The vapour pressure of pure liquid solvent is $0.50 \ atm$. When a non-volatile solute $B$ is added to the solvent,its vapour pressure drops to $0.30 \ atm$. The mole fraction of the solute $B$ is $......$

If the total vapour pressure of the liquid mixture $A$ and $B$ is given by the equation: $P = 180X_A + 90 \, mm \, Hg$,then the ratio of the vapour pressure of the pure liquids $A$ and $B$ is given by:

At $50^{\circ} C$, the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

The vapour pressure of pure benzene and toluene are $0.9 \ Bar$ and $0.85 \ Bar$ respectively. Calculate the vapour pressure of the solution prepared by dissolving $7.8 \ g$ of benzene in $180 \ g$ of toluene. (in $Bar$)

Which of the following will have the lowest vapor pressure?

Vapour pressure of a pure solvent is $550 \ mm$ of $Hg$. By addition of a non-volatile solute,it decreases to $510 \ mm$ of $Hg$. Calculate the mole fraction of the solute in the solution.