Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to homegenous solution. These are called colligative properties. Applications of colligative properties are very useful in day to day life. One of its exaples is the use of ethylene glycol and water mixtue as anti-freezing liquid in the radiators of automobiles. A solution M is prepared by mixing ehanol and water. The mole fraction of ethanol in the mixture is 0.9 Given.
`K_(f)"of water"=1.86 "K kg mol"^(-1)`
`K_(f)"of ethanol" = 2.0 "K kg mol"^(-1)`
`K_(b)"for water "=0.52 "K kg mol"^(-1)`
`K_(b)" of ethanol"=1.2 "K kg mol"^(-1)`
`T_(f)("water")=273 K`
`T_(f)("ethanol")=155.7 K`
`T_(b)^(@) ("water")=373 K`
`T_(b)^(@)("ethanol")=351.5 K`
`P^(@)("water")32.8 mm Hg.`
`P^(@)("ethanol")=40 mmHg`
`M("ethanol")==46 "g mol"^(-1)`
`M("water")=18"g mol"^(-1)`
In answering the following quesion, consider the solutions to be ideal dilute solutions ans solute to be non-voltile and non-dissoviative.
(7) water is added to the solution M such that the mole fraction so wter in solution becomes 0.9. The boiling point of solution is :

Options

A. `380.4 K`
B. `376.2 K`
C. 357.5 K`
D. 245.7 K`

Answer

b

Explanation

In this case, water is solvent and ethanol is solute
`"Molality of solution (m)"=("No. of moles ethanol")/("Mass of water in kg")`
`n_("ethanol")=0.1 `,
`W_("water")=((0.9mol)xx(18" g mol"^(-1)))/1000=0.0162 kg.`
`m=((0.1mol))/((0.0162 kg))=6.173 "mol kg"^(-1)`
=6.173m
`DeltaT_(b)=K_(b)xxm=(0.52"k kg mol"^(-1))xx(6.173" mol kg"^(-1))=3.2 K`
b.p. of solution = 373+3.2= 376.2 K