160. Derive the relationship between relative lowering of vapour pressure and mole fraction of a volatiles liquid.

Answer:

for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
p_A ∝ x_A

p_A x x_A where p_A is vapour pressure of solvent having mole fraction x_A.
PA = P⁰A x A
But x_A + x_B = L
∴ x_A = 1 – x_B
When x_B is mole fraction of non-voltile solute B
Pa = P⁰_A (1–x_B)
= p⁰A – p⁰_A x _B
Total vapour of solution is equal to p_A as nonvolatile solute does not have any vapour pressure.
i.e., ${\mathrm{p}}_{\mathrm{B}} = 0$   Total vapour pressure,
               $$\begin{gathered} \mathrm{p} = {\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}} = {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{A}}- {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{B}}\times \mathrm{B} \\ {\mathrm{x}}_{\mathrm{B}} = \frac{ {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{A}} - {\mathrm{p}}_{\mathrm{A}}}{{{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{A}}} \end{gathered}$$