College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 13 - Problems - Page 498: 39


There are $2.49\times 10^{19}$ molecules in $1.0~cm^3$ of air.

Work Step by Step

We can convert the average molecular mass to units of $kg$: $m = 29.0~u\times \frac{1.66\times 10^{-27}~kg}{1~u} = 4.814\times 10^{-26}~kg$ We can find the mass of $1.0~cm^3$ of air: $1.2~kg/m^3\times \frac{1~m^3}{10^6~cm^3} = 1.2\times 10^{-6}~kg/cm^3$ We can find the number of molecules in $1.0~cm^3$ of air: $\frac{1.2\times 10^{-6}~kg/cm^3}{4.814\times 10^{-26}~kg/molecule} = 2.49\times 10^{19}~molecules/cm^3$ There are $2.49\times 10^{19}$ molecules in $1.0~cm^3$ of air.
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