Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 41: 126


\[\underline{36.9\times {{10}^{-10}}}\]

Work Step by Step

Radius of neon atom is \[\text{69 pm}\]. \[\text{1 pm}=\text{1}{{\text{0}}^{-12}}\text{ m}\] Thus, radius of neon atom in decimeter is written as follows: \[r=69\times {{10}^{-12}}\text{ m}\] Assuming a spherical shape: \[V=\frac{4}{3}\pi {{r}^{3}}\] Thus, \[\begin{align} & V=\frac{4}{3}\pi {{\left( 69\times {{10}^{-12}}\text{ m} \right)}^{3}} \\ & =1.37\times {{10}^{-30}}\text{ }{{\text{m}}^{3}} \end{align}\] Now, \[\text{1 L}=\text{1}{{\text{0}}^{-3}}\text{ }{{\text{m}}^{3}}\] Thus, \[\begin{align} & V=1.37\times {{10}^{-30}}\,\times \,{{10}^{-3}}\text{ L} \\ & \,\,\,\,\text{=}\,1.37\times {{10}^{-33}}\,\,\text{L} \\ \end{align}\] There are \[2.69\,\times \,{{10}^{22}}\,Ne\,atoms\,\] present in \[1\,L\]of gaseous sample. Thus, the volume occupied by \[2.69\,\times \,{{10}^{22}}\,Ne\,atoms\,\]is : \[\begin{align} & 2.69\,\times \,{{10}^{22}}\,\,\,\times \,\,1.37\times {{10}^{-33}}\,\,\text{L}\, \\ & =\,3.68\times {{10}^{-11}}\,\,L\, \\ \end{align}\] The total volume of the gaseous sample is \[1\,L\]. The fraction of space occupied by the atoms is the ratio of volume occupied by the atoms to total volume of the gaseous sample. Thus, \[\begin{align} & \text{Fraqction of space}=\frac{3.69\times {{10}^{-11}}\,\,\text{L}}{1\text{ L}} \\ & =3.69\times {{10}^{-11}}\,\approx \,\,36.9\times {{10}^{-10}} \end{align}\] The fraction of the space occupied by the atoms is \[\underline{\,\,36.9\times {{10}^{-10}}}\].
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