$X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ at different concentrations:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is:

  • A
    $p_1 < p_2 < p_3$
  • B
    $p_3 < p_2 < p_1$
  • C
    $p_3 < p_1 < p_2$
  • D
    $p_2 < p_1 < p_3$

Explore More

Similar Questions

The vapour pressure of two pure liquids $(A)$ and $(B)$ are $100 \ torr$ and $80 \ torr$ respectively. The total pressure of the solution obtained by mixing $2 \ mole$ of $(A)$ and $3 \ mole$ of $(B)$ would be ........ $torr$.

At $300 \ K$,the vapor pressures of two pure liquids $A$ and $B$ are $150 \ mm \ Hg$ and $100 \ mm \ Hg$,respectively. If the mole fractions of $A$ and $B$ in the solution are equal,then the mole fraction of $B$ in the vapor phase at the same temperature is:

Why is the vapour pressure of an aqueous solution of glucose lower than that of pure water?

If two substances $A$ and $B$ have $P_A^0 : P_B^0 = 1 : 2$ and have mole fraction in solution $1 : 2$,then what is the mole fraction of $A$ in the vapour phase?

At $T$ $(K)$,the vapour pressures of pure liquids $A$ and $B$ are $100 \ mm$ and $160 \ mm$ respectively. An ideal solution is formed by mixing $2 \ moles$ of $A$ and $3 \ moles$ of $B$ at the same temperature. The mole fractions of $A$ and $B$ in the vapour state respectively are

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