Chapter 3, Problem 3.55QE

(a)

Interpretation Introduction

Interpretation:

Number of moles present in $\text{3}{\text{.44×10}}^{\text{24}}{\text{ molecules of O}}_{\text{2}}$ has to be stated.

Explanation of Solution

  $\text{1mol=6}{\text{.022×10}}^{\text{23}}\text{molecules}$.

$\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}$ is known as Avogadro’s number.

From Avogadro’s number, it is clear that ${\text{1mol O}}_2\text{= 6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{O}}_2$.

Therefore, number of moles of ${\text{O}}_{\text{2}}$ is:

    $\begin{array}{l}{\text{Amount O}}_{\text{2}}\text{= 3}{\text{.44×10}}^{\text{24}}{\text{ molecules of O}}_{\text{2}}\text{×}\left(\frac{{\text{1mol O}}_{\text{2}}}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{O}}_{\text{2}}}\right)\\ \text{ = 5}{\text{.71 mol O}}_{\text{2}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Number of moles present in $\text{1}{\text{.11×10}}^{\text{22}}\text{ atoms of Na}$ has to be stated.

Explanation of Solution

  $\text{1mol=6}{\text{.022×10}}^{\text{23}}\text{atoms}$.

$\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}$ is known as Avogadro’s number.

From Avogadro’s number, it is clear that $\text{1mol Na= 6}{\text{.022×10}}^{\text{23}}\text{atoms}\text{Na}$.

Therefore, number of moles of $\text{Na}$ is:

    $\begin{array}{l}\text{Amount Na = 1}{\text{.11×10}}^{\text{22}}\text{ atoms of Na ×}\left(\frac{\text{1mol Na}}{\text{6}{\text{.022×10}}^{\text{23}}\text{atoms of Na}}\right)\\ \text{ = 0}\text{.018 mol Na}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Number of moles present in $\text{5}{\text{.57×10}}^{\text{30}}{\text{ molecules of C}}_{\text{2}}{\text{H}}_{\text{6}}$ has to be stated.

Explanation of Solution

  $\text{1mol=6}{\text{.022×10}}^{\text{23}}\text{molecules}$.

$\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}$ is known as Avogadro’s number.

From Avogadro’s number, it is clear that ${\text{1mol C}}_{\text{2}}{\text{H}}_{\text{6}}\text{= 6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$.

Therefore, number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is:

    $\begin{array}{l}{\text{Amount C}}_{\text{2}}{\text{H}}_{\text{4}}\text{= 5}{\text{.57×10}}^{30}{\text{ molecules of C}}_{\text{2}}{\text{H}}_{\text{6}}\text{×}\left(\frac{{\text{1mol C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\right)\\ \text{ = 9}{\text{.24×10}}^6{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{6}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Number of moles present in $\text{1}{\text{.66×10}}^{\text{24}}\text{ molecules of CO}$ has to be stated.

Explanation of Solution

  $\text{1mol=6}{\text{.022×10}}^{\text{23}}\text{molecules}$.

$\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}$ is known as Avogadro’s number.

From Avogadro’s number, it is clear that $\text{1mol CO= 6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{CO}$.

Therefore, number of moles of $\text{CO}$ is:

    $\begin{array}{l}\text{Amount CO= 1}{\text{.66×10}}^{\text{24}}\text{ molecules of CO×}\left(\frac{\text{1mol CO}}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{CO}}\right)\\ \text{ = 2}\text{.75 mol CO}\end{array}$