Calculate the solubility (in $\text{moles/litre}$) of a saturated aqueous solution of $Ag_3PO_4$ if the vapour pressure of the solution becomes $750 \ torr$ at $373 \ K$. (Assume vapour pressure of pure water at $373 \ K$ is $760 \ torr$ and density of water is $1 \ g/mL$)

For a dilute solution containing $2.5 \ g$ of a non-volatile non-electrolyte solute in $100 \ g$ of water,the elevation in boiling point at $1 \ atm$ pressure is $2^{\circ} C$. Assuming the concentration of solute is much lower than the concentration of solvent,the vapour pressure ($mm$ of $Hg$) of the solution is (take $K_{b}=0.76 \ K \ kg \ mol^{-1}$)

An aqueous solution contains $6\%$ and $4\%$ by weight of a weak acid $(HA)$ and urea respectively. If the depression in freezing point is $\frac{13.6}{3} \, ^{\circ}C$,calculate the $K_a$ of $HA$. [Given: $K_f = 1.8 \, K \, kg/mol$,Molecular weight of $HA = 60$,$d_{solution} = 1 \, g/mL$]

$PbCl_2$ 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) $..... ^oC$

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,$(K_b / K_f)_{\text{water}} = 0.3$.

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