Answer
30.0 g/mol
Work Step by Step
We find:
$P= 725\,Torr\times\frac{1\,atm}{760\,Torr}=0.953947\,atm$
$M=\frac{dRT}{P}=\frac{(0.841\,g/L)(0.08206\,L\,atm\,K^{-1}mol^{-1})(415\,K)}{0.953947\,atm}$
$=30.0\,g/mol$
General Chemistry: Principles and Modern Applications (10th Edition)
by Petrucci, Ralph H.; Herring, F. Geoffrey; Madura, Jeffry D.; Bissonnette, Carey
Published by
Pearson Prentice Hal
ISBN 10:
0132064529
ISBN 13:
978-0-13206-452-1
Chapter 6 - Gases - Exercises - Determining Molar Mass - Page 233: 36
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