JEE Advanced 2017 · Paper 1 · Chemistry Q2

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ChemistryMultiple correct+4 / -2Question 20 of 54

For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here $x_\mathrm{L}$ and $x_\mathrm{M}$ represent mole fractions of L and M, respectively, in the solution. The correct statement(s) applicable to this system is(are)

Plot of p_L (vapour pressure of L) against x_M, with x_M running from 1 at the left to 0 at the right. The curve starts at zero at x_M = 1, rises steeply, flattens into a shallow plateau with a slight dip, and rises again to the point Z at x_M = 0 on the right axis.

The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed from

x_\mathrm{L} = 0

x_\mathrm{L} = 1

The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when

x_\mathrm{L} \to 1

The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed when

x_\mathrm{L} \to 0

Attractive intermolecular interactions between L-L in pure liquid L and M-M in pure liquid M are stronger than those between L-M when mixed in solution

Answer and worked solution

Official answer: B, D

Read the axes. $x_M$ runs from 1 at the left to 0 at the right. So at Z, $x_M=0$ : pure L, and $p_L=p_L^\circ$ .

Deviation. The curve lies above the straight Raoult line joining $(x_M=1,\,0)$ and Z. That is a positive deviation.

(B) Z is pure L. Near $x_L\to1$ the curve meets the Raoult line, so the solvent L obeys Raoult's law there. True.
(A) and (C) say Z is pure M. False.
(D) A positive deviation means the L-M attractions are weaker than the L-L and M-M attractions. True.

(B), (D)

Topics:Ideal and non-ideal solutions Raoult's law and vapour pressure

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JEE Advanced 2017 · Paper 1, Chemistry question 2