In which solution,solute is liquid and solvent is gas?

On dissolving $0.5 \ g$ of a non-volatile non-ionic solute to $39 \ g$ of benzene,its vapor pressure decreases from $650 \ mm \ Hg$ to $640 \ mm \ Hg$. The depression of freezing point of benzene (in $K$) upon addition of the solute is. . . . .
(Given data : Molar mass and the molal freezing point depression constant of benzene are $78 \ g \ mol^{-1}$ and $5.12 \ K \ kg \ mol^{-1}$,respectively)

Solid Lead nitrate is dissolved in $1 \, L$ of water. The solution was found to boil at $100.15^{\circ}C$. When $0.2 \, mol$ of $NaCl$ is added to the resulting solution,it was observed that the solution froze at $-0.8^{\circ}C$. The solubility product of $PbCl_2$ formed is $........... \times 10^{-6}$ at $298 \, K$. (Nearest integer) Given: $K_b = 0.5 \, K \, kg \, mol^{-1}$ and $K_f = 1.8 \, K \, kg \, mol^{-1}$. Assume molality to be equal to molarity in all cases.

$A$ solution at $20\,^oC$ is composed of $1.5\,mol$ of benzene and $3.5\,mol$ of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are $74.7\,torr$ and $22.3\,torr,$ respectively,then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be,respectively

$A$ solution of urea in water has a boiling point of $100.18^{\circ} C$. What is the freezing point of the same solution,if $K_{f}$ and $K_{b}$ of water are $1.86$ and $0.52 \ K \ kg \ mol^{-1}$,respectively (in $^{\circ} C$)? (Boiling point of water $= 100^{\circ} C$ )

The elevation in boiling point for $1 \ m$ solution of non-volatile solute $A$ is $3 \ K$. The depression in freezing point for $2 \ m$ solution of $A$ in the same solvent is $6 \ K$. The ratio of $K_{b}$ and $K_{f}$ i.e.,$K_{b} / K_{f}$ is $1 : X$. The value of $X$ is [nearest integer].