The vapour pressure of benzene at $90\,^oC$ is $1020 \, \text{torr}$. $5 \, g$ of a solute is dissolved in $58.5 \, g$ of benzene. The vapour pressure of the solution is $990 \, \text{torr}$. What is the molar mass of the solute?

At $T \ K$,the vapour pressure of an aqueous solution of a non-volatile solute,whose mole fraction is $0.02$,is found to be $34.65 \ mm \ Hg$. What is the vapour pressure (in $mm \ Hg$) of pure water at the same temperature?

At a given temperature,the vapour pressure of pure water is $25 \ mm$ and the vapour pressure of a dilute solution of urea is $24.5 \ mm$. The molality of the solution is:

Two $5 \ molal$ solutions are prepared by dissolving a non-electrolyte,non-volatile solute separately in the solvents $X$ and $Y$. The molecular weights of the solvents are $M_X$ and $M_Y$,respectively,where $M_X = \frac{3}{4} M_Y$. The relative lowering of vapour pressure of the solution in $X$ is $m$ times that of the solution in $Y$. Given that the number of moles of solute is very small in comparison to that of solvent,the value of $m$ is

$A$ solution of a non-volatile solute is obtained by dissolving $2 \ g$ of solute in $50 \ g$ of benzene. Calculate the vapour pressure of the solution if the vapour pressure of pure benzene is $640 \ mmHg$ at $25^{\circ} C$. [Molar mass of benzene $= 78 \ g \ mol^{-1}$,Molar mass of solute $= 64 \ g \ mol^{-1}$] (in $mm \ Hg$)

In a solution,the mole ratio of pentane $(A)$ and hexane $(B)$ is $1:4$. At $20^o C$,the vapor pressures of these pure hydrocarbons are $440 \ mm$ and $120 \ mm$ respectively. The mole fraction of pentane in the vapor phase is.....