Calculate the osmotic pressure in $atm$ of an aqueous solution of urea at $37\,^oC$,which has a freezing point of $0.52\,^oC$. Assume molality and molarity are numerically equal. $(K_f = 1.86\,^oC\, m^{-1})$

An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:
Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.

At $T(K)$,the vapour pressure of pure benzene (molar mass $= 78 \ g \ mol^{-1}$) is $0.85 \ bar$. When $2.0 \ g$ of a non-volatile,non-electrolyte solute is added to $39 \ g$ of benzene,the vapour pressure of the solution at $T(K)$ is $0.83 \ bar$. The elevation in boiling point (in $K$) of the same solution is: ($K_b$ of benzene is $2.6 \ K \ kg \ mol^{-1}$)

$PbCl_2$ is dissolved in water to make its saturated solution. What will be the freezing point of this solution?
Given: $K_f (H_2O) = 2 \ K \ kg \ mol^{-1}$,$K_{sp} (PbCl_2) = 4 \times 10^{-6}$
(Assume molarity to be equal to molality) $..... ^oC$

At $100\, ^oC$ the vapour pressure of a solution of $6.5\, g$ of a solute in $100\, g$ water is $732\, mm.$ If $K_b = 0.52,$ the boiling point of this solution will be .........$^oC$

$1 \, \text{mole}$ of liquid $A$ and $2 \, \text{moles}$ of liquid $B$ make a solution having an observed vapour pressure of $42 \, \text{torr}$. The vapour pressures of pure $A$ and pure $B$ are $45 \, \text{torr}$ and $36 \, \text{torr}$ respectively. The described solution: