In a solution,the mole fraction of solute is $0.2$ when the lowering in vapour pressure is $10 \text{ mm Hg}$. To achieve a lowering of vapour pressure of $20 \text{ mm Hg}$,the mole fraction of the solute in the solution should be:

If the total vapour pressure of the liquid mixture $A$ and $B$ is given by the equation: $P = 180X_A + 90 \, mm \, Hg$,then the ratio of the vapour pressure of the pure liquids $A$ and $B$ is given by:

The boiling points of $C_6H_6, CH_3OH, C_6H_5NH_2$ and $C_6H_5NO_2$ are $80 \ ^oC, 65 \ ^oC, 184 \ ^oC$ and $212 \ ^oC$ respectively. Which will show the highest vapour pressure at room temperature?

Calculate the relative lowering of vapour pressure of a solution containing $3 \ g$ urea in $50 \ g$ water. [ Molar mass of urea $= 60 \ g \ mol^{-1} ]$

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 $atm$)

If two substances $A$ and $B$ have $P_A^o : P_B^o = 2 : 1$ and have mole fraction in solution $1 : 3$,then the mole fraction of $A$ in the vapour phase is: