Chapter 4, Problem 4CP

(a)

Interpretation Introduction

Interpretation:

The number of molecules present in $\text{4}\text{×}{\text{10}}^{\text{5}}\text{tons}$ of ${\text{SO}}_{\text{2}}$ has to be found.

Explanation of Solution

Given,

The mass of ${\text{SO}}_{\text{2}}$ is $\text{4}\text{×}{\text{10}}^{\text{5}}\text{tons}$.

  $\begin{array}{l}\text{1}\text{ton}\text{=}\text{1000}\text{kg}\\ \text{4}{\text{×10}}^{\text{5}}\text{tons}\text{=}\text{4}{\text{×10}}^{\text{5}}\text{×}\text{1000}\text{=}{\text{4×10}}^{\text{8}}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Mass}\text{in}\text{gram}}{\text{Molar}\text{mass}}\\ \text{Molar}\text{mass}\text{of}{\text{SO}}_{\text{2}}\text{=}\text{(32}\text{.065}\text{g/mol}\text{S)}\text{+}\text{(2}\text{×}\text{15}\text{.9994}\text{g/mol}\text{O)}\\ \text{=}\text{64}\text{.0638}\text{g/mol}\\ \text{Mass}\text{=}\text{4}\text{×}{\text{10}}^{\text{8}}\text{kg}\text{=}\text{4}\text{×}{\text{10}}^{\text{11}}\text{g}\\ \text{Number}\text{of}\text{moles}\text{=}\frac{\text{4}\text{×}{\text{10}}^{\text{11}}\text{g}}{\text{64}\text{.0638}\text{g/mol}}\text{=}6.24\text{×}{\text{10}}^{\text{9}}\text{moles}\\ \text{Number}\text{of}\text{molecules}\text{=}\text{6}\text{.24×}{\text{10}}^{\text{9}}\text{moles}\text{×}\text{6}{\text{.022×10}}^{\text{23}}\\ \text{=}\text{3}{\text{.76×10}}^{\text{33}}\text{molecules}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The number of molecules present in $\text{25}\text{.0}\text{lb}$ of ${\text{SO}}_{\text{2}}$ has to be found.

Explanation of Solution

Given,

The mass of ${\text{SO}}_{\text{2}}$ is $\text{25}\text{.0}\text{lb}$.

  $\begin{array}{l}\text{1 lb = 0}\text{.45359 kg}\\ \text{Mass}\text{in}\text{kg}\text{=}25.01\times \text{0}\text{.45359=}\text{11}\text{.344}\text{kg = 11344}\text{g}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Mass}\text{in}\text{gram}}{\text{Molar}\text{mass}}\\ \text{Molar}\text{mass}\text{of}{\text{SO}}_{\text{2}}\text{=}\text{(32}\text{.065}\text{g/mol}\text{S)}\text{+}\text{(2}\text{×}\text{15}\text{.9994}\text{g/mol}\text{O)}\\ \text{=}\text{64}\text{.0638}\text{g/mol}\\ \text{Mass}\text{=}\text{11344}\text{g}\\ \text{Number}\text{of}\text{moles}\text{=}\frac{\text{11344}\text{g}}{\text{64}\text{.0638}\text{g/mol}}\text{=}\text{177}\text{.07}\text{moles}\\ \text{Number}\text{of}\text{molecules}\text{=}\text{177}\text{.07}\text{moles}\text{×}\text{6}{\text{.022×10}}^{\text{23}}\\ \text{=}\text{1}{\text{.066×10}}^{\text{26}}\text{molecules}\end{array}$