The vapour pressure lowering caused by the addition of $100 \, g$ of sucrose (molecular mass $= 342$) to $1000 \, g$ of water if the vapour pressure of pure water at $25 \, ^oC$ is $23.8 \, mm \, Hg$.

The vapour pressure of a solvent decreases by $20 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.5$. What should be the mole fraction of the solvent for the decrease in the vapour pressure to be $10 \ mm$ of $Hg$?

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 ........... $torr$.

What is the vapour pressure of a solution containing $1.8 \ g$ of glucose in $16.2 \ g$ of water,if the vapour pressure of pure water is $32 \ mm \ Hg$ (in $mm \ Hg$)?

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:

$9 \ g$ anhydrous oxalic acid (mol. $Wt. = 90$) was dissolved in $9.9 \ moles$ of water. If vapour pressure of pure water is $P_1^o$,the vapour pressure of solution is (in $P_1^o$)