$A$ mixture of $3.0 \ mol$ of $Na_2O$ and $1.5 \ mol$ of $KO_2$ is dissolved in $1000 \ mL$ of water. The vapour pressure of the solution in $Torr$, at $100^{\circ}C$ is:

  • A
    $740$
  • B
    $760$
  • C
    $580$
  • D
    $608$

Explore More

Similar Questions

$34.2 \ g$ of cane sugar is dissolved in $180 \ g$ of water. The relative lowering of vapour pressure will be

What will be the molar mass of a non-volatile solute if the vapour pressure of pure benzene is $450 \ mm \ Hg$ and it decreases to $400 \ mm \ Hg$ when $1.5 \ g$ of the solute is added to $30 \ g$ of benzene? (Atomic mass: $C=12, H=1$)

$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.

Difficult
View Solution

The vapour pressure in $mm$ of $Hg$ of an aqueous solution obtained by adding $18 \ g$ of glucose $(C_6H_{12}O_6)$ to $180 \ g$ of water at $100^{\circ}C$ is:

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

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