MCQ
A solution of two miscible liquids showing negative deviation from Raoult's law will have :
  • A
     increased vapour pressure, increased boiling point
  • B
    increased vapour pressure, decreased boiling point
  • C
    decreased vapour pressure, decreased boiling point
  • decreased vapour pressure, increased boiling point

Answer

Correct option: D.
decreased vapour pressure, increased boiling point
d
Solution with negative deviation has

$ \mathrm{P}_{\mathrm{T}}<\mathrm{P}_{\mathrm{A}^0} \mathrm{X}_{\mathrm{A}} +\mathrm{P}_{\mathrm{B}^0} \mathrm{X}_{\mathrm{B}} $

$ \mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{A}^0} \mathrm{X}_{\mathrm{A}} $

$ \mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{B}^0} \mathrm{X}_{\mathrm{B}} $

If vapour pressure decreases so boiling point increases.

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