   # A solution is prepared by mixing 50.0 mL toluene (C 6 H 5 CH 3 , d = 0.867 g/cm 3 ) with 125 mL benzene (C 6 H 6 , d = 0.874 g/cm 3 ). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene. ### 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 34E
Textbook Problem
1739 views

## A solution is prepared by mixing 50.0 mL toluene (C6H5CH3, d = 0.867 g/cm3) with 125 mL benzene (C6H6, d = 0.874 g/cm3). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

Interpretation Introduction

Interpretation:

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

### Explanation of Solution

To calculate the mass of Benzene and Toluene:

Record the given info

Volume of Toluene = 50.0mL

Density of Toluene =0.867gcm-3

Volume of Benzene = 125mL

Density of Benzene = 0.874gcm-3

Mass of Benzene = 125mLbenzene×0.874gmL

=109g

Mass of Toluene = 50.0mL×0.867gmL

=43.4g

Mass of Benzene = 109g

Mass of Toluene = 43.4g

To calculate the mass percent of Toluene

Total mass of the compound = 109g+43.4g=152.4g

Mass of Toluene                  = 43.4g

Masspercentage(%)=43.4g152.4g×100

= 28.5%

Mass percentage of Toluene = 28.5%

To calculate the moles of Toluene and Benzene:

Record the given data

Mass of Benzene = 109g

Mass of Toluene = 43.4g

Molar mass of Benzene = 78.11g

Molar mass of Toluene =92.13g

Moles of Toluene = 43.4gtoluene×1mol92.13g

=0.471mol

Moles of Benzene = 109gbenzene×1mol78.11g

= 1

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