   # A solution is prepared by mixing 25 mL pentane (C 5 H 12 , d = 0.63 g/cm 3 ) with 45 mL hexane (C 6 H 14 , d = 0.66 g/cm 3 ). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the pentane. ### 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 33E
Textbook Problem
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## A solution is prepared by mixing 25 mL pentane (C5H12, d = 0.63 g/cm3) with 45 mL hexane (C6H14, d = 0.66 g/cm3). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the pentane.

Interpretation Introduction

Interpretation:

The mass percent, molality, molarity and mole fraction of pentane has to be calculated.

### Explanation of Solution

To calculate the mass of Pentane and Hexane:

Record the given data

Volume of Pentane = 25.0mL

Density of Pentane=0.63gcm-3

Volume of Hexane = 45mL

Density of Hexane = 0.66gcm-3

Mass of Pentane = 25mLC5H12×0.63gmL

= 16g

Mass of Hexane = 45mLC6H14×0.66gmL

= 30g

Mass of Pentane = 16g

Mass of Hexane = 30g

To calculate the mass percent of Pentane:

Total mass of the compound = 16g+30g=46g

Mass of Pentane                  = 16g

Masspercentage(%)=16g46g×100

= 35%

Mass percentage of Pentane = 35%

To calculate the moles of Pentane and Hexane:

Record the given info

Mass of Pentane = 16g

Mass of Hexane = 30g

Molar mass of Pentane= 72.15g

Molar mass of Hexane =86.17g

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

=0

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