At $T$ $(K)$,the vapour pressure of pure benzene is $0.85$ bar. $A$ non-volatile,non-electrolyte substance weighing $0.5$ $g$ when added to $39$ $g$ of benzene,the vapour pressure of the solution is $0.845$ bar. The molar mass (in $g$ $mol^{-1}$) of the substance is

At $25\, ^oC$,benzene and toluene form an ideal solution. What will be the vapor pressure (in $mm \, of \, Hg$) of a solution containing benzene and toluene in a weight ratio of $1:2$? The vapor pressures of pure benzene and pure toluene are $75\, mm \, of \, Hg$ and $30\, mm \, of \, Hg$,respectively.

At $T$ $(K)$,the vapour pressures of pure liquids $A$ and $B$ are $100 \ mm$ and $160 \ mm$ respectively. An ideal solution is formed by mixing $2 \ moles$ of $A$ and $3 \ moles$ of $B$ at the same temperature. The mole fractions of $A$ and $B$ in the vapour state respectively are

When $18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water,the vapour pressure of the aqueous solution at $100 \ ^\circ C$ is .......... $torr$.

At a certain temperature,when some amount of glucose is dissolved in water,it is observed that the lowering in vapour pressure is $0.6 \ mm \ Hg$ for this solution. What will be the vapour pressure of the same glucose solution if its molality is $(\frac{1}{18}) \ mol \ kg^{-1}$ (in $mm \ Hg$)?

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: