Explanation:
Raoult's law for volatile solutes states that at a given temperature vapor pressure of a component is equal to the mole fraction of that component component in the solution multiplied to the vapor pressure of the component in the pure state.
Mathematically, [tex]p_{A} = p^{o}_{A} \times x_{A}[/tex]
[tex]p_{B} = p^{o}_{B} \times x_{B}[/tex]
P = [tex]p_{A} + p_{B}[/tex]
= [tex]p^{o}_{A} \times x_{A} + p^{o}_{B} \times x_{B}[/tex]
Since, [tex]x_{A} + x_{B}[/tex] = 1
[tex]x_{A}[/tex] = 1 - [tex]x_{B}[/tex]
hence, P = [tex]p^{o}_{A} \times (1 - x_{B}) + p^{o}_{B} \times x_{B}[/tex]
P = [tex](p^{o}_{B} - p^{o}_{A})x_{B} + p^{o}_{A}[/tex]
On the other hand, Raoult's law for non-volatile solutes states that relative lowering of vapor pressure of a solution that has non-volatile solute is equal to the mole fraction of the solute in the solution.
Mathematically, [tex]\frac{p^{o} - P_{s}}{P^{o}}[/tex] = [tex]\frac{n_{2}}{n_{1} + n_{2}}[/tex] = [tex]x_{2}[/tex]
where, [tex]x_{2}[/tex] = mole fraction of solute
[tex]n_{1}[/tex] = moles of solvent
[tex]n_{2}[/tex] = moles of solute
[tex]P_{s}[/tex] = vapor pressure of the solution
[tex]p^{o}[/tex] = vapor pressure of pure solvent
