Question

State and explain Raoult’s law.

Answer 1

Raoult’s law deals with the vapor pressure of the solution and it is decided on the factors like mole fraction and vapor pressure in the pure state. This law was given in 1886 by F. M. Raoult.

Complete step by step answer:
We know that the solution is made up of two components, solute, and solvent, so when the solution of liquid in a liquid having both the components volatile, there will be a vapor pressure above the solution. On reaching equilibrium, both the components in the solution will exert pressure, known as its partial pressure and this partial pressure whose value depends on the mole fraction of the component in the solution. So by studying all these factors, French Chemist, F. M. Raoult in 1886, gave the Raoult’s law statement as:
"In a solution, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapor pressure of that component in the pure state."
According to Raoult’s law, the equation will be:
$p={{p}^{\circ }}x$
In which p represents the partial vapor pressure of the solution, ${{p}^{\circ }}$ is the vapor pressure in the pure state, and x is the mole fraction.
Suppose a solution is made up of two components A and B, the partial pressure of component A will be:
${{p}_{A}}=p_{A}^{\circ } \times {{x}_{A}}$
The partial pressure of component B will be:
${{p}_{B}}=p_{B}^{\circ }\times {{x}_{B}}$
So by combining, the equation will be:
$P=p_{A}^{\circ }\text{ }{{x}_{A}}+p_{B}^{\circ }{{x}_{B}}$
Where P is the total partial pressure of the solution.

Note: This equation is used only when both the components of the solution are volatile. So when any of the components in the solution is non-volatile, then there will be a lowering of the total partial pressure because molecules will convert into a vapor state.