## Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning

# Chapter Review - Exercises - Page 27e: 58

959 g

#### Work Step by Step

Volume of ethanol = 100. mL Density of ethanol = 0.789 $g/cm^{3}$ = 0.789 g/mL $m = V\times D$ Then, Mass of ethanol = $100. mL \times 0.789 g/mL$ = 78.9 g Volume of benzene = 1.00 L = 1000 mL Density of benzene = 0.880 g/mL Mass of benzene = $1000 mL\times 0.880 g/mL$ = 880. g Mass of the mixture = mass of benzene + mass of ethanol = 880. g + 78.9 g = 959 g

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