   # The vapor in equilibrium with a pentane-hexane solution at 25°C has a mole fraction of pentane equal to 0.15 at 25°C. What is the mole fraction of pentane in the solution? (See Exercise 57 for the vapor pressures of the pure liquids.) ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 10, Problem 123CP
Textbook Problem
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## The vapor in equilibrium with a pentane-hexane solution at 25°C has a mole fraction of pentane equal to 0.15 at 25°C. What is the mole fraction of pentane in the solution? (See Exercise 57 for the vapor pressures of the pure liquids.)

Interpretation Introduction

Interpretation: The mole fraction of Pentane in solution has to be calculated.

Concept Introduction: Raoult’s law for ideal solution states that the mole fraction of the solvent is directly proportional to the vapor pressure of an ideal solution. Raoult’s law can be expressed by the equation,

Psolutionsolventsolvent

Where, Psolution   = observed vapor pressure of the solution

χsolvent    = mole fraction of solvent

solvent = vapor pressure of pure solvent

### Explanation of Solution

Record the given info

Vapor in equilibrium = 25°C

Mole fraction of Pentane = 0.15 at 25°C

To mole fraction of Pentane in solution

Vapor pressure of Pentane= 511torr

Vapor pressure Hexane = 145torr

χPenV=0.15=PPenPtotalPPenPenLPenPtotal=PPen+PHexPenL(511)+χHexL(150.)Since,χHexL=1.000-χPenLPtotalPenL(511)+(1

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