Chapter 4, Problem 4.11QP

(a)

Interpretation Introduction

Interpretation:

The average mass (in $\text{amu}$) of $\text{Hg}$ has to be calculated.

Explanation of Solution

The abundance of $\text{Hg}$ in nature is ${}^{\text{196}}\text{Hg 0}\text{.15\%, 10}{\text{.04\% }}^{\text{198}}\text{Hg, 16}{\text{.94\% }}^{\text{199}}\text{Hg, 23}{\text{.14\% }}^{\text{200}}\text{Hg, 13}\text{.17}\%{}^{\text{201}}\text{Hg, 29}{\text{.74\% }}^{\text{202}}\text{Hg, 6}{\text{.82\%}}^{\text{204}}\text{Hg }$

The average atomic mass of Mercury is calculated as follows,

  $\begin{array}{l}\text{Average atomic mass is,}\\ \text{=}\frac{\text{196(0}\text{.15)+198(10}\text{.04)+199(16}\text{.94)+200(23}\text{.14)+201(13}\text{.17)+202(29}\text{.74)+204(6}\text{.82)}}{100}\\ \\ \text{=}\text{200}\text{.59}\text{amu}\end{array}$

The average mass of $\text{Hg}$ is $\text{200}\text{.59 amu}$.

(b)

Interpretation Introduction

Interpretation:

The average mass (in $\text{amu}$) of $\text{Kr}$ has to be calculated.

Explanation of Solution

The abundance of $\text{Kr}$ in nature is $\text{ 0}{\text{.36\% }}^{\text{78}}\text{Kr, 2}{\text{.29\% }}^{\text{80}}\text{Kr, 11}{\text{.59\% }}^{82}\text{Kr, 11}{\text{.50\% }}^{83}\text{Kr, 56}\text{.99}\%{}^{84}\text{Kr, 17}{\text{.28\% }}^{86}\text{Kr}$

The average atomic mass of Krypton is calculated as follows,

  $\begin{array}{l}\text{Average atomic mass is,}\\ \text{=}\frac{\text{ 78(0}\text{.36)+80(2}\text{.29)+82(11}\text{.59)+83(11}\text{.50)+84(56}\text{.99)+86(17}\text{.28)}}{100}\\ =83.798\text{amu}\end{array}$

The average mass of $\text{Kr}$ is $83.798\text{amu}$.

(c)

Interpretation Introduction

Interpretation:

The average mass (in $\text{amu}$) of $\text{Mg}$ has to be calculated.

Explanation of Solution

The abundance of $\text{Mg}$ in nature is $\text{79}{\text{.0\% }}^{24}\text{Mg, 10}{\text{.0\% }}^{25}\text{Mg, 11}{\text{.0\% }}^{26}\text{Mg}$

The average atomic mass of Magnesium is calculated as follows,

  $\begin{array}{l}\text{Average atomic mass is,}\\ \text{=}\frac{\text{24(79)+25(10)+26(11)}}{100}\\ \text{=}\text{24}\text{.305}\text{amu}\end{array}$

The average mass of $\text{Mg}$ is $\text{24}\text{.305}\text{amu}$.