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. Thus moel fraction of ethanol in the mixture is `0.9`.
Given: Freezing point depression constant of water
`(K_(f)^("water")) = 1.86 K kg mol^(-1)`
Freezing point depression constant of ethanol
`(K_(f)^("ethanol")) = 2.0 K kg mol^(-1)`
Boiling point elevation constant of water
`(K_(b)^("water")) = 0.52 K kg mol^(-1)`
Boiling point elevation constant of ethanol
`(K_(b)^("ethanol")) = 1.2 K kg mol^(-1)`
Standard freezing point of water `= 273 K`
Standard freezing point of ethanol `= 155.7K`
Standard boiling point of water `= 373 K`
Standard boiling point of ethanol `= 351.5 K`
vapour pressure of pure water `= 32.8 mm Hg`
Vapour pressure of pure ethanol `= 40 mm Hg`
Molecualr weight of water `= 18 g mol^(-1)`
Molecular weight of ethanol `= 46 g mol^(-1)`
In asweering 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 `M` is :

Answer

D

Explanation

Solution `M` is a mixture of ethanol and water, where ethanol is the solvent because the mole fraction of ethanol is `0.9`.
The mole fraction of water =`0.1`
Molarity of `H_(2)O(m_(H2O))=(chi_(2) xx 1000)/(chi_(1) xx Mw_(1))`
`=(0.1 xx 1000)/(0.9 xx 46) =2.415`
`DeltaT_(f)=K_(f) xx m =2 xx 2.415 =4.83`
Freezing point =`155.7-4.83 =150.87 K`