The vapour pressure of a pure solvent is $0.70 \ atm.$ When a non-volatile solute $A$ is dissolved in it,the vapour pressure of the solution becomes $0.35 \ atm.$ The mole fraction of solute $A$ will be$:-$

The vapour pressure of water at room temperature is lowered by $5\%$ by dissolving a solute in it. Then the approximate molality of the solution is

The vapour pressure of pure liquids $A$ and $B$ are $450 \ mm \ Hg$ and $700 \ mm \ Hg$ respectively at $350 \ K$. Find out the composition of the liquid mixture if total vapour pressure is $600 \ mm \ Hg$. Also,find the composition of the vapour phase.

For a solution of volatile liquids,the partial vapour pressure of each component in the solution is directly proportional to:

If vapour pressure of pure solvent and solution are $240 \ mm Hg$ and $216 \ mm Hg$ respectively,then the mole fraction of solvent in the solution is:

The vapour pressure of $30 \%$ $(w/v)$ aqueous solution of glucose is $...... \ mm \ Hg$ at $25^{\circ} \ C$. [Given : The density of $30 \%$ $(w/v)$ aqueous solution of glucose is $1.2 \ g \ cm^{-3}$ and vapour pressure of pure water is $24 \ mm \ Hg$.] (Molar mass of glucose is $180 \ g \ mol^{-1}$.)