At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 150 - 60 X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and pure $B$ at the same temperature are respectively:

  • A
    $120, 80$
  • B
    $90, 150$
  • C
    $150, 90$
  • D
    $80, 40$

Explore More

Similar Questions

At $25^\circ C$,the vapor pressure of $CCl_4$ is $143 \ mm \ Hg$. If $0.5 \ g$ of a non-volatile solute (molar mass $= 65 \ g/mol$) is dissolved in $100 \ mL$ of $CCl_4$,the vapor pressure of the resulting solution is $...... \ mm \ Hg$. (Density of $CCl_4 = 1.58 \ g/cm^3$)

Difficult
View Solution

Which of the following is an incorrect form of Raoult's law?

For a dilute solution,Raoult's law states that

When $25 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of water,the vapour pressure is lowered by $2.25 \times 10^{-1} \ mm$. If the vapour pressure of water at $20^{\circ}C$ is $17.5 \ mm$,what is the molecular weight of the solute?

Vapour pressure of a solution and of a pure solvent are $P_1$ and $P_1^0$ respectively. If $\frac{P_1}{P_1^0}$ is $0.15$,find the mole fraction of the solute.

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