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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

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(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.

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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.

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Most popular questions from this chapter

At \(300 \mathrm{~K}\), the vapour pressure of a solution containing 1 mole of \(n\) hexane and 3 moles of \(n\)-heptane is \(550 \mathrm{~mm}\) of \(\mathrm{Hg}\). At the same temperature, if one more mole of \(n\)-heptane is added to this solution, the vapour pressure of the solution increases by \(10 \mathrm{~mm}\) of \(\mathrm{Hg}\). What is the vapour pressure in \(\mathrm{mm} \mathrm{Hg}\) of \(n\)-heptane in its pure state _____ ?

The vapour pressures of pure liquids \(\mathrm{A}\) and \(\mathrm{B}\) are 400 and \(600 \mathrm{mmHg}\), respectively at \(298 \mathrm{~K}\). On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid \(\mathrm{B}\) is \(0.5\) in the mixture. The vapour pressure of the final solution, the mole fractions of components \(\mathrm{A}\) and \(\mathrm{B}\) in vapour phase, respectively are: (a) \(450 \mathrm{mmHg}, 0.4,0.6\) (b) \(500 \mathrm{mmHg}, 0.5,0.5\) (c) \(450 \mathrm{mmHg}, 0.5,0.5\) (d) \(500 \mathrm{mmHg}, 0.4,06\)

An azeotropic solution of two liquids has boiling point lower than either of them when it (a) shows negative deviation from Raoult's law (b) shows no deviation from Raoult's law (c) shows positive deviation from Raoult's law (d) is saturated

At room temperature, a dilute solution of urea is prepared by dissolving \(0.60 \mathrm{~g}\) of urea in \(360 \mathrm{~g}\) of water. If the vapour pressure of pure water at this temperature is \(35 \mathrm{mmHg}\), lowering of vapour pressure will be : (molar mass of urea \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(0.027 \mathrm{mmHg}\) (b) \(0.028 \mathrm{mmHg}\) (c) \(0.017 \mathrm{mmHg}\) (d) \(0.031 \mathrm{mmHg}\)

Dissolving \(120 \mathrm{~g}\) of urea (mol. wt. 60 ) in \(1000 \mathrm{~g}\) of water gave a solution of density \(1.15 \mathrm{~g} / \mathrm{mL}\). The molarity of the solution is (a) \(1.78 \mathrm{M}\) (b) \(2.00 \mathrm{M}\) (c) \(2.05 \mathrm{M}\) (d) \(2.22 \mathrm{M}\)
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