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 pure benzene at a certain temperature is $640 \ mm \ Hg$. $A$ non-volatile non-electrolyte solid weighing $2.175 \ g$ is added to $39.0 \ g$ of benzene. The vapour pressure of the solution is $600 \ mm \ Hg$. What is the molar mass of the solid substance?

At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be:

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 $Hg$?

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

At $298 \ K$,the vapour pressure of a solution of $7.5 \ g$ of non-volatile solute in $90 \ g$ of water is $2.8 \ kPa$. If $18 \ g$ of water is added to this solution,the vapour pressure becomes $2.81 \ kPa$ at the same temperature. The molar mass of the solute in $g \ mol^{-1}$ is: