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 :

Options

A. 268.7K
B. 268.5K
C. 234.2K
D. 150.9K

Answer

D

Explanation

`W_("ethanol") gt W_("water")`
`therefore` Ethanol acts as solvent
`DeltaT_(f)=K_(f)("ethanol")xx(n_("water")xx1000)/(W_("ethanol")`
`=0.2 xx (0.1xx1000)/(0.9xx46)`
=4.83
`T_(f)("solution")=155.70-4.83K`
=150.87 K
`~~150.9 K`