Consider a binary solution of two volatile liquid components, 1 and 2. x₁ and y₁ are the mole fractions of component 1 in liquid and vapour phases, respectively. The slope and intercept of the linear plot of $\frac{1}{x_{1}} vs \frac{1}{y_{1}}$ Are given respectively as :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{P_{1}^{0}}{P_{2}^{0}}, \frac{P_{1}^{0} - P_{2}^{0}}{P_{2}^{0}}$

$\frac{P_{2}^{0}}{P_{1}^{0}}, \frac{P_{2}^{0} - P_{1}^{0}}{P_{2}^{0}}$

$\frac{P_{1}^{0}}{P_{2}^{0}}, \frac{P_{2}^{0} - P_{1}^{0}}{P_{2}^{0}}$

$\frac{P_{2}^{0}}{P_{1}^{0}}, \frac{P_{1}^{0} - P_{2}^{0}}{P_{2}^{0}}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$∵$ For liquid solution of two liquids ‘1’ and ‘2’

$\begin{array}{l} P_{1} = P_{T} y_{1} = P_{1}^{\circ} x_{1} \\ ∴ \frac{P_{T}}{x_{1}} = \frac{P_{1}^{\circ}}{y_{1}} \\ ∴ \frac{P_{2}^{\circ} + x_{1} (P_{1}^{\circ} - P_{2}^{\circ})}{x_{1}} = \frac{P_{1}^{\circ}}{y_{1}} \\ \begin{array}{l} ∴ \frac{P_{2}^{\circ}}{x_{1}} + (P_{1}^{\circ} - P_{2}^{o}) = \frac{P_{1}^{\circ}}{y_{1}} \\ ∴ \frac{1}{x_{1}} = (\frac{P_{1}^{o}}{P_{2}^{o}}) (\frac{1}{y_{1}}) + (\frac{P_{2}^{o} - P_{1}^{o}}{P_{2}^{\circ}}) \\ ∴ Slope = (\frac{P_{1}^{o}}{P_{2}^{o}}) \\ ∴ Intercept = (\frac{P_{2}^{o} - P_{1}^{o}}{P_{2}^{o}}) \end{array} \end{array}$