Chemistry: An Introduction to General, Organic, and Biological Chemistry (12th Edition)

Published by Prentice Hall
ISBN 10: 0321908449
ISBN 13: 978-0-32190-844-5

Chapter 8 - Section 8.7 - Partial Pressure (Dalton's Law) - Additional Questions and Problems - Page 279: 8.69

Answer

370 torr

Work Step by Step

$P_{O_{2}}= 0.60\,atm= 0.60\,atm\times\frac{760\,torr}{1\,atm}$ $=456\,torr$ $P_{Ar}=425\, mmHg=425\, mmHg\times\frac{1\,torr}{1\, mmHg}$ $=425\,torr$ $P_{total}=1250\,torr$ By using Dalton's law, we have $P_{total}= P_{O_{2}}+P_{Ar}+P_{N_{2}}$ $\implies P_{N_{2}}= P_{total}-P_{O_{2}}-P_{Ar}$ $=1250\,torr-456\,torr-425\,torr$ $=369\,torr$ As there are only two significant figures in 0.60 atm, reporting the final answer in two significant figures is better. That is, $P_{N_{2}}=370\,torr$
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