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 $24 \ mm \ Hg$ )

If two substances $A$ and $B$ have $P_{A}^{\circ} : P_{B}^{\circ} = 1 : 2$ and have mole fraction in solution in the ratio $1 : 2$,then mole fraction of $A$ in vapour phase is:-

$A$ solution of liquids $A$ and $B$ is given. The mole fraction of $A$ in the vapor phase is $x_1$ and in the liquid phase is $x_2$. If $P_A^o$ and $P_B^o$ are the vapor pressures of pure $A$ and $B$ respectively,then the total vapor pressure $P_{total}$ will be:

In a liquid mixture of $A$ and $B$,what is the mole fraction of $A$ in the vapor phase $(X_A = 0.4)$? Given: $P_A^o = 100 \, mm \, Hg$ and $P_B^o = 200 \, mm \, Hg$.

An aqueous solution of $2\%$ non-volatile solute exerts a pressure of $1.004 \, bar$ at the normal boiling point of the solvent. The molar mass of the solute is $......... \, g/mol$.

At $50^{\circ} C$, the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)