Question

What is Raoult's Law?
Derive Raoult's Law for solution which contain non-volatile solute.

Answer 1

Raoult's Law: According to this law 'The vapour pressure of solution containing non-volatile solute is directly proportional to the mole fraction of the solvent'.
For a solution of two components $A$ (Volatile solvent) and $B$ (non-volatile solute)
Vapour pressure of solution $=$ Vapour pressure of solvent $\propto$ Mole fraction of solvent
Or $p=p_A\propto X_a$
Pr $P_A=K{X}_A$ ....(i)
Where, $k=$ Proportionality constant
Apply equation (i) for pure solvent for this purpose, put (i) $X_a=1$ and instead of $p_A$ put $p_A^0$
where $P_A^0=$ vapour pressure of pure solvent
$\therefore$ $P_A^0=K{X}_1$
Or $P_A^0=K$ ....(ii)
According to equation (ii) the value of proportionality constant $k$ is equal to the vapour pressure of pure solvent by putting the value of $k$ in equation (i) we get
$P_A=P_A^0{X}_A$
Or $p$ solution $=p$ pure solvent $\times$ mole fraction of solvent.
Derive it mathematically: Mathematically, if $P$ is the vapour pressure of the pure solvent and $P_S$, that of the solution, $X_S$ is the mole fraction of the solvent in the solution, $n$ are number of moles of the solute and $N$ are the number of moles of the solvent in the solution, then
$P_S\propto X_S$ or $P_S=K{X}_S$
(Where $X_S=\frac{N}{n+N}$ and $K$ is constant ....(i)
In case of pure solvent $n=0$ hence
Mole fraction of solvent,
$X_S=\frac{N}{n+N}=\frac{N}{0+N}=1$ and $P_S=p$
$P=k\times 1=k$
From equation (i) $P_S=P\times X_S$ ...(ii)
Thus, the vapour pressure of the solvent in the solution is equal to the product of its mole fraction and its vapour pressure in the pure state at the same temperature.