The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be

At $80\,^oC,$ the vapour pressure of pure liquid $'A'$ is $520\, mm\,Hg$ and that of pure liquid $'B'$ is $1000\, mm\,Hg.$ If a mixture solution of $'A'$ and $'B'$ boils at $80\,^oC$ and $1\, atm$ pressure,the amount of $'A'$ in the mixture is ........... $mol$ percent $(1\, atm = 760\, mm\,Hg).$

Vapour pressure in $mm \ Hg$ of $0.1 \ mole$ of urea in $180 \ g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 \ mm \ Hg$ )

The vapour pressure of pure liquid solvent is $0.50 \ atm$. When a non-volatile solute $B$ is added to the solvent,its vapour pressure drops to $0.30 \ atm$. The mole fraction of the solute $B$ is $......$

At a given temperature,the vapour pressure of pure water is $25 \ mm$ and the vapour pressure of a dilute solution of urea is $24.5 \ mm$. The molality of the solution is:

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 150 - 60 X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and pure $B$ at the same temperature are respectively: