Chapter F, Problem E.18E

(a)

Interpretation Introduction

Interpretation:

Mass in micrograms has to be calculated for $3.27\times {10}^{16}\text{Hg}\text{atoms}$.

Answer to Problem E.18E

Mass of $3.27\times {10}^{16}\text{Hg}\text{atoms}$ is $\text{10}\text{.9}\mu \text{g}$.

Explanation of Solution

Number of atoms of mercury is given as $3.27\times {10}^{16}\text{atoms}$.  The number of moles of mercury in $3.27\times {10}^{16}\text{atoms}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}(n)=\frac{N}{N_A}\\ =\frac{3.27\times {10}^{16}\text{atoms}}{6.0221\times {10}^{23}{\text{mol}}^{-1}}\\ =\text{5}\text{.43}\times {\text{10}}^{\text{-8}}\text{mol}\end{array}$

Thus, $\text{3}\text{.27}\times {\text{10}}^{\text{26}}\text{atoms}$ of mercury contains $\text{5}\text{.43}\times {\text{10}}^{\text{-8}}\text{mol}$.

Mass of mercury atoms can be calculated as follows;

    $\begin{array}{c}\text{Mass}\text{of}\text{Hg}=\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}\text{of}\text{Hg}\\ \text{=5}\text{.43}\times {\text{10}}^{\text{-8}}\text{mol}\times 200.59\text{g}\cdot {\text{mol}}^{-1}\\ =1089.2\times {\text{10}}^{\text{-8}}\text{g}\\ \text{=10}\text{.9}\times {\text{10}}^{\text{-6}}\text{g}\\ \text{=10}\text{.9}\mu \text{g}\end{array}$

Thus the mass of $3.27\times {10}^{16}\text{Hg}\text{atoms}$ is $\text{10}\text{.9}\mu \text{g}$.

(b)

Interpretation Introduction

Interpretation:

Mass in micrograms has to be calculated for $\text{963}\text{nmol}\text{Hf}\text{atoms}$.

Answer to Problem E.18E

Mass of $\text{963}\text{nmol}\text{Hf}\text{atoms}$ is $\text{171}\text{.9}\mu \text{g}$.

Explanation of Solution

Number of moles of hafnium is given as $\text{963}\text{nmol}$.  This can be expressed in $\text{mol}$ using the conversion factor as shown below;

    $\begin{array}{c}\text{963}\text{nmol=963}\text{nmol}\times \frac{{\text{10}}^{\text{-9}}\text{ mol}}{\text{1}\text{nmol}}\\ =963\times {10}^{-9}\text{mol}\end{array}$

Mass of hafnium atoms can be calculated as follows;

    $\begin{array}{c}\text{Mass}\text{of}\text{Hf}=\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}\text{of}\text{Hf}\\ \text{=}963\times {10}^{-9}\text{mol}\times 178.49\text{g}\cdot {\text{mol}}^{-1}\\ =171885.87\times {\text{10}}^{\text{-9}}\text{g}\\ \text{=171}\text{.9}\times {\text{10}}^{\text{-6}}\text{g}\\ \text{=171}\text{.9}\mu \text{g}\end{array}$

Thus the mass of $\text{963}\text{nmol}\text{Hf}\text{atoms}$ is $\text{171}\text{.9}\mu \text{g}$.

(c)

Interpretation Introduction

Interpretation:

Mass in micrograms has to be calculated for $\text{5}\text{.50}\mu \text{mol}\text{Gd}\text{atoms}$.

Answer to Problem E.18E

Mass of $\text{5}\text{.50}\mu \text{mol}\text{Gd}\text{atoms}$ is $\text{864}\text{.9}\mu \text{g}$.

Explanation of Solution

Number of moles of gadolinium is given as $\text{5}\text{.50}\mu \text{mol}$.  This can be expressed in $\text{mol}$ using the conversion factor as shown below;

    $\begin{array}{c}\text{5}\text{.50}\mu \text{mol=5}\text{.50}\mu \text{mol}\times \frac{{\text{10}}^{\text{-6}}\text{ mol}}{\text{1}\mu \text{mol}}\\ =\text{5}\text{.50}\times {10}^{-6}\text{mol}\end{array}$

Mass of gadolinium atoms can be calculated as follows;

    $\begin{array}{c}\text{Mass}\text{of}\text{Gd}=\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}\text{of}\text{Gd}\\ \text{=5}\text{.50}\times {10}^{-6}\text{mol}\times 157.25\text{g}\cdot {\text{mol}}^{-1}\\ =864.875\times {\text{10}}^{\text{-6}}\text{g}\\ \text{=864}\text{.9}\times {\text{10}}^{\text{-6}}\text{g}\\ \text{=864}\text{.9}\mu \text{g}\end{array}$

Thus the mass of $\text{5}\text{.50}\mu \text{mol}\text{Gd}\text{atoms}$ is $\text{864}\text{.9}\mu \text{g}$.

(d)

Interpretation Introduction

Interpretation:

Mass in micrograms has to be calculated for $6.02\times {10}^{25}\text{Sb}\text{atoms}$.

Answer to Problem E.18E

Mass of $6.02\times {10}^{25}\text{Sb}\text{atoms}$ is $\text{1}\text{.21}\times {\text{10}}^{\text{10}}\mu \text{g}$.

Explanation of Solution

Number of atoms of antimony is given as $6.02\times {10}^{25}\text{atoms}$.  The number of moles of antimony in $6.02\times {10}^{25}\text{atoms}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}(n)=\frac{N}{N_A}\\ =\frac{6.02\times {10}^{25}\text{atoms}}{6.0221\times {10}^{23}{\text{mol}}^{-1}}\\ =\text{0}\text{.99}\times {\text{10}}^{\text{2}}\text{mol}\end{array}$

Thus, $6.02\times {10}^{25}\text{atoms}$ of antimony contains $\text{0}\text{.99}\times {\text{10}}^{\text{2}}\text{mol}$.

Mass of antimony atoms can be calculated as follows;

    $\begin{array}{c}\text{Mass}\text{of}\text{Sb}=\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}\text{of}\text{Sb}\\ \text{=0}\text{.99}\times {\text{10}}^{\text{2}}\text{mol}\times 121.76\text{g}\cdot {\text{mol}}^{-1}\\ =120.5424\times {\text{10}}^{\text{2}}\text{g}\\ \text{=1}\text{.21}\times {\text{10}}^{\text{10}}\mu \text{g}\end{array}$

Thus the mass of $6.02\times {10}^{25}\text{Sb atoms}$ is $\text{1}\text{.21}\times {\text{10}}^{\text{10}}\mu \text{g}$.