Chapter F, Problem E.22E

(a)

Interpretation Introduction

Interpretation:

Mass of $\text{3}\text{.60}\text{kg}$ of water has to be converted into moles of water and also into number of molecules.

Answer to Problem E.22E

The total number of molecules and moles of water present in $\text{3}\text{.60}\text{kg}$ of water is $1203.7\times {10}^{23}\text{ molecules}$ and $\text{199}\text{.8}\text{mol}$ respectively.

Explanation of Solution

Given mass of water is $\text{3}\text{.60}\text{kg}$.  Molar mass of water is $18.01\text{g}\cdot {\text{mol}}^{-1}$.  Therefore, the number of molecules present in $\text{3}\text{.60}\text{kg}$ of water can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{molecules}=\frac{6.0221\times {10}^{23}\text{molecules}\text{per}\text{mole}}{18.01\text{g}\cdot {\text{mol}}^{-1}}\times 3600\text{g}\\ =1203.7\times {10}^{23}\text{ molecules}\end{array}$

Therefore, $\text{3}\text{.60}\text{kg}$ of water contains $1203.7\times {10}^{23}\text{ molecules}$ of water.

One mole of water is $\text{18}\text{.01}\text{g}$.  Therefore, the number of moles that is present in $\text{3}\text{.60}\text{kg}$ of water is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{water}}{\text{Molar}\text{mass}\text{of}\text{water}}\\ =\frac{3600\text{g}}{18.01\text{g}\cdot {\text{mol}}^{-1}}\\ =199.8\text{mol}\end{array}$

Thus, $\text{3}\text{.60}\text{kg}$ of water contains $1203.7\times {10}^{23}\text{ molecules}$ and $\text{199}\text{.8}\text{mol}$

(b)

Interpretation Introduction

Interpretation:

Mass of $\text{91}\text{kg}$ of benzene has to be converted into moles of benzene and also into number of molecules.

Answer to Problem E.22E

The total number of molecules and moles of benzene present in $\text{91}\text{kg}$ of benzene is $1203.7\times {10}^{23}\text{ molecules}$ and $\text{199}\text{.8}\text{mol}$ respectively.

Explanation of Solution

Given mass of benzene is $\text{91}\text{kg}$.  Molar mass of benzene is $78.11\text{g}\cdot {\text{mol}}^{-1}$.  Therefore, the number of molecules present in $\text{91}\text{kg}$ of benzene can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{molecules}=\frac{6.0221\times {10}^{23}\text{molecules}\text{per}\text{mole}}{78.11\text{g}\cdot {\text{mol}}^{-1}}\times 91000\text{g}\\ =7015.9\times {10}^{23}\text{ molecules}\end{array}$

Therefore, $\text{91}\text{kg}$ of benzene contains $7015.9\times {10}^{23}\text{ molecules}$ of benzene.

One mole of benzene is $\text{78}\text{.11}\text{g}$.  Therefore, the number of moles that is present in $\text{91}\text{kg}$ of benzene is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{benzene}}{\text{Molar}\text{mass}\text{of}\text{benzene}}\\ =\frac{91000\text{g}}{78.11\text{g}\cdot {\text{mol}}^{-1}}\\ =1165\text{mol}\end{array}$

Thus, $\text{91000}\text{kg}$ of benzene contains $7015.9\times {10}^{23}\text{ molecules}$ and $1165\text{mol}$

(c)

Interpretation Introduction

Interpretation:

Mass of $\text{350}\text{.0}\text{g}$ of phosphorus as $\text{P}$ atoms and as ${\text{P}}_{\text{4}}$  has to be converted into moles of phosphorus and also into number of molecules.

Answer to Problem E.22E

The total number of atoms and moles of phosphorus atom present in $\text{350}\text{.0}\text{g}$ of phosphorus is $68.06\times {10}^{23}\text{ atoms}$ and $11.3\text{mol}$ respectively.  The total number of molecules and moles of ${\text{P}}_{\text{4}}$ present in $\text{350}\text{.0}\text{g}$ of phosphorus is $17.01\times {10}^{23}\text{ molecules}$ and $\text{2}\text{.83}\text{mol}$ respectively.

Explanation of Solution

For phosphorus atoms:

Given mass of phosphorus is $\text{350}\text{.0}\text{g}$.  Molar mass of phosphorus is $30.97\text{g}\cdot {\text{mol}}^{-1}$.  Therefore, the number of atoms present in $\text{350}\text{.0}\text{g}$ of phosphorus atoms can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{atoms}=\frac{6.0221\times {10}^{23}\text{atoms}\text{per}\text{mole}}{30.97\text{g}\cdot {\text{mol}}^{-1}}\times 350\text{g}\\ =68.06\times {10}^{23}\text{ atoms}\end{array}$

Therefore, $\text{350}\text{.0}\text{g}$ of phosphorus contains $68.06\times {10}^{23}\text{ atoms}$ of phosphorus.

One mole of phosphorus is $\text{30}\text{.97}\text{g}$.  Therefore, the number of moles that is present in $\text{350}\text{.0}\text{g}$ of phosphorus atoms is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{phosphorus atom}}{\text{Molar}\text{mass}\text{of}\text{phosphorus}\text{atom}}\\ =\frac{350.\text{g}}{30.97\text{g}\cdot {\text{mol}}^{-1}}\\ =11.3\text{mol}\end{array}$

Thus, $\text{350}\text{g}$ of phosphorus atoms contains $68.06\times {10}^{23}\text{ atoms}$ and $11.3\text{mol}$

For phosphorus molecules:

Given mass of phosphorus is $\text{350}\text{.0}\text{g}$.  Molar mass of phosphorus molecule is $123.88\text{g}\cdot {\text{mol}}^{-1}$.  Therefore, the number of molecules present in $\text{350}\text{.0}\text{g}$ of phosphorus molecule can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{molecule}=\frac{6.0221\times {10}^{23}\text{molecules}\text{per}\text{mole}}{123.88\text{g}\cdot {\text{mol}}^{-1}}\times 350\text{g}\\ =17.01\times {10}^{23}\text{ molecules}\end{array}$

Therefore, $\text{350}\text{.0}\text{g}$ of phosphorus molecule contains $17.01\times {10}^{23}\text{ molecules}$ of phosphorus.

One mole of phosphorus molecule is $\text{123}\text{.88}\text{g}$.  Therefore, the number of moles that is present in $\text{350}\text{.0}\text{g}$ of phosphorus molecule is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{phosphorus molecule}}{\text{Molar}\text{mass}\text{of}\text{phosphorus}\text{molecule}}\\ =\frac{350.0\text{g}}{123.88\text{g}\cdot {\text{mol}}^{-1}}\\ =2.83\text{mol}\end{array}$

Thus, $\text{350}\text{g}$ of phosphorus molecule contains $17.01\times {10}^{23}\text{ molecules}$ and $2.83\text{mol}$

(d)

Interpretation Introduction

Interpretation:

Mass of $\text{1}\text{.2}\text{g}$ of carbon dioxide has to be converted into moles of carbon dioxide and also into number of molecules.

Answer to Problem E.22E

The total number of molecules and moles of ${\text{CO}}_{\text{2}}$ present in $\text{1}\text{.2}\text{g}$ of carbon dioxide is $0.164\times {10}^{23}\text{ molecules}$ and $\text{0}\text{.027}\text{mol}$ respectively.

Explanation of Solution

Given mass of carbon dioxide is $\text{1}\text{.2}\text{g}$.  Molar mass of carbon dioxide is $44.01\text{g}\cdot {\text{mol}}^{-1}$.  Therefore, the number of molecules present in $\text{1}\text{.2}\text{g}$ of carbon dioxide can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{molecule}=\frac{6.0221\times {10}^{23}\text{molecules}\text{per}\text{mole}}{44.01\text{g}\cdot {\text{mol}}^{-1}}\times 350\text{g}\\ =0.164\times {10}^{23}\text{ molecules}\end{array}$

Therefore, $\text{1}\text{.2}\text{g}$ of carbon dioxide molecule contains $0.164\times {10}^{23}\text{ molecules}$ of carbon dioxide.

One mole of carbon dioxide molecule is $\text{44}\text{.01}\text{g}$.  Therefore, the number of moles that is present in $\text{1}\text{.2}\text{g}$ of carbon dioxide is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{carbon dioxide molecule}}{\text{Molar}\text{mass}\text{of}\text{carbon dioxide}\text{molecule}}\\ =\frac{1.2\text{g}}{44.01\text{g}\cdot {\text{mol}}^{-1}}\\ =0.027\text{mol}\end{array}$

Thus, $\text{1}\text{.2}\text{g}$ of carbon dioxide molecule contains $0.164\times {10}^{23}\text{ molecules}$ and $0.027\text{mol}$

(e)

Interpretation Introduction

Interpretation:

Mass of $\text{0}\text{.37}\text{g}$ of nitrogen dioxide has to be converted into moles of nitrogen dioxide and also into number of molecules.

Answer to Problem E.22E

The total number of molecules and moles of ${\text{NO}}_{\text{2}}$ present in $\text{0}\text{.37}\text{g}$ of nitrogen dioxide is $17.01\times {10}^{23}\text{ molecules}$ and $\text{2}\text{.83}\text{mol}$ respectively.

Explanation of Solution

Given mass of nitrogen dioxide is $\text{0}\text{.37}\text{g}$.  Molar mass of nitrogen dioxide is $46.01\text{g}\cdot {\text{mol}}^{-1}$.  Therefore, the number of molecules present in $\text{0}\text{.37}\text{g}$ of nitrogen dioxide can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{molecule}=\frac{6.0221\times {10}^{23}\text{molecules}\text{per}\text{mole}}{46.01\text{g}\cdot {\text{mol}}^{-1}}\times 0.37\text{g}\\ =0.048\times {10}^{23}\text{ molecules}\end{array}$

Therefore, $\text{0}\text{.37}\text{g}$ of nitrogen dioxide molecule contains $0.048\times {10}^{23}\text{ molecules}$ of nitrogen dioxide.

One mole of nitrogen dioxide molecule is $\text{46}\text{.01}\text{g}$.  Therefore, the number of moles that is present in $\text{0}\text{.37}\text{g}$ of nitrogen dioxide is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{nitrogen dioxide molecule}}{\text{Molar}\text{mass}\text{of}\text{nitrogen dioxide}\text{molecule}}\\ =\frac{0.37\text{g}}{46.01\text{g}\cdot {\text{mol}}^{-1}}\\ =0.008\text{mol}\end{array}$

Thus, $\text{0}\text{.37}\text{g}$ of nitrogen dioxide molecule contains $0.048\times {10}^{23}\text{ molecules}$ and $0.008\text{mol}$