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 vapour pressure of the solution `M` is:

Options

A. `724`
B. `740`
C. `736`
D. `718`

Answer

A

Explanation

`DeltaT_(b)=2^(@)C, m_(a)=2.5g`
`m_("solvent")=100g`
`K_(b)=0.76K. Kg. "mol"^(-1)`
`P_("solution")=?`
`DeltaT_(b)=K_(b)xxm`
`2=0.76xxm thereforem=(2)/(0.76)`
`(P^(0)-P)/(P)=mxxMMxx10^(-3)` therefore(760-P)/(P)=(2)/(0.76)xx18xx(10^(-3)`
`760-P=(36)/(760)P therefore 760=(796)/(760)P`
`thereforeP=760((796)/(760))"torr"=725.6` torr `=724` torr