What is Raoult's law

What is Raoult's law for Volatile and Non-volatile solute

Explain Raoult's law for volatile and non-volatile solute.


Raoult’s law for Volatile solutes

Partial vapour pressure of a solvent in a solution is equal to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
PSolution = XSolvent × P'Solvent
where, PSolution = vapour pressure of the solution
XSolvent = mole fraction of the solvent
PSolvent = vapour pressure of the pure solvent

For two component system-
P1 = X1 × P'1
and P2 = X2 × P'2
According to Dalton's law of partial pressures, the total pressure( Ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as-
Ptotal = P1 + P2
Ptotal = X1 × P'1 + X2 × P'2
or, Ptotal = (1 − X2) × P'1 + X2 × P'2
or, Ptotal = P'1 − (X2 × P'1) + X2 × P'2
or, Ptotal = P'1 − (P'2 − P'1) X2


From Raoult's law, it is evident that as the mole fraction of a component reduces, its partial pressure also reduces in the vapour phase. The graph of pressure vs mole fraction for component 1 and component 2 are shown below.
Raoult’s law Graph


Raoult’s law for Non-Volatile solutes

Relative lowering of vapour pressure is equal to the mole fraction of the solute is Raoult's law for non-volatile solutes.
If a dilute solution is prepared by dissolving n mol of a solute in N mol solvent then, the vapour pressure of solvent above the solution is proportional to mole fraction of solvent.
P ∝ N/(N + N)
or, P = K N/(n + N) -----Eq:1
where, K is proportionality constant.
For pure solvent, n = 0
so, Po = K N/N =K -----Eq:2
From Eq:1 and Eq:2, we have-
P/Po = N/(n + N)
or, 1 − (P/Po) = 1 − [N / (n + N)]
or, (Po − P) / Po = (n + N −N) / (n + N)
or, (Po − P) / Po = n / (n + N)
Thus, we can say that the relative lowering of vapour pressure is equal to the mole fraction of the solute.


Calculate the vapour pressure lowering caused by the addition of 100 g of sucrose (mol mass = 342) to 1000 g of water if the vapour pressure of pure water at 25°C is 23.8 mm Hg.

Answer

Hints:

Lowering of vapour pressure(ΔP) = 23.8 × (0.292 / 55.792) = 0.125mm Hg.