An aqueous solution of glucose was prepared by dissolving $18 \ g$ of glucose in $90 \ g$ of water. The relative lowering in vapour pressure is

At $25\,^oC$,the vapour pressure of pure liquid $A$ (mol. wt. $= 40$) is $100\,torr$,while that of pure liquid $B$ is $40\,torr$ (mol. wt. $= 80$). The vapour pressure at $25\,^oC$ of a solution containing $20\,g$ of each $A$ and $B$ is ........... $torr$.

The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

If a solution is prepared using the liquids mentioned in the previous question such that the mole fraction of $A$ is $0.8$,what will be the mole fraction of $A$ in the vapor phase?

Which solution will show the maximum vapour pressure at $300 \ K$?

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 $= 78 \ g \ mol^{-1}$)