General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 6 - Gases - Exercises - The Simple Gas Laws - Page 232: 17

Answer

0.132 g

Work Step by Step

We find: 1 mol gas occupies 22.711 L at STP. $\implies 75.0\,mL=75.0\times10^{-3}\,L\times\frac{1\,mol}{22.711\,L}$ $=0.00330\,mol $ Mass of argon= $0.00330\,mol\,Ar\times\frac{39.948\,g}{1\,mol\,Ar}$ $=0.132\,g$
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