Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 8 - Properties of Gases - Questions for Review and Thought - Topical Questions - Page 370a: 26a

Answer

42.8 mol

Work Step by Step

$V= 1.05\times10^{3}\,L$ $P=745\, mmHg\times\frac{1\,atm}{760\, mmHg}=0.980\,atm$ $T= (20+273)K= 293\,K$ $R=0.0821\,L\,atm\,mol^{-1}K^{-1}$ $PV=nRT$ (ideal gas law) $\implies n=\frac{PV}{RT}=\frac{0.980\,atm\times 1.05\times10^{3}\,L}{(0.0821\,L\,atm\,mol^{-1}K^{-1})(293\,K)}$ $=42.8\,mol$
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