At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 120 - 80 \, X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and $B$ at the same temperature are respectively:

The vapor pressure of a dilute solution of glucose at $373 \ K$ is $750 \ mm \ of \ Hg$. The molality of the solution is:

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in $1 \ kg$ of water is $0.002$. The molality of the solution is (in $m$)

$38.4 \ g$ of unknown substance (molar mass $384 \ g \ mol^{-1}$) and $116 \ g$ of acetone is used to prepare a solution at $313 \ K$. If vapour pressure of pure acetone (molar mass $58 \ g \ mol^{-1}$) is $0.842 \ atm$,what is the vapour pressure of the solution (in $atm$)?

The vapour pressure of acetone 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:

$A$ 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 $mm \ Hg$)