Raoult’s Law

Rault's law is an important principle in the field of chemistry basically physical chemistry it was discovered by the French scientist Francis Marie Raoult in the year 1887. This discovery of Raoult provided a description of how the vapour pressure of the solvent in the solution is related to the concentration of the solute. Raoult's law was discovered during the intense study of colligative properties of solution which mainly depend upon the number of solute particles in the given amount of solvent and they do not depend upon the nature of the solute they only depend upon the number.

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This Story also Contains

1. Raoult's Law
2. Recommended video
3. Some Solved Examples
4. Summary

The aim of the raoult's research is to understand how the non-volatile solutes affect the vapour pressure of the solvent. According to Raoult's work, the understanding of the solution was totally quantitative. results repeatedly investigate the various solutions and measure their vapour pressure in the different solutes. After doing all this research he found out that the vapour pressure of the solvent decrease is directly proportional to the concentration of the solute.

Initially, the results experiment was based on a simple solution and in these experiments he also found out that the vapour pressure of the solvent is directly proportional to the mole fraction of the solute in the solutions. This relationship is the basis of Raoult's law which states that the partial vapour pressure of each volatile compound in the solution is directly proportional to the vapour pressure of the pure component multiplied by the mole fraction in the solution

Raoult's Law

These solutions have vapour pressure less than that predicted by Raoult’s law for the entire range of composition.

This happens when the new solute-solvent interactions are stronger than the interactions in the pure components. Since the newly formed forces are stronger than the existing forces, heat is liberated. Hence, the enthalpy change in the mixing of these solutions is negative i.e. $\Delta H_{\operatorname{mix}}<0$.

The change in the volume during the mixing process is positive i.e.$\Delta V_{\operatorname{mix}}<0$. For example, if 1 litre solution of each of liquids A and B is mixed, then the solution obtained has a volume lesser than 2 litres.

The entropy of mixing is positive as new interactions are introduced into the solution which increases the randomness of the system hence, The mixing process is spontaneous and hence $\Delta G_{\operatorname{mix}}>0$

These solutions have vapour pressure greater than that predicted by Raoult’s law.

$\begin{aligned} & P_A<P_A^o X_A \\ & P_B<P_B^o X_B \\ & \quad P_T=P_A+P_B<P_A^o X_A+P_B^o X_B\end{aligned}$

Examples of solutions showing negative deviation:

1. Acetone + Chloroform

2. Nitric acid HNO3 + water

3. Acetic acid + pyridine

4. Phenol + Aniline

Solutions showing positive deviation from Raoult’s law

These solutions have vapour pressure greater than that predicted by Raoult’s law for the entire range of composition.

This happens when the new interactions are weaker than the interactions in the pure component (A-B < A-A or B-B interactions). Since the newly formed forces are weaker than the existing forces, heat has to be supplied to break old bonds and form new ones. Hence, the enthalpy change in the mixing of these solutions is positive i.e. $\Delta H_{\operatorname{mix}}>0$.

The change in the volume during the mixing process is positive i.e.$\Delta V_{\operatorname{mix}}>0$ For example if 1 litre of solutions each of liquid A and B are mixed, then the solution obtained has a volume greater than 2 litres.

The entropy of mixing is positive as new interactions are introduced into the solution which increases the randomness of the system and hence $\Delta S_{\operatorname{mix}}>0$. The mixing process is spontaneous and hence $\Delta G_{\operatorname{mix}}<0$

These solutions have vapour pressure greater than that predicted by Raoult’s law.
$\begin{aligned} & P_A>P_A{ }^o X_A \\ & P_B>P_B{ }^o X_B \\ & P_T=P_A+P_B>P_A{ }^o X_A+P_B{ }^o X_B\end{aligned}$

Examples:

1. C₂H₅OH + cyclohexane

2. Acetone + carbon disulphide

3. Acetone + benzene

4. Acetone + Ethyl alcohol

5. Carbon tetrachloride + chloroform or Toluene

6. Methyl alcohol + water

7. Water + Ethyl alcohol

Recommended video

Some Solved Examples

Example.1

1. Two liquids A and B on mixing produce a warm solution. Which type of deviation from Raoult's law does it show?

1) (correct)Negative deviation

2)Positive deviation

3)Can't say anything

4)Data insufficient

Solution

Warming up of the solution means that the process of mixing is exothermic, i.e.
This implies that the solution shows a negative deviation.
Hence, the answer is the option (1).

Example.2

2. A solution of sulphuric acid in water exhibits:

1) (correct)Negative deviations from Raoult’s law

2)Positive deviations from Raoult’s law

3)Ideal Properties

4)The applicability of Henry’s law

Solution

There is a hydrogen bond between two molecules of sulphuric acid and between two molecules of water, but there is an ion-dipole interaction between sulphuric acid and water. Ion dipole interaction is stronger than hydrogen bonds. So solution shows a negative deviation from Raoult’s law.
Hence, the answer is the option (1).

Example.3

3. Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapor pressures of pure A and pure B are$7 \times 10^3 \mathrm{~Pa}$ and $12 \times 10^3 \mathrm{~Pa}$ , respectively. The composition of the vapour in equilibrium with a solution containing 40 mol per cent of A at this temperature is :

1)$x_A=0.37 ; x_B=0.63$

2) (correct)$x_A=0.28 ; x_B=0.72$

3)$x_A=0.4 ; x_B=0.6$

4)$x_A=0.76 ; x_B=0.24$

Solution

We know that

$\mathrm{y}_{\mathrm{a}}=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {total }}}=\frac{\mathrm{P}_{\mathrm{A}}^0 \times \mathrm{X}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^0 \times \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^0 \times \mathrm{X}_{\mathrm{B}}}$

$\begin{aligned} & \mathrm{y}_{\mathrm{A}}=\frac{7 \times 10^3 \times 0.4}{7 \times 10^3 \times 0.4+12 \times 10^3 \times 0.6} \\ & \mathrm{y}_{\mathrm{A}}=\frac{2.8}{10}=0.28\end{aligned}$

$\mathrm{y}_{\mathrm{B}}=0.72$

Hence, the answer is the option (2).

Example.4

4. The vapour pressure of the solution of two liquids A(Pº = 80 mm) and B(Pº = 120 mm) is found to be 100 mm when $X_A$ = 0.4. The result shows that:

1)solution exhibits ideal behaviour

2) The solution shows a positive deviation

3) (correct)solution shows negative deviation

4) The solution will show positive deviations for lower concentrations and negative deviations for higher concentrations

Solution

Negative deviation means lower vapour pressure, it suggests a high boiling point, thus the resultant intermolecular force should be stronger than the individual one.

Hence, the answer is the option (4).

Example.5

5. In mixtures A and B components show negative deviation as

(a) $\Delta V_{\text {mix }}>0$

(b) $\Delta V_{\text {mix }}>0$

(c) $\mathrm{A}-\mathrm{B}$ interaction is weaker than $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$ interaction

(d) $A-B$ interaction is stronger than $A-A$ and $B-B$ interaction

The correct statement is/are:

1)(a) only

2)(b) only

3)(a) and (d) only

4) (correct)(b) and (d) only

Solution

Negative deviation means lower vapour pressure, it suggests a high boiling point, thus the resultant intermolecular force should be stronger than the individual one.

Hence, the answer is the option (4).

Summary

Raoult's law is used to describe the relationship between the vapour pressure of solution and the concentration of solute. raoult's law has various benefits in chemistry such that it is used to understand the behaviour of the solution as it predicts how the presence of solutes affects the vapour pressure of the solution which is important for understanding the properties of the solution. Raoults law is also used to determine the molecular weight of the solutions in such a way that the methods used are ebullioscopy and cryoscopy which is used to measure the change in boiling and freezing point.

Raoults law has more application in the field of pharmaceutical industries in such a way that it drug formulated it help to check the solubility of drugs according to different solvents and concentration. it also has use in the liquid-vapour equilibrium separation technique and also has a role in environmental science in such a way that it can help to understand the behaviour of pollutants and their evaporation rates in the water systems and as solutions are also used to make alloys so the raoult's law also assists their formation and mixing of different substance which need precise control on their vapour pressure.