   # Pentane (C 5 H 12 ) and hexane (C 6 H 14 ) form an ideal solution. At 25°C the vapor pressures of pentane and hexane are 511 and 150. torr, respectively. A solution is prepared by mixing 25 mL pentane (density, 0.63 g/mL) with 45 mL hexane (density, 0.66 g/mL). a. What is the vapor pressure of the resulting solution? b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? ### 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 57E
Textbook Problem
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## Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25°C the vapor pressures of pentane and hexane are 511 and 150. torr, respectively. A solution is prepared by mixing 25 mL pentane (density, 0.63 g/mL) with 45 mL hexane (density, 0.66 g/mL).a. What is the vapor pressure of the resulting solution?b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?

a)

Interpretation Introduction

Interpretation: The vapor pressure of solution along with mole fraction composition of Pentane in the vapor 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 pressure of Pentane= 511torr

Vapor pressure Hexane = 145torr

Density of Pentane = 0.63gml-1

Volume of Pentane = 25mL

Density of Hexane = 0.66gml-1

Volume of Pentane = 45mL

To calculate the total moles of solution

Molecular weight of Pentane = 72.15g

Molecular weight of Hexane = 86.17g

Moles of Pentane = 25mLC5H12×0.63gmL×1mol72.15g

= 0.22mol

Moles of Hexane = 45mLC6H14×0.66gmL×1mol86.17g

= 0.34mol

Total moles = 0.34mol+0.22mol

= 0.56mol

Total moles in solution = 0.56mol

To calculate the mole fraction of Pentane and Hexane

Mole fraction of Pentane = MolepentaneinsolutionTotalmolesinsolution

= 0.22mol0.56mol

= 0

b)

Interpretation Introduction

Interpretation: The vapor pressure of solution along with mole fraction composition of Pentane in the vapor 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

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