Answer :

Raoult’s law for volatile solutes states that in a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in the pure state.

Mathematically:

The vapour pressure P of a solution containing two components A and B is

P_{A} = P^{o}_{A} × X_{A}

P_{B} = P^{0}_{B ×} X_{B}

P = P_{A} + P_{B} = P^{o}_{A} × X_{A} + P^{0}_{B ×} X_{B}

*where*, is the vapour pressure of component A

is the vapour pressure of component B

is the vapour pressure of component A in its pure state

is the vapour pressure of component B in its pure state

is the mole fraction of component A

is the mole fraction of component B

Now we know that,

X_{A +} X_{B} = 1

X_{A} = 1- X_{B}

P = (1-X_{B})P^{0}_{A} + X_{B}P^{0}_{B}

P = P^{0}_{A} + X_{B}(P_{B}^{0} – P^{0}_{A})

For a solution to be able to obey Raoult’s law at all concentrations, it should be an *ideal solution*. For a solution to be called ideal, it should have the following two characteristics:

•The enthalpy of mixing of the pure components to form the solution should be zero, that is, . It means that no heat is absorbed or released during mixing.

•The volume of mixing the pure components should be zero, that is, . This means that the volume of the solution would be equal to the sum of volumes of the two components.

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