The relative lowering of vapor pressure is equal to the mole fraction of the non-volatile solute. This statement was given by whom?

If the vapor pressure of a solution formed by dissolving $1 \ mol$ of a non-volatile solute in $4 \ mol$ of a solvent is $24.0 \ kPa$,what will be the vapor pressure of the pure solvent in $kPa$?

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).

The vapour pressure of a solvent $A$ is $0.80 \, atm$. When a non-volatile substance $B$ is added to this solvent its vapour pressure drops to $0.6 \, atm$. What is the mole fraction of $B$ in the solution?

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 $mm \ Hg$)?

Derive an equation for a solution that shows the relation between total pressure and the mole fraction of a volatile solute and a volatile solvent,and explain it by plotting a graph.