The vapour pressure of pure benzene and toluene are $0.9 \ Bar$ and $0.85 \ Bar$ respectively. Calculate the vapour pressure of the solution prepared by dissolving $7.8 \ g$ of benzene in $180 \ g$ of toluene. (in $Bar$)

The vapor pressures of two liquids $P$ and $Q$ are $80 \ torr$ and $60 \ torr$,respectively. The total vapor pressure of a solution formed by mixing $3 \ moles$ of $P$ and $2 \ moles$ of $Q$ is .......... $torr$.

The correct option for the value of vapour pressure of a solution at $45^{\circ} C$ with benzene to octane in molar ratio $3 : 2$ is ...... $mm$ of $Hg$. [At $45^{\circ} C$,vapour pressure of benzene is $280 \ mm \ Hg$ and that of octane is $420 \ mm \ Hg$. Assume ideal solution.]

The relative lowering of the vapour pressure is equal to the ratio between the number of

Two liquids $A$ and $B$ form a solution at $306 \ K$ that follows the equation $P \ (\text{in } atm) = 0.172X_A + 0.215$,where $P$ is the total vapor pressure of the solution and $X_A$ is the mole fraction of $A$. The vapor pressure of pure $B$ is ............ . (in $atm$)

What will be the molar mass of a non-volatile solute if the vapour pressure of pure benzene is $450 \ mm \ Hg$ and it decreases to $400 \ mm \ Hg$ when $1.5 \ g$ of the solute is added to $30 \ g$ of benzene? (Atomic mass: $C=12, H=1$)