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?

$A$ set of solutions is prepared using $180 \ g$ of water as a solvent and $10 \ g$ of different non-volatile solutes $A, B$ and $C$. The relative lowering of vapour pressure in the presence of these solutes are in the order
[Given,molar mass of $A = 100 \ g \ mol^{-1}; B = 200 \ g \ mol^{-1}; C = 10,000 \ g \ mol^{-1}]$

The vapour pressure of a solvent decreases by $10 \ mm \ Hg$ if the mole fraction of a non-volatile solute is $0.2$. Calculate the vapour pressure of the pure solvent. (in $mm \ Hg$)

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

"Relative lowering in vapour pressure of a solution containing a non-volatile solute is directly proportional to the mole fraction of the solute". The above statement is:

When a non-volatile solute is added to a pure solvent,the vapor pressure decreases by $11.5 \ torr$. If the mole fraction of the solute is $0.2$,what will be the vapor pressure of the pure solvent in $torr$?