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?

For a dilute solution,Raoult's law states that

The vapour pressure of acetone $(CH_3COCH_3)$ at $20\,^{\circ}C$ is $185\, torr$. When $1.2\, g$ of a non-volatile substance was dissolved in $100\, g$ of acetone at $20\,^{\circ}C$,its vapour pressure was $183\, torr$. The molar mass $(g\, mol^{-1})$ of the substance is

"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

In $108 \, g$ of water,$18 \, g$ of a non-volatile compound is dissolved. At $100^{\circ} C$,the vapour pressure of the solution is $750 \, mm \, Hg$. Assuming that the compound does not undergo association or dissociation,the molar mass of the compound in $g \, mol^{-1}$ is

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 150 - 60 X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and pure $B$ at the same temperature are respectively: