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. $(K_f = 1.86\,^oC\, m^{-1})$

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 .........$^oC$

An aqueous solution of a solute $AB$ has a $b.p.$ of $101.08^\circ C$ ($AB$ is $100\%$ ionized at the boiling point of the solution) and freezes at $-1.80^\circ C$. Given $K_b / K_f = 0.3$,the solute $AB$:

Elevation in boiling point for $1.5 \, m$ solution of glucose in water is $4 \, K$. The depression in freezing point for $4.5 \, m$ solution of glucose in water is $4 \, K$. The ratio of molal elevation constant to molal depression constant $(K_{b} / K_{f})$ is $....$

If the freezing point of an aqueous solution of salt $MX_4$ $(\alpha = 90\%)$ is $-3\ ^oC$,then its boiling point will be ........ $^oC$. (Assume it is $100\%$ ionized at the boiling temperature and $k_f = 1.86\ K\ kg\ mol^{-1}$,$k_b = 0.52\ K\ kg\ mol^{-1}$)