The ionization constant of a monobasic acid $HA$ is to be determined. If a $0.025 \ m$ aqueous solution of the acid freezes at $-0.060 \ ^{\circ}C$,calculate the ionization constant $K_a$. (Assume molality = molarity and $K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$).

$A$ non-volatile solute is dissolved in water. The $\Delta T_b$ of the resultant solution is $0.052 \ K$. What is the freezing point of the solution (in $K$)?
($K_b$ of water $= 0.52 \ K \ kg \ mol^{-1}$; $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$,Freezing point of water $= 273 \ K$)

The solution from the following with the highest depression in freezing point (lowest freezing point) is:

$A$ non-volatile, non-electrolyte solid solute when dissolved in $40 \text{ g}$ of a solvent, the vapour pressure of the solvent decreased from $760 \text{ mm Hg}$ to $750 \text{ mm Hg}$. If the same solution boils at $320 \text{ K}$, then the number of moles of the solvent present in the solution is . . . . . . . (Nearest integer) [Given: boiling point of the pure solvent = $319.5 \text{ K}$, $K_b$ of the solvent = $0.3 \text{ K kg mol}^{-1}$]

An aqueous solution of urea (molar mass $60 \ g \ mol^{-1}$) boils at $100.18 \ ^\circ C$ at $1 \ atm$ pressure. If $K_f$ and $K_b$ for water are $1.86$ and $0.512 \ K \ kg \ mol^{-1}$ respectively,at what temperature will this solution freeze (in $K$)?

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)