JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Solutions

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles

A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9

Given : Freezing point depression constant of water $\left({\mathrm{K}}_{\mathrm{f}}^{\text{water }}\right)=1.86\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$
Freezing point depression constant of ethanol $\left({\mathrm{K}}_{\mathrm{f}}^{\text{ethanol }}\right)=2.0\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$
Boiling point elevation constant of water $\left({\mathrm{K}}_{\mathrm{b}}^{\text{water }}\right)=0.52\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$
Boiling point elevation constant of ethanol $\left({\mathrm{K}}_{\mathrm{b}}^{\text{ethanol }}\right)=1.2\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$
Standard freezing point of water = 273 K
Standard freezing point of ethanol = 155.7 K
Standard boiling point of water = 373 K
Standard boiling point of ethanol = 351.5 K
Vapour pressure of pure water $=32.8\mathrm{mm}\mathrm{Hg}$
Vapour pressure of pure ethanol $=40\mathrm{mm}\mathrm{Hg}$
Molecular weight of water $=18\mathrm{g}{\mathrm{mol}}^{-1}$
Molecular weight of ethanol $=46\mathrm{g}{\mathrm{mol}}^{-1}$

In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.

The freezing point of the solution $\mathbf{M}$ is
(A) 268.7 K
(B) 268.5 K
(C) 234.2 K
(D) 150.9 K