Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An azeotropic mixture of two liquids has a boiling point higher than either of them when it: (a) shows positive deviation from Raoult's law (b) shows negative deviation from Raoult's law (c) shows ideal behaviour (d) is saturated

Short Answer

Expert verified
The boiling point of an azeotropic mixture is higher than either of its components when it shows positive deviation from Raoult's law (option a).

Step by step solution

01

Understanding Azeotropes

An azeotropic mixture is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. This mixture exhibits a constant boiling point and composition during boiling. An azeotrope that boils at a higher temperature than either of its constituents is known as a positive azeotrope.
02

Identifying Azeotrope Type

A positive azeotrope shows a positive deviation from Raoult's law. This means that the vapor pressure of the mixture is higher than that predicted by Raoult's law. Consequently, the boiling point of the mixture must be higher to reach the ambient pressure, which is characteristic of a positive azeotrope.
03

Analyzing the Options

Option (a) shows a positive deviation from Raoult's law, which is associated with a boiling point higher than either of the pure components. Options (b), (c), and (d) are not associated with an azeotropic mixture that boils at a higher temperature than either of its components.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Positive Deviation from Raoult's Law
Have you ever wondered why some liquid mixtures don't behave as you'd expect when it comes to their boiling points? It turns out there's a concept in physical chemistry that explains this phenomenon, known as positive deviation from Raoult's law.

Raoult's law states that the partial vapor pressure of each component in an ideal solution is directly proportional to its mole fraction. Ideally, the total vapor pressure of the solution would be the sum of the partial pressures of each component. However, some mixtures exhibit a higher vapor pressure than predicted, which is known as a positive deviation.

This occurs because the molecules of the different liquids interact less favorably compared to the interactions in the pure liquids. With weaker intermolecular forces at play, the molecules escape into the vapor phase more readily, resulting in a higher vapor pressure. As a result, such mixtures require a higher temperature to reach the ambient atmospheric pressure, hence they have an elevated boiling point, a characteristic property of an azeotrope exhibiting positive deviation.
  • Less favorable intermolecular forces
  • Increased vapor pressure over predicted
  • Higher boiling temperature to reach ambient pressure
Boiling Point Elevation in Azeotropic Mixtures
Understanding boiling point elevation is crucial when you're diving into physical chemistry and studying for exams like the JEE. In the context of azeotropic mixtures, this term describes a scenario where the boiling point of a mixture is higher than that of the pure components.

For an azeotropic mixture demonstrating a positive deviation from Raoult's law, the boiling point elevation happens due to the weaker intermolecular forces between the different liquid molecules in the mixture. Since these forces are weaker than those in the pure components, more energy (in the form of heat) is needed to overcome the atmospheric pressure, thus elevating the boiling point of the mixture.

The practical relevance of this is seen in distillation processes where azeotropes can be stubborn, as their boiling points do not change with composition. This is why breaking an azeotropic mixture into its individual components through standard distillation is a challenge.
  • Higher boiling point than pure components
  • Weak intermolecular forces require more heat
  • Significance in distillation processes
Navigating Physical Chemistry for JEE
For students preparing for competitive exams such as the JEE (Joint Entrance Examination), grasping the nuances of physical chemistry can make a significant difference. Physical chemistry often deals with the behavior of materials on an atomic or molecular scale, and understanding concepts like azeotropic mixtures and their boiling points is essential.

When tackling problems on azeotropic mixtures for JEE, it's important to recognize the characteristics of mixtures with positive deviation from Raoult's law, as such questions are commonly featured. Mastery of these topics not only helps in answering multiple-choice questions but also in explaining the underlying concepts in a descriptive format.

The JEE syllabus demands a conceptual understanding that goes beyond rote learning, and connecting phenomena such as boiling point elevation to concrete examples of azeotropes is a great way to solidify your grasp of the subject.
  • Understanding atomic or molecular behavior
  • Recognize characteristics of azeotropic mixtures
  • Conceptual understanding for descriptive explanations

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(M_{\text {normal }}\) is the normal molecular mass and \(\alpha\) is the degree of ionization of \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), then the abnormal molecular mass of the complex in the solution will be: (a) \(M_{\text {normal }}(1+2 \alpha)^{-1}\) (b) \(M_{\text {normal }}(1+3 \alpha)^{-1}\) (c) \(M_{\text {normal }}(1+\alpha)^{-1}\) (d) equal to \(M_{\text {normal }}\)

Consider \(0.1 \mathrm{M}\) solutions of two solutes \(X\) and \(Y\). The solute \(X\) behaves as a univalent electrolyte while the solute \(Y\) dimerises in solution. Which of the following statements are correct regarding these solutions? (1) The boiling point of the solution of \(X\) will be higher than that of \(Y\) (2) The osmotic pressure of the solution of \(Y\) will be lower than that of \(X\) (3) The freezing point of the solution of \(X\) will be lower than that of \(Y\) (4) The relative lowering of vapour pressure of both the solutions will be the same Select the correct answer from the option given below : (a) 1,2 and 3 (b) 2,3 and 4 (c) 1,2 and 4 (d) 1,3 and 4

\(1 \mathrm{~kg}\) of water under a nitrogen pressure of 1 atmosphere dissolves \(0.02 \mathrm{gm}\) of nitrogen at 293K. Calculate Henry's law constant : (a) \(7.7 \times 10^{4}\) atm (b) \(7.7 \times 10^{3}\) atm (c) \(2 \times 10^{-5} \mathrm{~atm}\) (d) \(2 \times 10^{-2} \mathrm{~atm}\)

An ideal solution is formed by mixing two volatile liquids \(A\) and \(B . X_{A}\) and \(X_{B}\) are the mole fractions of \(A\) and \(B\) respectively in the solution and \(Y_{A}\) and \(Y_{B}\) are the mole fractions of \(A\) and \(B\) respectively in the vapour phase. A plot of \(\frac{1}{Y_{A}}\) along \(y\) -axis against \(\frac{1}{X_{1}}\) along \(x\) -axis gives a straight line. What is the slope of the straight line? (a) \(\frac{P_{B}^{\circ}}{P_{A}^{\circ}}\) (b) \(\frac{P_{A}^{\circ}}{P_{B}^{\circ}}\) (c) \(P_{B}^{\circ}-P_{A}^{\circ}\) (d) \(P_{A}^{\circ}-P_{B}^{\circ}\)

According to Henry's law, the partial pressure of gas \(\left(P_{g}\right)\) is directly proportional to mole fraction of gas in liquid solution, \(P_{\text {gas }}=K_{H} \cdot X_{\text {gas }}\), where \(K_{H}\) is Henry's constant. Which is incorrect ? (a) \(K_{H}\) is characteristic constant for a given gas-solvent system (b) Higher is the value of \(K_{H}\), lower is solubility of gas for a given partial pressure of gas (c) \(K_{H}\) has temperature dependence (d) \(K_{H}\) decreases with increase of temperature

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free