rm{RT}}\)

Reverse Osmosis

If the pressure on the solution is greater than the osmotic pressure, then the solvent starts passing from the solution to the solvent. This phenomenon is generally used for the purification of seawater and hard water.

Isotonic, Hypotonic, and Hypertonic Solutions

1. Isotonic solutions are solutions that have equal osmotic pressure at a given temperature.
2. Hypotonic solutions are those that have a lower osmotic pressure than other solutions.
3. Hypertonic solutions are those that have a higher osmotic pressure than other solutions.

Importance of Osmosis

1. Shrivelling of raw mangoes to pickle.
2. Revival of wilted flowers and limped carrots- When wilted flowers and limped carrots are put in water, they regain their freshness by the process of osmosis.
3. Preservation of meat- This is done by salting the meat.
4. Absorption of water by plants- Plants absorb water from the soil by the process of osmosis.

Van’t Hoff Factor

It is the ratio between the actual concentration of the particles produced when the substance is dissolved and the concentration of the substance as calculated from its mass. It is denoted by \({\rm{i}}{\rm{.}}\)

Van’t Hoff Factor represents the extent to which a solute associates or dissociates in a solution.

Practical Applications of Colligative Properties: Colligative Properties Examples

1. Antifreeze solutions: To avoid the difficulty of waterzing in radiators, a solution of ethylene glycol with water is used in radiators. This solution has azing point lesser than zero.
2. Clearing of ice from the roads on hills: Salts such as \({\rm{NaCl}}\) and \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) are scattered on the icy roads, which helps in the melting of the ice.

Solved Questions on Colligative Properties

Q.1: Calculate the osmotic pressure at \(273\;{\rm{K}}\) of a \(5\,\% \) solution of urea. \(\left( {{\rm{Mol}}.\,{\rm{Mass}}\,{\rm{ = }}\,{\rm{60}}} \right){\mkern 1mu} \left( {{\rm{R}}\,{\rm{ = }}\,{\rm{0}}.{\rm{021}}\,{\rm{L}}\,{\rm{atm}}\,{{\rm{K}}^{ – 1}}\,{\rm{mole}}} \right).\)
Ans: \(5\,\% \) solution of urea means that it contains \({\rm{5}}\,{\rm{g}}\) of urea per \({\rm{100}}\,{\rm{c}}{{\rm{m}}^{ – 3}}\) of the solution, i.e.,
\({{\rm{w}}_{\rm{2}}}\,{\rm{ = }}\,{\rm{5}}\,{\rm{g}},\;{\rm{V}}\,{\rm{ = }}\,{\rm{100}}\;{\rm{c}}{{\rm{m}}^{{\rm{ – 3}}}}\,{\rm{ = }}\,\frac{{\rm{1}}}{{{\rm{10}}}}\,{\rm{litre}}\,{\rm{ = }}\,{\rm{0}}.1\,{\rm{litre}}.\)
\({{\rm{M}}_{\rm{2}}}{\rm{ = 60}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}{\rm{,R = 0}}{\rm{.0821}}\;{\rm{L}}\;{\rm{atm}}\;{{\rm{K}}^{{\rm{ – 1}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}{\rm{,}}\;{\rm{T = 273}}\;{\rm{K}}\)
\({\rm{\pi }}\,{\rm{ = }}\,\frac{{{{\rm{w}}_2}{\rm{RT}}}}{{{{\rm{M}}_2}{\rm{V}}}}{\rm{ = }}\frac{{{\rm{5}}\;{\rm{g}}\; \times {\rm{0}}.{\rm{0821}}{\mkern 1mu} {\rm{L}}\;{\rm{atm}}\;{{\rm{K}}^{{\rm{ – 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}} \times {\rm{273}}\;{\rm{K}}\;}}{{{\rm{60}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}} \times {\rm{0}}.{\rm{1}}\;{\rm{L}}}}{\rm{ = 18}}.{\rm{68}}\;{\rm{atm}}.\)
The osmotic pressure is \({\rm{18}}{\rm{.68}}{\mkern 1mu} \,{\rm{atm}}{\rm{.}}\)

Q.2: Calculate the boiling point of a solution containing \(0.456\;{\rm{g}}\) of camphor (mol. mass\( = 152\)) dissolves in \(31.4\,{\rm{g}}\) of acetone \(\left( {{\rm{b}}{\rm{.p = 56}}{\rm{.3}}{{\rm{0}}^{\rm{o}}}{\rm{C}}} \right){\rm{,}}\) if the molecular elevation constant per \(100\,{\rm{g}}\) of acetone is \({{{17.2}^{\rm{o}}}{\rm{C}}}\).
Ans: Here, \({{\rm{w}}_2}{\rm{ = }}\,{\rm{0}}.{\rm{456}}\;{\rm{g}},{{\rm{M}}_2}{\rm{ = 152}},{{\rm{w}}_1}{\rm{ = 31}}.{\rm{4}}\;{\rm{g}},{{\rm{T}}_0}{\rm{ = 56}}.{\rm{30}}{\,^{\rm{o}}}{\rm{C}},{{\rm{k}}_{\rm{b}}}{\rm{ = 17}}.{\rm{2}}{\,^{\rm{o}}}{\rm{C/100}}\;{\rm{g}}.\)
\({\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\,{\rm{ = }}\,{{\rm{k}}_{\rm{b}}}{\rm{m}}\)
\({\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\,{\rm{ = }}\,\frac{{{\rm{100}}{{\rm{k}}_{\rm{b}}}{{\rm{w}}_{\rm{2}}}}}{{{{\rm{w}}_{\rm{1}}}{{\rm{M}}_{\rm{2}}}}}\,{\rm{ = }}\,\frac{{{\rm{100 \times 17}}{\rm{.2 \times 0}}{\rm{.456}}}}{{{\rm{31}}{\rm{.4 \times 152}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.16}}\,^\circ {\rm{C}}\)
Therefore, the Boiling point of the solution \(\left( {{{\rm{T}}_{\rm{b}}}} \right)\,{\rm{ = }}\,{{\rm{T}}_{\rm{b}}}{\rm{^\circ  + \Delta }}{{\rm{T}}_{\rm{b}}}\,{\rm{ = }}\,{\rm{56}}{\rm{.30 + 0}}{\rm{.16}}\,{\rm{ = }}\,{\rm{56}}{\rm{.46}}\,{\rm{^\circ C}}.\)

Q.3: \(1\;{\rm{g}}\) of non-electrolyte solute dissolves in \(50\;{\rm{g}}\) of benzene lowered thezing point of benzene by \(0.40\;{\rm{K}}.\) Thezing point depression constant of benzene is \(5.12{\mkern 1mu} {\rm{k}}\,{\rm{kg}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}.\) Find the molar mass of the solute.
Ans: Here, we are given \({{\rm{w}}_2}{\rm{ = }}\,{\rm{1}}\,{\rm{g}},\,{{\rm{w}}_1}\,{\rm{ = }}\,{\rm{50}}\,{\rm{g}},\,\Delta {{\rm{T}}_{\rm{f}}}\,{\rm{ = }}\,{\rm{0}}.{\rm{40K}},\)
\({{\rm{k}}_{\rm{f}}}{\rm{ = }}\,{\rm{5}}.{\rm{12}}\,{\rm{K}}\,{\rm{kg}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}.\)
Putting these values in the formula \({{\rm{M}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{1000}}{{\rm{k}}_{\rm{f}}}{{\rm{w}}_{\rm{2}}}}}{{{{\rm{w}}_{\rm{1}}}{\rm{\Delta }}{{\rm{T}}_{\rm{f}}}}}\)
\({{\rm{M}}_{\rm{2}}}\,{\rm{ = }}\,{\mkern 1mu} \frac{{{\rm{1000}} \times {\rm{5}}.{\rm{12}} \times {\rm{1}}}}{{{\rm{50}} \times {\rm{0}}.{\rm{40}}}}\,{\mkern 1mu} {\rm{ = }}{\mkern 1mu} \,{\rm{256}}\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}.\)

Q.4: The vapour pressure of a \({\rm{5}}\,{\rm{\% }}\) aqueous solution of a non-volatile organic substance at \({\rm{373}}\,{\rm{K}}\) is \({\rm{745}}\,{\rm{mm}}{\rm{.}}\) Calculate the molar mass of the solute.
Ans: \({\rm{5}}\,{\rm{\% }}\) aqueous solution of the solute implies that \(5\,{\rm{g}}\) of the solute are present in \(100\,{\rm{g}}\) of the solution.
i.e., Mass of the solute \(\left( {{{\rm{w}}_{\rm{2}}}} \right){\rm{ = 5}}\;{\rm{g}}\) Mass of the solution \( = 100\,{\rm{g}}\)
Therefore, Mass of the solvent \(\left( {{{\rm{w}}_{\rm{1}}}} \right){\rm{ = 100 – 5 = 95}}\;{\rm{g}}\)
Further, as the solution is aqueous, it means that the solvent is water, and we know that.
Vapour pressure of pure water at \({\rm{373}}\;{\rm{K}}\left( {{{\rm{p}}^{\rm{o}}}} \right){\rm{ = 760}}\;{\rm{mm}}\)
Vapour pressure of the solution at \({\rm{373}}\;{\rm{K}}\left( {{{\rm{p}}_{\rm{s}}}} \right){\rm{ = 745}}\;{\rm{mm}}\) (Given)
Molar mass of solvent, \({{\rm{M}}_{\rm{1}}}{\rm{ = 18\;g\;mo}}{{\rm{l}}^{{\rm{ – 1}}}}\)
The molar mass of solute, \({{\rm{M}}_2}{\rm{ = }}\) To be calculated
Using the formula for the dilute solution, \(\frac{{{{\rm{p}}^{\rm{o}}}{\rm{ – p}}}}{{\rm{p}}}{\rm{ = }}\frac{{\rm{n}}}{{{\rm{n + N}}}}\)
\(\frac{{{{\rm{p}}^{\rm{o}}}{\rm{ – p}}}}{{\rm{p}}}{\rm{ = }}\frac{{{{\rm{n}}_2}}}{{{{\rm{n}}_1}}}{\rm{ = }}\frac{{\frac{{{{\rm{w}}_2}}}{{{{\rm{M}}_2}}}}}{{\frac{{{{\rm{w}}_1}}}{{{{\rm{M}}_1}}}}}{\rm{ = }}\frac{{{{\rm{w}}_2}{{\rm{M}}_1}}}{{{{\rm{w}}_1}{{\rm{M}}_2}}}\)
we get, \(\frac{{{\rm{760 – 745}}}}{{{\rm{760}}}}{\rm{ = }}\frac{{{\rm{5}}\;{\rm{g \times 18}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}{{{\rm{95}}\;{\rm{g \times }}{{\rm{M}}_{\rm{2}}}}}\)
\({{\rm{M}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{5 \times 18 \times 760}}}}{{{\rm{15 \times 95}}}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}{\rm{ = 48}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\)
\({{\rm{M}}_{\rm{2}}}{\rm{ = 48}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\)

Summary

Through this article, we studied colligative properties and their types. We also studied the expressions for each of the colligative properties and some of the practical applications of depression ofzing point and osmotic pressure. 

FAQs on Colligative Properties of Water and Other Solutions

The frequently asked questions on Colligative properties are:

Q.1: What are the four colligative properties?
Ans: The four colligative properties are- Relative lowering of vapour pressure, the elevation of boiling point, depression inzing point, osmotic pressure.

Q.2: What is colligative property? Explain all colligative properties.
Ans: Dilute solution of non-volatile solutes exhibit a certain set of properties that are related to the number of solute and solvent particles present in the solution and do not depend upon the nature of the solute. This property is called colligative property.

Q.3: What does colligative property depend on?
Ans: Colligative properties depend only on the number of solute particles present and not on the type of solute present.

Q.4: What are the factors affecting the colligative properties?
Ans: The colligative property factor is molality, as the properties depend on the solute’s concentration

Q5. How are colligative properties used in everyday life?
Ans: Some uses of the colligative properties in everyday life are as follows:

• Antifreeze solutions- To avoid the difficulty of waterzing in radiators, a solution of ethylene glycol with water is used in radiators. This solution has azing point lesser than zero.
• Clearing of ice from the roads on hills- Salts such as \({\rm{NaCl}}\) and \({\rm{CaC}}{{\rm{l}}_2}\) are scattered on the icy roads, which helps in the melting of the ice.
• Shrivelling of raw mangoes to pickle.
• Revival of wilted flowers and limped carrots- When wilted flowers and limped carrots are put in water, they regain their freshness by the process of osmosis.
• Preservation of meat- This is done by salting the meat.
• Absorption of water by plants- Plants absorb water from the soil by the process of osmosis.