JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Solutions

Consider a binary solution of two volatile liquid components 1 and

2 x_{1}

and

y_{1}

are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of

\frac{1}{x _{1}}

\frac{1}{y _{1}}

are given respectively as:

(A) $\frac{P _{1}^{0}}{P _{2}^{0}}, \frac{P _{2}^{0} - P _{1}^{0}}{P _{2}^{0}}$ ✓ Correct
(B) $\frac{P _{2}^{0}}{P _{1}^{0}}, \frac{P _{1}^{0} - P _{2}^{0}}{P _{2}^{0}}$
(C) $\frac{P _{1}^{0}}{P _{2}^{0}}, \frac{P _{1}^{0} - P _{2}^{0}}{P _{2}^{0}}$
(D) $\frac{P _{2}^{0}}{P _{1}^{0}}, \frac{P _{2}^{0} - P _{1}^{0}}{P _{2}^{0}}$

Answer: (A)

Solution

For an ideal binary solution obeying Raoult's law,

P_{1} = P_{1}^{\circ} x_{1}

and also

P_{1} = P_{T} y_{1}

, so

P_{1}^{\circ} x_{1} = P_{T} y_{1}

. The total pressure is

P_{T} = P_{2}^{\circ} + x_{1} (P_{1}^{\circ} - P_{2}^{\circ})

. Substituting and rearranging in terms of

1/ x_{1}

and

1/ y_{1}

\frac{1}{x _{1}} = (\frac{P _{1}^{\circ}}{P _{2}^{\circ}}) \frac{1}{y _{1}} + (\frac{P _{2}^{\circ} - P _{1}^{\circ}}{P _{2}^{\circ}})

Comparing with

y = m x + c

, the slope of the

1/ x_{1}

1/ y_{1}

plot is

P_{1}^{\circ} / P_{2}^{\circ}

and the intercept is

(P_{2}^{\circ} - P_{1}^{\circ}) / P_{2}^{\circ}

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option