/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Mixture(s) showing positive devi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 30

Mixture(s) showing positive deviation from Raoult's law at \(35{ }^{\circ} \mathrm{C}\) is (are) (a) carbon tetrachloride + methanol (b) carbon disulphide + acetone (c) benzene \(+\) toluene (d) phenol + aniline

Short Answer

Expert verified
(a) and (b) show positive deviation from Raoult's law.

Step by step solution

01

Understanding Positive Deviation

A mixture shows positive deviation from Raoult's law if the interactions between the molecules of the different components are weaker than the interactions between the like molecules. This results in a higher vapor pressure than predicted.
02

Analyzing Options for Interaction

For option (a) carbon tetrachloride + methanol, the interactions between non-polar carbon tetrachloride and polar methanol are weaker than those within each component, leading to positive deviation. For option (b) carbon disulphide + acetone, similar weaker interactions occur between non-polar and polar components. Option (c), benzene and toluene, both non-polar, show no such deviation due to comparable interactions. Option (d), phenol and aniline, both polar, also show no deviation as their interactions are similar.
03

Identifying the Correct Mixtures

Options (a) and (b) contain mixtures where weaker interactions between different molecules lead to positive deviation. Therefore, carbon tetrachloride + methanol and carbon disulphide + acetone demonstrate positive deviation.

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 91影视!

Key Concepts

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

Positive Deviation
Positive deviation from Raoult's law occurs when a mixture has a vapor pressure higher than expected. This happens because the interactions between the molecules of the two different components in the mixture are weaker than the interactions within each pure component.
When mixed, the lack of strong molecular attraction allows more molecules to escape into the vapor phase, resulting in increased vapor pressure.
Hence, the "positive" in positive deviation implies that the actual vapor pressure is more than what Raoult's law predicts.
  • Molecules in a mixture escape more easily.
  • Results in higher vapor pressure.
  • Weaker interactions between different kinds of molecules.
Molecular Interactions
The strength and nature of molecular interactions are crucial to understanding deviations in vapor pressure.
In mixtures that show positive deviation, the interactions between unlike molecules are weaker than those in pure substances.
Consider a mixture of non-polar carbon tetrachloride and polar methanol; these different properties result in weaker interaction forces when mixed.
  • Like molecules have stronger interactions (e.g., hydrogen bonding in methanol).
  • Unlike molecules, typically weak van der Waals forces, cause higher vapor pressure.
  • Polar and non-polar interactions are often weaker.
As a result, unlike molecules in these mixtures have lesser attraction, leading to positive deviation.
Vapor Pressure
Vapor pressure is determined by the tendency of molecules to escape from a liquid to a vapor state.
In mixtures with positive deviation, the vapor pressure is higher because the weaker molecular interactions allow more molecules to vaporize.
Raoult's law provides a theoretical framework for vapor pressure in ideal mixtures, but real solutions often deviate due to molecular interaction discrepancies. Here is how it works:
  • Raoult's law assumes similar intermolecular forces.
  • Positive deviation arises due to weaker intermolecular forces when unlike molecules mix.
  • Increased vapor pressure means more molecules in the gas phase.
Consequently, these weakly interacting mixtures lead to an excess of vapor pressure, going beyond the ideal value.

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

At \(35{ }^{\circ} \mathrm{C}\), the vapour pressure of \(\mathrm{CS}_{2}\) is \(512 \mathrm{~mm} \mathrm{Hg}\) and that of acetone is \(344 \mathrm{~mm} \mathrm{Hg}\). A solution of \(\mathrm{CS}_{2}\) in acetone has a total vapour pressure of \(600 \mathrm{~mm} \mathrm{Hg}\). The false statement amongst the following is: (a) Raoult's law is not obeyed by this system (b) a mixture of \(100 \mathrm{~mL} \mathrm{CS}_{2}\) and \(100 \mathrm{~mL}\) acetone has a volume \(<200 \mathrm{~mL}\) (c) \(\mathrm{CS}_{2}\) and acetone are less attracted to each other than to themselves (d) heat must be absorbed in order to produce the solution at \(35^{\circ} \mathrm{C}\)

The Henry's law constant for the solubility of \(\mathrm{N}_{2}\) gas in water at \(298 \mathrm{~K}\) is \(1.0 \times 10^{5}\) atm. The mole fraction of \(\mathrm{N}_{2}\) in air is \(0.8\). The number of moles of \(\mathrm{N}_{2}\) from air dissolved in 10 moles of water at \(298 \mathrm{~K}\) and 5 atm pressure is (a) \(4.0 \times 10^{-4}\) (b) \(4.0 \times 10^{5}\) (c) \(5.0 \times 10^{-4}\) (d) \(4.0 \times 10^{-6}\)

\(12 \mathrm{~g}\) of a nonvolatile solute dissolved in \(108 \mathrm{~g}\) of water produces the relative lowering of vapour pressure of \(0.1 .\) The molecular mass of the solute is : (a) 80 (b) 60 (c) 20 (d) 40

Elevation in the boiling point for 1 molal solution of glucose is \(2 \mathrm{~K}\). The depression in the freezing point for 2 molal solution of glucose in the same solvent is \(2 \mathrm{~K}\). The relation between \(\mathrm{K}_{\mathrm{b}}\) and \(\mathrm{K}_{\mathrm{f}}\) is: (a) \(\mathrm{K}_{\mathrm{b}}=1.5 \mathrm{~K}_{\mathrm{f}}\) (b) \(\mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{f}}\) (c) \(\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K}_{\mathrm{f}}\) (d) \(\mathrm{K}_{\mathrm{b}}=2 \mathrm{~K}_{\mathrm{f}}\)

Vapour pressure of pure benzene is 119 torr and that of toluene is \(37.0\) torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene \(0.50\), will be : (a) \(0.137\) (b) \(0.237\) (c) \(0.435\) (d) \(0.205\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation
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.