Chapter 3, Problem 3.56QE

(a)

Interpretation Introduction

Interpretation:

Number of moles present in $\text{1}{\text{.33×10}}^{\text{26}}{\text{ molecules of Br}}_{\text{2}}$ 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 ${\text{1mol Br}}_{\text{2}}\text{= 6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{Br}}_{\text{2}}$.

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

    $\begin{array}{l}{\text{Amount Br}}_{\text{2}}\text{= 1}{\text{.33×10}}^{\text{26}}{\text{ molecules of Br}}_{\text{2}}\text{×}\left(\frac{{\text{1mol Br}}_{\text{2}}}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{Br}}_{\text{2}}}\right)\\ \text{ = 2}{\text{.20×10}}^{\text{2}}{\text{ mol Br}}_{\text{2}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Number of moles present in $\text{7}{\text{.71×10}}^{\text{26}}{\text{ molecules of C}}_{\text{5}}{\text{H}}_{\text{12}}$ 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 ${\text{1mol C}}_{\text{5}}{\text{H}}_{\text{12}}\text{= 6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$.

Therefore, number of moles of ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$ is:

    $\begin{array}{l}{\text{Amount C}}_{\text{5}}{\text{H}}_{\text{12}}\text{ = 7}{\text{.71×10}}^{\text{26}}{\text{ molecules of C}}_{\text{5}}{\text{H}}_{\text{12}}\text{ ×}\left(\frac{{\text{1mol C}}_{\text{5}}{\text{H}}_{\text{12}}}{\text{6}{\text{.022×10}}^{\text{23}}{\text{atoms of C}}_{\text{5}}{\text{H}}_{\text{12}}}\right)\\ \text{ = 1}{\text{.3×10}}^{\text{2}}{\text{ mol C}}_{\text{5}}{\text{H}}_{\text{12}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Number of moles present in $\text{2}{\text{.34×10}}^{23}{\text{ molecules of B}}_{\text{2}}{\text{H}}_6$ 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 ${\text{1mol B}}_{\text{2}}{\text{H}}_6\text{= 6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{B}}_{\text{2}}{\text{H}}_6$.

Therefore, number of moles of ${\text{B}}_{\text{2}}{\text{H}}_6$ is:

    $\begin{array}{l}{\text{Amount B}}_{\text{2}}{\text{H}}_6\text{= 2}{\text{.34×10}}^{23}{\text{ molecules of B}}_{\text{2}}{\text{H}}_6\text{×}\left(\frac{{\text{1mol B}}_{\text{2}}{\text{H}}_6}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{B}}_{\text{2}}{\text{H}}_6}\right)\\ \text{ = }0.388{\text{ mol B}}_{\text{2}}{\text{H}}_6\end{array}$

(d)

Interpretation Introduction

Interpretation:

Number of moles present in $\text{7}{\text{.76 ×10}}^{\text{23}}\text{ molecules of Ne}$ 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 $\text{1mol Ne = 6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{Ne}$.

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

    $\begin{array}{l}\text{Amount Ne= 7}{\text{.76×10}}^{\text{23}}\text{ molecules of Ne×}\left(\frac{\text{1mol Ne}}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{Ne}}\right)\\ \text{ = 1}\text{.28 mol Ne}\end{array}$