Solutions Question 3
Question 3 - 2024 (27 Jan Shift 1)
A solution of two miscible liquids showing negative deviation from Raoult’s law will have :
(1) increased vapour pressure, increased boiling point
(2) increased vapour pressure, decreased boiling point
(3) decreased vapour pressure, decreased boiling point
(4) decreased vapour pressure, increased boiling point
Show Answer
Answer (4)
Solution
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.
