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?

Calculate the osmotic pressure in $atm$ at $0\,^oC$ of a $0.18\, m$ aqueous solution of $KCl$ which has a freezing point of $-0.68\,^oC$. Assume the volume of the solution is equal to the volume of pure water. $(K_f = 1.86\,^oC\, m^{-1})$

$2 \ \text{moles}$ 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 $pH$ of a $0.1 \ M$ monobasic acid is $2$. Its osmotic pressure at a given temperature $T \ (K)$ is: (Given that the effective concentration for osmotic pressure is $(1+\alpha) \times$ concentration of acid,where $\alpha$ is the degree of dissociation)

$A$ solution at $20\,^oC$ is composed of $1.5\,mol$ of benzene and $3.5\,mol$ of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are $74.7\,torr$ and $22.3\,torr,$ respectively,then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be,respectively

$1$ molal $K_{4}[Fe(CN)_{6}]$ solution has a degree of dissociation of $0.4$. Its boiling point is equal to that of another solution which contains $18.1$ weight percent of a non-electrolytic solute $A$. The molar mass of $A$ is $.......\, u$. (Round off to the Nearest Integer). [Density of water $= 1.0\, g\, cm^{-3}$]