An aqueous solution freezes at $-0.186\,^\circ C$ ($K_f = 1.86\, K\, kg\, mol^{-1}$; $K_b = 0.512\, K\, kg\, mol^{-1}$). What is the elevation in boiling point?

The elevation in the boiling point of an aqueous urea solution is $0.104 \ K$. What is its $\Delta T_{f}$ (in $K$) value? (Given for water: $K_{b} = 0.52 \ K \ kg \ mol^{-1}$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$)

At $T \ K$,the vapour pressure of pure benzene $(C_6H_6)$ and toluene $(C_7H_8)$ are $75 \ mm \ Hg$ and $22 \ mm \ Hg$ respectively. $23.4 \ g$ of benzene and $64.4 \ g$ of toluene are mixed to form an ideal solution. If the vapours are in equilibrium with the liquid mixture,the mole fraction of toluene in the vapour phase is (Atomic weight: $C=12, H=1$).

On mixing urea, the boiling point of $H_{2}O$ changed to $100.5^{\circ}C$. Calculate the freezing point of the solution, if $K_{f}$ of water is $1.87 \ K \cdot kg \cdot mol^{-1}$ and $K_{b}$ of water is $0.52 \ K \cdot kg \cdot mol^{-1}$. (in $^{\circ}C$)

Which of the following has the lowest freezing point?

The freezing point of an aqueous solution is $-0.186^oC$. If the molal elevation constant and molal depression constant of the solvent are $0.512$ and $1.86$ respectively,then the elevation in boiling point is .......... $^oC$.