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# Raoult’s law: principle and formula, examples, exercises

This law is also used to analyze and describe the composition of volatile solvents in the gas phase, located in space on a solution that presents a mixture of them. The law is named after its creator, François-Marie Rauolt (1830-1901). Raoult’s law diagrams. Line 1 represents the ideal behavior, while the red and blue lines correspond to the positive and negative deviations, respectively. Source: Joanna Kośmider / Public domain

Rauolt’s law applies to ideal solutions that meet some characteristics, including the following: the intermolecular forces between equal molecules (cohesive forces) must be equal to the intermolecular forces between different or dissimilar molecules (adhesive forces).

Many of the solutions are not ideal, which explains the deviations from Rauolt’s law observed in some volatile solvent mixtures. For example, the mixture of chloroform (CH 3 Cl) and acetone (CH 3 COCH 3 ), presents a negative deviation from Raoult’s law.

The vapor pressure in the gas phase in such cases is less than that predicted by law, which can be explained by the formation of hydrogen bonds between the components of the mixture.

## Principle and formula

Rauolt’s law indicates that the partial vapor pressure exerted by a volatile component or solvent of the gaseous mixture, above the solution, is related to the vapor pressure exerted by the pure volatile component or solvent, and their respective molar fractions.

The following equation summarizes the above:

sv = P sv º · X sv

Where P sv is the partial pressure of the volatile solvent in the gas mixture, P sv º the pressure of the pure volatile solvent, and X sv the mole fraction in the solution of the volatile solvent.

### Volatile solvent mixture

If you have a mixture of two volatile solvents (A and B) in the solution, you can calculate the vapor pressure that they originate in the gas phase, above the solution. This will be a sum of the partial pressures exerted by gases A and B:

A = X A  ·  A º

B = X B · P B º

Therefore, adding the pressures of A and B we obtain the total pressure P:

P = X A  · P A º + X B · P B º

Where P is the vapor pressure of the gas mixture above the solution, X A and X B the mole fractions of the volatile solvents A and B in the mixture, and P A º and P B º the vapor pressures of the solvents pure volatiles A and B.

The partial pressure of a volatile solvent in the gas phase is given by the expression:

P = P A º · X A

In the presence of a solute B in the solution, the mole fraction of B is expressed as follows:

B = 1 – X A

Then, by means of a simple mathematical treatment, we arrive at the expression:

ΔP = P A º · X B  (1)

Where ΔP is the decrease in the partial pressure of the solvent in the gas phase.

The mathematical expression (1) indicates the decrease in the vapor pressure of a solvent due to the presence of a non-volatile B solute in the solution. The decrease in vapor pressure of the solvent has been explained by the location of the solute B molecules on the surface of the solution.

The presence of molecules of solute B would produce a decrease in the concentration of molecules of solvent A on the surface of the solution, limiting their evaporation; and thus being explained, the decrease in its vapor pressure in the gas phase.

## Examples

Raoult’s law is used to calculate the vapor pressure of a volatile component of a solution, such as ethanol, benzene, toluene, ethane, propane, etc., in space above the solution.

It can be used to calculate the vapor pressure that is generated in space over a solution, as a consequence of the mixture of volatile liquids, be it benzene and toluene, ethane and propane, acetone and ethanol, etc.

Likewise, with this law it is possible to determine what the decrease in vapor pressure will be if, for example, sucrose were dissolved in water, being a non-volatile solute.

## Solved exercises

### Exercise 1

Calculate the vapor pressure of a solution made by dissolving 60 g of sodium chloride (NaCl) in 400 g of water (H 2 O). The pressure of water vapor (P H2O º) at 37 ºC is 47.1 mmHg. Molecular weight H 2 O = 18 g / mol and molecular weight NaCl = 58.5 g / mol.

We first calculate the moles of water and sodium chloride in order to determine their mole fractions:

Moles of H 2 O = grams of H 2 O / PM H 2 O

= 400 g / (18 g / mol)

= 22.22 moles

Moles of NaCl = g of NaCl / pm NaCl

= 60 g / (58.5 g / mol)

= 1.03 moles

NaCl is an electrolytic compound that dissociates into Na + + Cl  . Therefore, 1.03 moles of NaCl dissociates into 1.03 moles of Na + and 1.03 moles of Cl  .

We have the expression:

v = X H2O · P H2O º

We therefore lack the mole fraction of water:

H2O = moles of H 2 O / (moles of H 2 O + moles of Na +    + moles of Cl  )

= 22.2 moles / 22.22 moles + 1.03 moles + 1.03 moles

= 0.896

And we calculate P v :

v  = 47.1 mmHg 0.896

v  = 42.20 mmHg

Being the decrease in vapor pressure due to the presence of sodium chloride:

ΔP v = 47.1 mmHg – 42.20 mmHg

= 4.9 mmHg

### Exercise 2

At -100 ° C ethane (CH 3 CH 3 ) and propane (CH 3 CH 2 CH 3 ) are liquid. At this temperature, the vapor pressure of pure ethane (P ethane º) is 394 mmHg, while the vapor pressure of pure propane (P propane º) is 22 mmHg. What will be the vapor pressure over a solution containing equimolar amounts of both compounds?

The problem statement states that the solution contains equimolar amounts of the compounds. This implies that the mole fraction of the compounds ethane and propane is necessarily equal to 0.5.

Again, the solution comes from the expression:

v   = P ethane   + P propane

We calculate the partial pressures of both ethane and propane:

ethane = P ethane º · X ethane

= 394 mmHg 0.5

= 197 mmHg

propane  = P propane º · X propane

= 22 mmHg0.5

= 11 mmHg

And so we finally calculate P v :

v   = P ethane  + P propane

= 197 mmHg + 11 mmHg

= 208 mmHg