Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 13 Solutions and Their Behavior - Study Questions - Page 505e: 83


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Work Step by Step

From Henry's Law: $S=k_HP=2.4\times10^{-2}\ mol/\times1.00\ bar=2.4\times10^{-2}\ mol/kg$ Mass of water: $500.\ mL\times 1.00\ g/mL=500\ g$ N. of moles of N2O: $2.4\times10^{-2}\ mol/kg\times 500\ g\div 1000\ g/kg=0.012\ mol$ Mass of N2O: $0.012\ mol\times 44.01\ g/mol=0.528\ g$ ppm concentration: $0.528/(500+0,528)\times10^6\ ppm=1055\ ppm$
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