Raoult’s Law: Statement, Definition, Formula & Equation

Raoult’s Law: What’s the best way to create lemonade? Is it true that only solids dissolve in liquids? When we open a cold drink bottle, what is the fizzy sound we hear? With the right amount of sugar and salt in the water, we make lemonade. The solutes are the salt or sugar added to the water, which acts as a solvent. The dissolved carbon dioxide gas escapes with a fizzing sound when a cold drink bottle is opened.

We all know that the major component of a solution is the solvent (i.e. water, air), while the minor component of a solution is the solute (sugar, carbon dioxide, etc.). The nature of the solute added to the solvent plays a vital role in predicting the properties of a solution. The relationship between a solute, a solvent and the solution was given by a French Chemist named Francois Marte Raoult in the year 1886. Let’s learn more about it in this article.

Discovery of Raoult’s Law

Raoult’s Law was established in \(1887\). In \(1886,\) French chemist François-Marie Raoult discovered that when a solute is dissolved in a solvent, the vapour pressure of the resulting solution will generally decrease. His observation was based on two factors:

1. Mole fraction of the dissolved solute
2. Vapor pressure of the pure solvent

Based on his observations, Raoult made the following conclusions:

1. At a given temperature, there is a pressure at which the vapour formed above the resultant solution is in dynamic equilibrium with its liquid or solid form. This is the vapour pressure of the substance at that temperature.
2. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate at which the gas is condensing back to its original form.
3. All solids and liquids possess a particular vapour pressure, and this pressure is constant regardless of how much of the substance is present.

Raoult’s Law Formula

Francois Marte Raoult proposed a quantitative relationship between partial pressure and the mole fraction of volatile liquids. This relationship is known as Raoult’s Law which states that for any solution (obtained by dissolving a solute in a solvent) the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. It is expressed by the formula:

\({{\text{P}}_{{\text{solution}}}}{\text{=}}{{\text{X}}_{{\text{solvent}}}}\times{\text{P}}{{{^\circ }}_{{\text{solvent}}}}\)

Where solute is non volatile and,

\({\text{P}}_{\text{solution}}\) is the vapour pressure of the solution

\({\text{X}}_{\text{solvent}}\) is the mole fraction of the solvent

\({\text{P}}{{^\circ }}_{\text{solvent}}\) is the vapour pressure of the pure solvent

In a binary solution consisting of a volatile solute and a volatile solvent, the partial pressure of each component gets added to the total pressure.

\({{\text{P}}_{{\text{solution}}}}{\text{=}}\left({{{\text{X}}_{{\text{solvent}}}}\times\text{ P}_{{\text{solvent}}}^{\text{o}}} \right){\text{ + }}\left({{{\text{X}}_{{\text{solute}}}}\times \text{P}_{{\text{solute}}}^{\text{o}}} \right)\)

Raoult’s Law is akin to the ideal gas law, except as it relates to the properties of a solution.

Raoult’s Law assumes solutions to be ideal in which the intermolecular forces between dissimilar molecules (solute-solvent) of a solution equal forces between similar molecules (solute-solute and solvent-solvent).

The solutions that abide by the rules of Raoult’s Law at every concentration range and at all temperatures are called Ideal Solutions.

1. An ideal solution is obtained by mixing two identical liquids \(\rm{A}\) (solvent) and \(\rm{B}\) (solute), that have similar molecular size, structure and identical intermolecular forces. This solution will experience intermolecular forces of attraction as below-
   a. \(\rm{A – A}\) intermolecular forces of attraction (solvent-solvent)
   b. \(\rm{B – B}\) intermolecular forces of attraction (solute-solute)
   c. \(\rm{A – B}\) intermolecular forces of attraction (solvent-solute)
2. In an ideal solution, the intermolecular forces of attraction between \(\rm{A – A}\), \(\rm{B – B}\) and \(\rm{A – B}\) are nearly equal, i.e., \(\rm{A – A ≈ B – B ≈ A – B}\)

Ideal Solutions

According to Raoult’s Law, the partial vapour pressure of two components of the solution may be given as:

\({{\text{P}}_{\text{A}}}{\text{=}}{{\text{X}}_{\text{A}}} \times \text{P}{{{^\circ }}_{\text{A}}}\)

\({{\text{P}}_{\text{B}}}{\text{=}}{{\text{X}}_{\text{B}}} \times \text{P}{{{^\circ }}_{\text{B}}}\)

Total pressure \(\rm{P}\) is given by-

\({\text{P=}}{{\text{P}}_{\text{A}}}{{ + }}{{\text{P}}_{\text{B}}}\)

\({\text{P=}}{{\text{X}}_{\text{A}}}{ \times \text{P}}_{\text{A}}^{\text{o}}{\text{ + }}{{\text{X}}_{\text{B}}}{ \times \text{P}}_{\text{B}}^{\text{o}}\)

We know the total mole fraction of a solution is equal to \(1\). So,

\({{\text{X}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{\text{=1}}\)

\({{\text{X}}_{\text{A}}}{\text{=1}} – {{\text{X}}_{\text{B}}}\)

Hence, total pressure \(\rm{P}\) is given by-

\({\text{P=}}\left({{\text{1-}}{{\text{X}}_{\text{B}}}} \right){\text{P}}_{\text{A}}^{\text{o}}{\text{ + }}{{\text{X}}_{\text{B}}}{ \times \text{P}}_{\text{B}}^{\text{o}}\)

\({\text{P=P}}_{\text{A}}^{\text{o}}{\text{-}}{{\text{X}}_{\text{B}}}\left( {{\text{P}}_{\text{A}}^{\text{o}}{\text{-P}}_{\text{B}}^{\text{o}}} \right)\)

The partial pressure of components \(\rm{A}\) and \(\rm{B}\) in an ideal solution is given by-

\({{\rm{P}}_{\rm{A}}}{\rm{ = }}{{\rm{X}}_{\rm{A}}}{\rm{P}}_{\rm{A}}^{\rm{o}}\) and \({{\rm{P}}_{\rm{B}}}{\rm{ = }}{{\rm{X}}_{\rm{B}}}{\rm{P}}_{\rm{B}}^{\rm{o}}\)

where \({\text{P}}{{^\circ }}_{\text{A}}\) and \({\text{P}}{{^\circ }}_{\text{B}}\) are respective vapour pressures in pure form. \(\rm{X}_\rm{A}\) and \(\rm{X}_\rm{B}\) are respective mole fractions of components \(\rm{A}\) and \(\rm{B}\).

a. In an ideal solution, no heat is released or absorbed during the mixing of two pure components. Hence, \(Δ_\rm{mix} \rm{H} = 0\) which means that the enthalpy of mixing of two components should be zero.

b. No contraction or expansion of volume takes place in the formation of an ideal solution. Hence, \(Δ_\rm{mix} \rm{V} = 0\).

c. The interactions between similar molecules (solute-solute) \(\rm{B-B}\) interaction and (solvent-solvent) \(\rm{A-A}\) interaction is nearly equal to dissimilar molecules (solute-solvent) \(\rm{A-B}\) interaction.

No solution is ideal in nature; however, few solutions approach the ideal behaviour to a great extent. Some of the examples are-

1. \(\rm{n}-\)hexane and \(\rm{n}-\)heptane
2. Bromoethane and Chloroethane
3. Benzene and Toluene
4. \(\rm{CCl}_4\) and \(\rm{SiCl}_4\)
5. Chlorobenzene and Bromobenzene
6. Ethyl Bromide and Ethyl Iodide
7. n-Butyl Chloride and \(\rm{n}-\)Butyl Bromide

Non-Ideal Solutions

1. The solutions that do not abide by the rules of Raoult’s Law at every concentration range and at all temperatures are called Non-Ideal Solutions. These solutions show deviations from Raoult’s Law.
2. The resulting solution obtained by mixing two liquids, \(\rm{A}\) (solvent) and \(\rm{B}\) (solute), will experience several intermolecular forces of attractions as summarised below:
   a. \(\rm{A – A}\) intermolecular forces of attraction (solvent-solvent)
   b. \(\rm{B – B}\) intermolecular forces of attraction (solute-solute)
   c. \(\rm{A – B}\) intermolecular forces of attraction (solvent-solute)
3. In a non-ideal solution, the intermolecular forces of attraction between \(\rm{A – A}\), \(\rm{B – B}\) and \(\rm{A – B}\) are not equal, i.e., \(\rm{A – A ≠B – B ≠A – B.}\)
   
4. In non-ideal solutions, the partial pressure of components \(\rm{A}\) and \(\rm{B}\) is given by-
   \({{\text{P}}_{\text{A}}} \ne {{\text{X}}_{\text{A}}}{\text{P}}_{\text{A}}^{\text{o}}\) and \({{\text{P}}_{\text{B}}} \ne {{\text{X}}_{\text{B}}}{\text{P}}_{\text{B}}^{\text{o}}\)
   where \({\text{P}}_{\text{A}}^{\text{o}}\) and \({\text{P}}_{\text{B}}^{\text{o}}\) are respective vapour pressures in pure form. \(\rm{X}_\rm{A}\) and \(\rm{X}_\rm{B}\) are respective mole fractions of components \(\rm{A}\) and \(\rm{B}\).
5. In non-ideal solutions, the heat might be released or absorbed during the mixing of two pure components. Hence, \({\Delta _{{\text{mix}}}}{\text{H}} \ne 0\).
   Heat might be released, resulting in negative enthalpy of mixing \(\left( {{\Delta _{{\text{mix}}}}{\text{H < }}0} \right)\), Heat might be absorbed, resulting in positive enthalpy \(\left( {{\Delta _{{\text{mix}}}}{\text{H > }}0} \right)\).
6. Contraction or expansion of volume takes place in the formation of a non ideal solution. Hence, \(\left({{\Delta _{{\text{mix}}}}{\text{V}} \ne 0} \right).\)
   If the dissolution of liquids results in expansion of the resultant volume, then \(\left({{\Delta _{{\text{mix}}}}{\text{V}} > 0} \right).\)
   If the dissolution of liquids results in contraction of the resultant volume, then \(\left({{\Delta _{{\text{mix}}}}{\text{V}} < 0} \right).\)
7. The interactions between similar molecules (solute-solute ) \(\rm{B-B}\) interaction and (solvent-solvent ) \(\rm{A-A}\) interaction is different from that of dissimilar molecules (solute-solvent ) \(\rm{A-B}\) interaction.

The Non-ideal solutions deviate from Raoult’s Law and can occur in two ways:

1. When the intermolecular forces of attraction between dissimilar molecules (solvent-solute or \(\rm{A-B}\)) are weaker than between similar (solute-solute or solvent-solvent) molecules, this results in a positive deviation from Raoult’s Law.
2. When the intermolecular forces of attraction between dissimilar molecules (solvent-solute or \(\rm{A-B}\)) are greater than between similar (solute-solute or solvent-solvent) molecules, this results in a negative deviation from Raoult’s Law.

Positive Deviation from Raoult’s Law

Solutions exhibit Positive Deviation from Raoult’s Law when the vapour pressure of its components is greater than what is expected in Raoult’s Law.

1. In anon-ideal liquid-liquid binary solution showing positive deviation, the vapour pressure of \(\rm{A}\) can be represented as-
   \({{\text{P}}_{\text{A}}}{\text{ > }}{{\text{X}}_{\text{A}}}{{\text{P}}^{\text{o}}}_{\text{A}}\)
   And vapour pressure of \(\rm{B}\) can be represented as-
   \({{\text{P}}_{\text{B}}}{\text{ > }}{{\text{X}}_{\text{B}}}{{\text{P}}^{\text{o}}}_{\text{B}}\)
   where \({{\text{P}}^{\text{o}}}_{\text{A}}\) and \({{\text{P}}^{\text{o}}}_{\text{B}}\) are respective vapour pressures in pure form. \(\rm{X}_\rm{A}\) and \(\rm{X}_\rm{B}\) are respective mole fractions of components \(\rm{A}\) and \(\rm{B}\).
   Hence, the total vapour pressure of the non-ideal solution showing positive deviation is given by-
   \({\text{P > }}{{\text{X}}_{\text{A}}}{{\text{P}}^{\text{o}}}_{\text{A}} + {{\text{X}}_{\text{B}}}{{\text{P}}^{\text{o}}}_{\text{B}}\)
2. The intermolecular forces of attraction between dissimilar molecules (solvent-solute, \(\rm{A-B}\)) are weaker than similar molecules (solute-solute, \(\rm{B-B}\) or solvent-solvent, \(\rm{A-A}\)), i.e., \(\rm{A – B < A – A}\) or \(\rm{B – B.}\) Due to the weak \(\rm{A-B}\) attraction, the escaping tendency of the \(\rm{A-B}\) type of molecules is more. This leads to an increase in the vapour pressure of the solution.
3. The heat released to form new molecular interactions is less than the heat absorbed on breaking of original molecular interactions in such solutions. Hence, the enthalpy of mixing is positive, that is, \({\Delta _{{\text{mix}}}}{\text{H}} > 0.\)
4. As heat is absorbed, these solutions are endothermic in nature.
5. Expansion of volume takes place in such solutions. Due to the weak intermolecular force of attraction, the molecules are held loosely to each other. Hence, \({\Delta _{{\text{mix}}}}{\text{V}} > 0.\)

Example:

A non-ideal solution of carbon disulphide and acetone exhibits positive deviation from Raoult’s Law. Carbon disulfide is a non-polar molecule, whereas acetone is a polar molecule. As carbon disulfide is non-polar, the intermolecular forces of attraction are London dispersion forces. These forces are weak compared to other types of intermolecular forces. At the same time, acetone being polar has dipole-dipole forces, which are very strong compared to London dispersion forces.

Putting these two types of intermolecular forces together in a mixture results in dipole-induced dipole interactions. But dipole-induced dipole forces are not nearly as strong as the dipole-dipole interactions between the acetone molecules. Hence, the carbon disulfide-acetone solution is a  non-ideal solution showing positive deviation.

Few more examples of non-ideal solutions showing positive deviation:

1. Acetone and Carbon disulphide
2. Acetone and Benzene
3. Carbon Tetrachloride and Toluene or Chloroform
4. Methyl Alcohol and Water
5. Acetone and Ethanol
6. Ethanol and Water

Negative Deviation from Raoult’s Law

Solutions exhibit Negative Deviation from Raoult’s Law when the vapour pressure of its components is less than what is expected in Raoult’s Law.

1. In a non-ideal liquid-liquid binary solution showing positive deviation, the vapour pressure of A can be represented as-
   \({{\text{P}}_{\text{A}}}{\text{ < }}{{\text{X}}_{\text{A}}}{{\text{P}}^{\text{o}}}_{\text{A}}\)
   And vapour pressure of \(\rm{B}\) can be represented as-
   \({{\text{P}}_{\text{B}}}{\text{ < }}{{\text{X}}_{\text{B}}}{{\text{P}}^{\text{o}}}_{\text{B}}\)
   where \({{\text{P}}^{\text{o}}}_{\text{A}}\) and \({{\text{P}}^{\text{o}}}_{\text{B}}\) are respective vapour pressures in pure form. \(\rm{X}_\rm{A}\) and \(\rm{X}_\rm{B}\)  are respective mole fractions of components \(\rm{A}\) and \(\rm{B}\).
   Hence, the total vapour pressure of the non-ideal solution showing positive deviation is given by-
   \({\text{P < }}{{\text{X}}_{\text{A}}}{{\text{P}}^{\text{o}}}_{\text{A}} + {{\text{X}}_{\text{B}}}{{\text{P}}^{\text{o}}}_{\text{B}}\)
2. The intermolecular forces of attraction between dissimilar molecules (solvent-solute, \(\rm{A-B}\)) are stronger than similar molecules (solute-solute, \(\rm{B-B}\) or solvent-solvent, \(\rm{A-A}\)), i.e., \(\rm{A – B > A – A}\) or \(\rm{B – B.}\) Due to the strong \(\rm{A-B}\) attraction, the escaping tendency of the \(\rm{A-B}\) type of molecules is less. This leads to a decrease in the vapour pressure of the solution.
3. In such solutions, heat is released to form new molecular interactions. Hence, the enthalpy of mixing is negative, that is \({\Delta _{{\text{mix}}}}{\text{H}} < 0.\)
4. As heat is released, these solutions are exothermic in nature.
5. Contraction of volume takes place in such solutions. Due to strong intermolecular forces, the molecules are held tightly to each other. Hence, \({\Delta _{{\text{mix}}}}{\text{V}} < 0.\)

Example:
A non-ideal solution of chloroform and acetone exhibits negative deviation from Raoult’s Law. The molecules of acetone and chloroform bond with each other through hydrogen bonding. The formation of hydrogen bonding reduces the escaping tendency of acetone-chloroform molecules. Therefore, the vapour pressure of the solution is less than that expected for an ideal solution.

Few more examples of non-ideal solutions showing positive deviation-

1. Chloroform and Benzene
2. Chloroform and Diether
3. Acetone and Aniline
4. Nitric Acid \((\rm{HNO}_3)\) and water
5. Acetic Acid and pyridine
6. Hydrochloric Acid \((\rm{HCl})\) and water

Raoult’s Law and Colligative Properties

Raoult’s law and colligative properties are explained below:

1. Relative Lowering of Vapour Pressure

The effect of Raoult’s Law is observed when a non-volatile solute is added to a volatile solvent. The vapour pressure of the resultant solution (solute \(+\) solvent) becomes lower than the vapour pressure of the pure solvent.

The Figure above shows the surface of a pure solvent compared to a solution (solute \(+\) solvent). Some of the surfaces of the solution are occupied by solute particles due to which less solvent molecules are available to escape the liquid state and enter the gaseous state. This results in less vapour pressure over the surface of the solution. Thus, the addition of non-volatile solute results in a lowering of the vapour pressure of the solvent.

According to Raoult’s Law, the partial vapour pressure of two components, solvent \((\rm{A})\) and solute \((\rm{B})\), of a solution may be given as:

\({{\text{P}}_{\text{A}}}{\text{=}}{{\text{X}}_{\text{A}}}{ \times \text{P}}_{\text{A}}^{\text{o}}\) …….eqn. \(\left( 1 \right)\)

\({{\text{P}}_{\text{B}}}{\text{=}}{{\text{X}}_{\text{B}}}{ \times \text{P}}_{\text{B}}^{\text{o}}\)

Total pressure \(\rm{P}\) is given by-

\(\rm{P} = \rm{P}_\rm{A} + \rm{P}_\rm{B}\)

\({\text{P=}}{{\text{X}}_{\text{A}}}{ \times \text{P}}_{\text{A}}^{\text{o}}{\text{ + }}{{\text{X}}_{\text{B}}}{ \times \text{P}}_{\text{B}}^{\text{o}}\)

Where \(\rm{P}_{\text{A}}^{\text{o}}\) and \(\rm{P}_{\text{B}}^{\text{o}}\) are respective vapour pressures in pure form and \(\rm{X}_\rm{A}\) and \(\rm{X}_\rm{B}\) are respective mole fractions of components \(\rm{A}\) and \(\rm{B}\).

The decrease in vapour pressure of the solvent \((Δ\rm{P}_\rm{A})\) is given by-

\({{\Delta }}{{\text{P}}_{\text{A}}}{\text{=P}}_{\text{A}}^{\text{o}}{{-}}{{\text{P}}_{\text{A}}}\) …….eqn \(\left( 2 \right)\)

Substituting eqn \(\left( 1 \right)\) in eqn \(\left( 2 \right),\) we get-

\({{\Delta }}{{\text{P}}_{\text{A}}}{\text{=P}}_{\text{A}}^{\text{o}}{{-}}{{\text{X}}_{\text{A}}}{\text{P}}_{\text{A}}^{\text{o}}\)

\({{\Delta }}{{\text{P}}_{\text{A}}}{\text{=P}}_{\text{A}}^{\text{o}}\left({{{1-}}{{\text{X}}_{\text{A}}}} \right)\) ……..eqn \(\left( 3 \right)\)

We know the total mole fraction of a solution is equal to \(1\). So,

\(\rm{X}_\rm{A} + \rm{X}_\rm{B} = 1\)

\(\rm{X}_\rm{B} = 1 – \rm{X}_\rm{A}\) ……..eqn \(\left( 4 \right)\)

Substituting eqn \(\left( 4 \right)\) in eqn \(\left( 3 \right),\) the decrease in vapour pressure of the solvent \((Δ\rm{P}_\rm{A})\) is given by-

\({{\Delta }}{{\text{P}}_{\text{A}}}{\text{=P}}_{\text{A}}^{\text{o}}{{\text{X}}_{\text{B}}}\)

Using eqn \(\left( 2 \right)\) in the above equation, we get,

\({\text{P}}_{\text{A}}^{\text{o}}{{-}}{{\text{P}}_{\text{A}}}{\text{=P}}_{\text{A}}^{\text{o}}{{\text{X}}_{\text{B}}}\)

\( \Rightarrow \frac{{{{\text{P}}^{\text{o}}}_{\text{A}}{\text{-}}{{\text{P}}_{\text{A}}}}}{{{\text{P}}_{\text{A}}^{\text{o}}}}{\text{=}}{{\text{X}}_{\text{B}}}\)

\( \Rightarrow \frac{{{\text{P}}_{\text{A}}^{\text{o}}{{-}}{{\text{P}}_{\text{A}}}}}{{{\text{P}}_{\text{A}}^{\text{o}}}}{\text{=}}\frac{{{{\text{n}}_{\text{B}}}}}{{{{\text{n}}_{\text{A}}}{\text{ + }}{{\text{n}}_{\text{B}}}}}\)

Where, \(\rm{n}_\rm{B}\) is the number of moles of the solute and \(\rm{n}_\rm{A}\) is the number of moles of the solvent.

For dilute solutions, \(\rm{n}_\rm{A} >> \rm{n}_\rm{B}\), hence, \(\rm{n}_\rm{A} + \rm{n}_\rm{B} ≈ \rm{n}_\rm{A}\)

The above equation can be written as-

\(\frac{{{\text{P}}_{\text{A}}^{\text{o}}{{-}}{{\text{P}}_{\text{A}}}}}{{{\text{P}}_{\text{A}}^{\text{o}}}}{\text{=}}\frac{{{{\text{n}}_{\text{B}}}}}{{{{\text{n}}_{\text{A}}}}}\)

2. Elevation in Boiling Point

According to Raoult’s Law, adding a non-volatile solute to a pure solvent lowers the vapour pressure of the solution. This causes more heat to be supplied to the solution to bring its vapour pressure up to the atmospheric pressure, thereby increasing the boiling point of the solution. The difference in temperature between the boiling point of the solution to that of the pure solvent gives the boiling point elevation. \(\Delta {{\text{T}}_{\text{b}}}\) represents the boiling point elevation in the graph below.

3. Depression inzing Point

According to Raoult’s Law, adding a non-volatile solute to a pure solvent lowers the vapour pressure of the solution. The lowering of the vapour pressure of the solution results in a decrease in thezing point of the solution compared to the solvent.zing point difference is the difference in temperature between thezing point of the pure solvent and that of the solution. On the graph, thezing point depression is represented by \(\Delta {{\text{T}}_{\text{f}}}\).

The vapour pressure of a solution (blue) is lower than the vapour pressure of a pure solvent (purple). As a result, thezing point of a solution decreases when a solute is dissolved into a solvent.

Limitations of Raoult’s Law

Raoult’s Law is applicable only to very dilute solutions. Raoult’s Law applies to solutions containing non-volatile solute only. It does not apply to solutes that dissociate or associate in a particular solution.

Summary

Raoult’s Law is vital to determine the properties of solutions. It measures the strength of bonding in liquids and lowering in vapour pressure of the solution due to the addition of non-volatile solute. Through this article, we learnt the formula of Raoult’s Law and its application. We also learnt how Raoult’s Law governs the colligative properties of solutions.

FAQs on Raoult’s Law

Q.1. What is Raoult’s Law equation?
Ans: Raoult’s Law states that for any solution (obtained by dissolving a non-volatile solute in a solvent) the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. It is expressed by the formula:
\({{\text{P}}_{{\text{solution}}}}{\text{=}}{{\text{X}}_{{\text{solvent}}}}{{ \times }}{{\text{P}}^{\text{o}}}_{{\text{solvent}}}\)
Where,
\({{\text{P}}_{{\text{solution}}}}\) is the vapour pressure of the solution
\({{\text{X}}_{{\text{solvent}}}}\) is the mole fraction of the solvent
\({{\text{P}}^{\text{o}}}_{{\text{solvent}}}\) is the vapour pressure of the pure solvent

Q.2. What is Raoult’s Law and its application?
Ans: Raoult’s Law states that for any solution (obtained by dissolving a non-volatile solute in a solvent), the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. It is expressed by the formula:
\({{\text{P}}_{{\text{solution}}}}{\text{=}}{{\text{X}}_{{\text{solvent}}}}{{ \times }}{{\text{P}}^{\text{o}}}_{{\text{solvent}}}\)
Applications of Raoult’s Law:
It is used to measure the strength of bonding in non-volatile solutes. It helps us to measure the vapour pressure lowering caused due to the addition of non-volatile solute to a volatile solvent.

Q.3. What is a positive deviation from Raoult’s Law?
Ans: Solutions exhibit Positive Deviation from Raoult’s Law when the vapour pressure of its components is greater than what is expected in Raoult’s Law. In a non-ideal liquid-liquid binary solution showing positive deviation, the vapour pressure of \(\rm{A}\) can be represented as-
\({{\text{P}}_{\text{A}}}{\text{ > }}{{\text{X}}_{\text{A}}}{{\text{P}}^{\text{o}}}_{\text{A}}\)
And vapour pressure of \(\rm{B}\) can be represented as-
\({{\text{P}}_{\text{B}}}{\text{ > }}{{\text{X}}_{\text{B}}}{{\text{P}}^{\text{o}}}_{\text{B}}\)
where \({{\text{P}}^{\text{o}}}_{\text{A}}\) and \({{\text{P}}^{\text{o}}}_{\text{B}}\) are respective vapour pressures in pure form. \(\rm{X}_\rm{A}\) and \(\rm{X}_\rm{B}\) are respective mole fractions of components \(\rm{A}\) and \(\rm{B}\).
Hence, the total vapour pressure of the non-ideal solution showing positive deviation is given by-
\({\text{P > }}{{\text{X}}_{\text{A}}}{\text{P}}_{\text{A}}^{\text{o}}{\text{ + }}{{\text{X}}_{\text{B}}}{\text{P}}_{\text{B}}^{\text{o}}\)

Q.4. How is Raoult’s Law a special case of Henry’s Law?
Ans: According to Henry’s Law: \({{\text{P}}_{{\text{solution}}}}{\text{=}}{{\text{K}}_{\text{H}}}{{ \times }}{{\text{X}}_{{\text{solvent}}}}\)
where \({{\text{P}}_{{\text{solution}}}}\) is the partial pressure of the solution and \(\rm{X}_\rm{solvent}\) is the mole fraction of the solvent. \({\text{K}}_{\text{H}}\) is the proportionality constant (Henry’s constant)
According to Raoult’s Law,
\({{\text{P}}_{{\text{solution}}}}{\text{=}}{{\text{X}}_{{\text{solvent}}}}{{ \times }}{{\text{P}}^{\text{o}}}_{{\text{solvent}}}\)
where \({{\text{P}}_{{\text{solution}}}}\) is the partial pressure of the solution and \({\text{X}}_{\text{solvent}}\) is the mole fraction of the solvent. \({\rm{P}}_{{\rm{solvent}}}^{\rm{o}}\) is the vapour pressure of the pure solvent.
When,
\({\text{K}}_{\text{H}} = {\text{P}}^{\text{o}}_{\text{solvent}}\)
Raoult’s law becomes equal to Henry’s Law. Hence, Raoult’s Law is a special case of Henry’s Law.

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