Explain the vapour pressure of solutions of solids in liquids.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Liquids at a given temperature vaporize,and under equilibrium conditions,the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure.
In a pure liquid,the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to form a solution,the vapour pressure of the solution is solely due to the solvent molecules.
This vapour pressure of the solution at a given temperature is found to be lower than the vapour pressure of the pure solvent at the same temperature. In the solution,the surface contains both solute and solvent molecules; thereby,the fraction of the surface covered by the solvent molecules is reduced. Consequently,the number of solvent molecules escaping from the surface is correspondingly reduced,thus,the vapour pressure is also reduced.

Explore More

Similar Questions

$A$ solution is prepared by mixing $8.5 \ g$ of $CH_2Cl_2$ and $11.95 \ g$ of $CHCl_3$. If the vapour pressures of pure $CH_2Cl_2$ and $CHCl_3$ at $298 \ K$ are $415 \ mm \ Hg$ and $200 \ mm \ Hg$ respectively,the mole fraction of $CHCl_3$ in the vapour phase is: (Molar mass of $Cl = 35.5 \ g \ mol^{-1}$)

At $298 \ K$,$0.714$ moles of liquid $A$ is dissolved in $5.555$ moles of liquid $B$. The vapour pressure of the resultant solution is $475 \ torr$. The vapour pressure of pure liquid $A$ at the same temperature is $280.7 \ torr$. What is the vapour pressure of pure liquid $B$ in $torr$?

If the vapour pressure of a solution containing a non-volatile solute is $2\%$ less than the vapour pressure of pure water,find the molality of the solution.

The relative lowering of the vapour pressure is equal to the ratio between the number of

$A$ set of solutions is prepared using $180 \ g$ of water as a solvent and $10 \ g$ of different non-volatile solutes $A, B$ and $C$. The relative lowering of vapour pressure in the presence of these solutes are in the order
[Given,molar mass of $A = 100 \ g \ mol^{-1}; B = 200 \ g \ mol^{-1}; C = 10,000 \ g \ mol^{-1}]$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo