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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. Application of colligative properties are very useful in day to day life.one example is the use of ethylene glycol and water mixture as antifreezing liquid in the radiators of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in water is 0.9. Given, freezing point depression constant of water `(K_(f) "water") = 1.86 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 `= 155.7 K` Standard boiling point of water `= 373 K` Standard boiling point of ethanol `= 351.5 K` Vapour pressure of pure water `= 32.8 mm Hg` Vapoure pressure of pure ethanol `= 40 mm Hg` Molecular weight of water `= 18 g mol^(-1)` Molecular weight of ethanol `= 46 g 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 solution M isA. `268.7 K`B. `268.5 K`C. `234.2 K`D. `150.9 K`

Answer» Correct Answer - D
Solution M is mixture of ethanol and water, Mole fraction of ethanol is 0.9 (Solvent is `C_(2)H_(5)OH`) Mole fraction water is 0.1 (`H_(2)O` is solute)
Molality of `H_(2)O=(n_(2)xx1000)/(n_(1)M_(1))`
`=(0.1xx1000)/(0.9xx46)=2.415` m
`DeltaT_(f)=K_(f)m=2xx2.2415=4.83`
Freezing point of solution
`= 155.7 - 4.83 = 150.87 K`
Total vapour pressure, `p=p_(A)^(@)xx A`.


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