If a solution is prepared using the liquids mentioned in the previous question such that the mole fraction of $A$ is $0.8$,what will be the mole fraction of $A$ in the vapor phase?

The vapour pressure of two liquids $P$ and $Q$ are $80$ and $60 \, torr$,respectively. The total vapour pressure of the solution obtained by mixing $3 \, mole$ of $P$ and $2 \, mole$ of $Q$ would be .......... $torr$.

If the total vapour pressure of the liquid mixture $A$ and $B$ is given by the equation: $P = 180X_A + 90 \, mm \, Hg$,then the ratio of the vapour pressure of the pure liquids $A$ and $B$ is given by:

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:

At $20^{\circ} C$,the vapour pressure of benzene is $70 \, torr$ and that of methyl benzene is $20 \, torr$. The mole fraction of benzene in the vapor phase at $20^{\circ} C$ above an equimolar mixture of benzene and methyl benzene is $..... \times 10^{-2}$ (Nearest integer).

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