One mole of a non-volatile solute is dissolved in two moles of water. What is the vapor pressure of this solution relative to that of pure water?

At $298 \ K$,vapour pressures of two pure liquids $A$ and $B$ are $200 \ mm \ Hg$ and $400 \ mm \ Hg$ respectively. If mole fractions of $A$ and $B$ in solution are $0.7$ and $0.3$ respectively,what is the mole fraction of $B$ in vapour phase?

Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene having mole fraction of toluene $0.50$. If vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature,mole fraction of toluene in vapour phase will be:

$18 \ g$ of glucose is dissolved in $90 \ g$ of water. The relative lowering of vapour pressure of the solution is equal to

$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.

The vapour pressure of pure liquids $A$ and $B$ are $450$ $mm$ $Hg$ and $700$ $mm$ $Hg$ respectively,at $350$ $K$. Find out the composition of the liquid mixture if the total vapour pressure is $600$ $mm$ $Hg$. Also,find the composition of the vapour phase.