Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 5 - Sections 5.1-5.10 - Exercises - Problems by Topic - Page 240: 55

Answer

$d=9.434\frac{g}{L}$

Work Step by Step

We know that $PV=nRT$ But $n=\frac{m}{M}$ Where $m$ and$M$ are mass in grams and molar mass respectively $\implies PV=\frac{mRT}{M}$ This can be rearranged as $PM=\frac{m}{V}RT$ But $d=\frac{m}{V}$ where $d$ represents density $\implies PM=dRT$ This can be rearranged as $d=\frac{PM}{RT}$.....eq(1) $P=1655psi\times \frac{1atm}{14.7psi}=112.585atm$ $T=20.0+273.15=293.15K$ We plug in the known values in eq(1) to obtain: $d=\frac{112.585\times 2.0158}{0.08206\times 293.15}=9.434\frac{g}{L}$
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