The vapour pressure of pure water is $23 \text{ mmHg}$. The vapour pressure of an aqueous solution,which contains $10$ mass per cent of solute '$A$' having molecular weight $50$ is (in $\text{ atm}$)

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 a solvent decreases by $10 \ mm \ Hg$ if the mole fraction of a non-volatile solute is $0.2$. Calculate the vapour pressure of the pure solvent. (in $mm \ Hg$)

The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?

The total vapour pressure of a solution of liquid $A$ and liquid $B$ is $600 \ torr$. The mole fraction of component $A$ in the liquid phase is $0.7$. If the mole fraction of $A$ in the vapour phase is $0.35$,what are the vapour pressures of pure $A$ and pure $B$ respectively?

The vapour pressures of two volatile liquids $A$ and $B$ at $25^{\circ} C$ are $50 \ Torr$ and $100 \ Torr,$ respectively. If the liquid mixture contains $0.3$ mole fraction of $A$,then the mole fraction of liquid $B$ in the vapour phase is $\frac{x}{17}$. The value of $x$ is $...$