Chapter 7, Problem 12PE

(a)

Interpretation Introduction

Interpretation:

The number of atoms in $\text{2}$ molecules of ${\text{CH}}_{\text{3}}\text{COOH}$ has to be given.

Answer to Problem 12PE

The number of atoms in $\text{2}$ molecules of ${\text{CH}}_{\text{3}}\text{COOH}$ is $\text{16}\text{atoms}\text{.}$

Explanation of Solution

Given,

The number of molecules of ${\text{CH}}_{\text{3}}\text{COOH}$ is $\text{2}\text{.}$

There are eight atoms in ${\text{CH}}_{\text{3}}\text{COOH}$ (two carbon atoms, two oxygen atoms and four hydrogen atoms).

The number of atoms in $\text{2}$ molecules of ${\text{CH}}_{\text{3}}\text{COOH}$ can be calculated as,

  $\left(\text{2}\text{molecules}\right)\text{×}\left(\frac{8\text{atoms}}{\text{1}\text{molcule}}\right)\text{=16}\text{atoms}$

The number of atoms in $\text{2}$ molecules of ${\text{CH}}_{\text{3}}\text{COOH}$ is $\text{16}\text{atoms}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The number of atoms in $0.75\text{mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ has to be given.

Answer to Problem 12PE

The number of atoms in $0.75\text{mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is $\text{36}{\text{.132×10}}^{\text{23}}\text{atoms}\text{.}$

Explanation of Solution

Given,

The number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is $0.75\text{mol}\text{.}$

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

There are eight atoms in ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{.}$

The number of atoms in $0.75\text{mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ can be calculated as,

  $\left(0.75\text{moles}\right)\text{×}\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}}\right)\left(\frac{8\text{atoms}}{\text{1}\text{molcule}}\right)\text{=36}{\text{.132×10}}^{\text{23}}\text{atoms}$

The number of atoms in $0.75\text{mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is $\text{36}{\text{.132×10}}^{\text{23}}\text{atoms}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The number of atoms in $25\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ has to be given.

Answer to Problem 12PE

The number of atoms in $25\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ is $\text{4}{\text{.5×10}}^{\text{25}}\text{atoms}\text{.}$

Explanation of Solution

Given,

The number of moles of ${\text{H}}_{\text{2}}\text{O}$ is $25\text{mol}\text{.}$

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

There are three atoms in ${\text{H}}_{\text{2}}\text{O}\text{.}$

The number of atoms in $25\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ can be calculated as,

$\left(25\text{moles}\right)\text{×}\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{1}\text{mol}}\right)\left(\frac{3\text{atoms}}{\text{1}\text{molcule}}\right)\text{=451}{\text{.65×10}}^{\text{23}}\text{atoms}=4.5{\text{×10}}^{\text{25}}\text{atoms}$

The number of atoms in $25\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ is $\text{4}{\text{.5×10}}^{\text{25}}\text{atoms}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The number of atoms in $92.5\text{g}$ of $\text{Au}$ has to be given.

Answer to Problem 12PE

The number of atoms in $92.5\text{g}$ of $\text{Au}$ is $\text{2}{\text{.83×10}}^{\text{23}}\text{ atoms}\text{.}$

Explanation of Solution

Given,

The molar mass of $\text{Au}$ is $197\text{g/mol}\text{.}$

The mass of $\text{Au}$ is $92.5\text{g}\text{.}$

The Avogadro’s number is $\text{6}{\text{.022×10}}^{\text{23}}\text{atoms}\text{.}$

The number of atoms in $92.5\text{g}$ of $\text{Au}$ can be calculated as,

  $\text{92}\text{.5}\text{g×}\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{atoms}}{\text{197}\text{g}}\right)\text{=2}{\text{.83×10}}^{\text{23}}\text{atoms}$

The number of atoms in $92.5\text{g}$ of $\text{Au}$ is $\text{2}{\text{.83×10}}^{\text{23}}\text{ atoms}\text{.}$

(e)

Interpretation Introduction

Interpretation:

The number of atoms in $75\text{g}$ of ${\text{PCl}}_{\text{3}}$ has to be given.

Answer to Problem 12PE

The number of atoms in $75\text{g}$ of ${\text{PCl}}_{\text{3}}$ is $\text{1}{\text{.3×10}}^{\text{24}}\text{ atoms}\text{.}$

Explanation of Solution

Given,

The molar mass of ${\text{PCl}}_{\text{3}}$ is $137.3\text{g/mol}\text{.}$

The mass of ${\text{PCl}}_{\text{3}}$ is $75\text{g}\text{.}$

The Avogadro’s number is $\text{6}{\text{.022×10}}^{\text{23}}\text{atoms}\text{.}$

There are four atoms in ${\text{PCl}}_{\text{3}}.$

The number of atoms in $75\text{g}$ of ${\text{PCl}}_{\text{3}}$ can be calculated as,

  $\text{75}\text{g×}\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{atoms×4}}{137.3\text{g}}\right)\text{=13}{\text{.15×10}}^{\text{23}}\text{atoms=1}{\text{.3×10}}^{\text{24}}\text{atoms}$

The number of atoms in $75\text{g}$ of ${\text{PCl}}_{\text{3}}$ is $\text{1}{\text{.3×10}}^{\text{24}}\text{ atoms}\text{.}$

(f)

Interpretation Introduction

Interpretation:

The number of atoms in $15\text{g}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ has to be given.

Answer to Problem 12PE

The number of atoms in $15\text{g}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is $\text{1}{\text{.2×10}}^{\text{24}}\text{ atoms}\text{.}$

Explanation of Solution

Given,

The molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is $180.2\text{g/mol}\text{.}$

The mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is $15\text{g}\text{.}$

The Avogadro’s number is $\text{6}{\text{.022×10}}^{\text{23}}\text{atoms}\text{.}$

There are $\text{24}$ atoms in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}.$

The number of atoms in $15\text{g}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ can be calculated as,

  $\text{15g×}\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{atoms×24}}{\text{180}\text{.2}\text{g}}\right)\text{=12}{\text{.03×10}}^{\text{23}}\text{atoms=1}{\text{.2×10}}^{\text{24}}\text{atoms}$

The number of atoms in $15\text{g}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is $\text{1}{\text{.2×10}}^{\text{24}}\text{ atoms}\text{.}$