Chapter 10, Problem 91A

Interpretation Introduction

Interpretation: If a typical virus has a length $5\times {10}^{-6}\text{cm}$ diameter, the length of the line needs to be calculated consisting of Avogadro’s number of virus particles arranged in a row.

Concept Introduction: Empirical formula of a compound shows the simplest ratio of atoms present in the compound, on the other hand, the molecular formula shows the actual number of moles of atoms present in the compound.

Explanation of Solution

The length of a virus is given $5\times {10}^{-6}\text{cm}$ . Now, the length of Avogadro’s number of the virus will be equal to the length of 1 virus multiplied by Avogadro’s number.

The length of the line will be:

  $L=l\times N_{\text{A}}$

Substitute the values,

  $\begin{array}{c}L=\frac{5\times {10}^{-6}\text{cm}}{1\text{ virus}}\times 6.022\times {10}^{23}\text{ virus}\\ =\text{30}\text{.11}\times {\text{10}}^{-6+23}\text{ cm}\\ =3.011\times {10}^{18}\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\left(\frac{{10}^{-3}\text{ m}}{1\text{ m}}\right)\\ =3.011\times {10}^{18-2-3}\\ =3.011\times {10}^{13}\text{ km}\end{array}$

Therefore, the length of Avogadro’s number of viruses arranged in a row is $3.011\times {10}^{13}\text{ km}$ .