Question

Relative lowering of vapor pressure is directly proportional to

Answer 1

Relative lowering in vapour pressure:

1. Relative lowering in vapour pressure is defined as the number of solvent molecules escaping into the vapour phase gets reduced as a result the pressure exerted by the vapour phase also gets decreased.
2. It is a colligative property of solutions.
3. Colligative properties are those that are affected by the mole fraction of solute.
4. Vapour pressure is the pressure exerted by a substance's vapour state on its liquid (or solid) state in a closed container at equilibrium conditions.
5. Raoult's law states that a solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
6. It is given by ${\mathrm{P}}_{\mathrm{solution}}={\mathrm{X}}_{\mathrm{solvent}}{{\mathrm{P}}^{°}}_{\mathrm{solvent}}$ where ${\mathrm{P}}_{\mathrm{solution}}$ is the vapour pressure of the solution, ${\mathrm{X}}_{\mathrm{solvent}}$ is the mole fraction of solvent, ${\mathrm{P}}_{\mathrm{solvent}}$ is the vapour pressure of the solvent.
7. It is suited for describing ideal solutions, that is, those in which the gas phase possesses thermodynamic parameters similar to a mixture of ideal gases.
8. Assume a binary solution in which the solute's mole fraction is ${\mathrm{x}}_2$, the solvent's mole fraction is${\mathrm{x}}_1$, the solvent's vapour pressure is ${\mathrm{p}}_1$, and the solvent's pure state vapour pressure is ${\mathrm{p}}_1^{°}$ Then according to Raoult's law ${\mathrm{p}}_1={\mathrm{x}}_1{\mathrm{p}}_1^{°}...\left(\mathrm{i}\right)$
9. The decrease in vapour pressure of the solvent ($∆{\mathrm{p}}_1$) is given by$={\mathrm{p}}_1^{°}-{\mathrm{p}}_1$. Putting value of ${\mathrm{p}}_1$ we get $∆{\mathrm{p}}_1={\mathrm{p}}_1^{°}-{\mathrm{p}}_1^{°}{\mathrm{x}}_1\Rightarrow {\mathrm{p}}_1^{°}(1-{\mathrm{x}}_1)$
10. Since we have assumed the solution to be a binary solution${\mathrm{x}}_2=1-{\mathrm{x}}_1$, $∆{\mathrm{p}}_1={\mathrm{p}}_1^{°}{\mathrm{x}}_2$or ${\mathrm{x}}_2=\frac{∆{\mathrm{p}}_1}{{\mathrm{p}}_1^{°}}$
11. Hence, the above equation relates the relative lowering of vapour pressure and mole fraction.

Therefore, the Relative lowering of vapour pressure is directly proportional to the mole fraction of solute.