Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 5 - Gases - Exercises: 80

Answer

$1.30\text { grams/Liter}$

Work Step by Step

We assume that we are dealing with only one mole of air. 1 mole of gas at STP = 22.42 Liters $0.78\text{ mol $N_2$}*\frac{28.0134 \text{ grams $N_2$}}{1 \text{ mol $N_2$}}=22\text{ grams $N_2$}$ $0.21\text{ mol $O_2$}*\frac{31.9988 \text{ grams $O_2$}}{1 \text{ mol $O_2$}}=6.7\text{ grams $O_2$}$ $0.010\text{ mol Ar}*\frac{39.948\text{ grams Ar}}{1 \text{ mol Ar}}=0.40\text{ grams Ar}$ Total mass = $22+6.7+0.40=29.1\text{ grams}$ Volume = 22.42 L Density = $\frac{\text{Mass}}{\text{ Volume}}=\frac{29.1}{22.42}=1.30\text { grams/Liter}$
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