$2 \ \text{moles}$ each of ethylene glycol and glucose are dissolved in $500 \ \text{g}$ of water. The boiling point of the resulting solution is $:$ (Given $:$ Ebullioscopic constant of water $= 0.52 \ \text{K kg mol}^{-1}$) (in $\text{K}$)

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 $K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$).

At $T(K)$, $2$ moles of liquid $A$ and $3$ moles of liquid $B$ are mixed. The vapour pressure of the ideal solution formed is $320 \ mm \ Hg$. At this stage, one mole of $A$ and one mole of $B$ are added to the solution. The vapour pressure is now measured as $328.6 \ mm \ Hg$. The vapour pressures of pure $A$ and pure $B$ (in $mm \ Hg$) are respectively:

$A$ 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 $= 273 \ K$)

Identify the false statement from the following.

The freezing point of an aqueous solution is $-0.186 \ ^oC$. The boiling point of the same solution is ........ $^oC$. (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$ and $K_b = 0.512 \ K \ kg \ mol^{-1}$)