Chapter 3, Problem 3.29E

Interpretation Introduction

Interpretation:

Equation to relate atomic radius and density of ccp solid of given molar mass has to be determined. Also, atomic radius of noble gases through constructed equation has to be determined.

Concept Introduction:

Density of substance is termed for mass of substance present in unit volume. It can be represented as follows:

  $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Explanation of Solution

Expression between volume of atom, mass, and density is as follows:

  ${\left(r\left(\text{cm}\right)\right)}^3=\left(k\right)\left(\frac{\text{Molar mass}}{\text{density}}\right)$        (1)

Here,

$r$ is radius of atom.

$k$ is proportionality constant.

Rearrange equation (1) to determine value of $k$.

  $k=\frac{\left(r^3\right)\left(\text{density}\right)}{\text{Molar mass}}$        (2)

Substitute $\text{1}\text{.20 g}\cdot {\text{cm}}^{-3}$ for density, $170\text{ pm}$ for $r$, and $20.18\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (2) to determine value of $k$ for $\text{Ne}$.

  $\begin{array}{c}k=\frac{\left({\left(\left(170\text{ pm}\right)\left(\frac{{10}^{-10}\text{ cm}}{1\text{ pm}}\right)\right)}^3\right)\left(\text{1}\text{.20 g}\cdot {\text{cm}}^{-3}\right)}{20.18\text{ g}\cdot {\text{mol}}^{-1}}\\ =2.92\times {10}^{-25}\text{ mol}\end{array}$

Equation (1) can be modified to determine value of $r$ in $\text{pm}$ unit as follows:

  ${\left(r\left(\text{pm}\right)\right)}^3=\left(k\right)\left(\frac{\text{Molar mass}}{\text{density}}\right){\left(\frac{{10}^{10}\text{ pm}}{\text{1 cm}}\right)}^3$        (3)

Rearrange equation (3) to determine value of $r$ in $\text{pm}$ unit.

  $r\left(\text{pm}\right)={\left(\left(k\right)\left(\frac{\text{Molar mass}}{\text{density}}\right){\left(\frac{{10}^{10}\text{ pm}}{\text{1 cm}}\right)}^3\right)}^{\text{1/3}}$        (4)

Substitute $2.92\times {10}^{-25}\text{ mol}$ for $k$ in equation (4).

  $r\left(\text{pm}\right)={\left(\left(2.92\times {10}^{-25}\text{ mol}\right)\left(\frac{\text{Molar mass}}{\text{density}}\right){\left(\frac{{10}^{10}\text{ pm}}{\text{1 cm}}\right)}^3\right)}^{\text{1/3}}$        (5)

Therefore, general relation between radius of atom $\left(\text{pm}\right)$ , density and molar mass of atom is as follows:

  $r\left(\text{pm}\right)={\left(\left(2.92\times {10}^{-25}\text{ mol}\right)\left(\frac{\text{Molar mass}}{\text{density}}\right){\left(\frac{{10}^{10}\text{ pm}}{\text{1 cm}}\right)}^3\right)}^{\text{1/3}}$        (5)

Substitute $\text{1}\text{.40 g}\cdot {\text{cm}}^{-3}$ for density, $39.95\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (5) to determine value of $r$ for $\text{Ar}$.

  $\begin{array}{c}r\left(\text{pm}\right)={\left(\left(2.92\times {10}^{-25}\text{ mol}\right)\left(\frac{39.95\text{ g}\cdot {\text{mol}}^{-1}}{\text{1}\text{.40 g}\cdot {\text{cm}}^{-3}}\right){\left(\frac{{10}^{10}\text{ pm}}{\text{1 cm}}\right)}^3\right)}^{\text{1/3}}\\ =202\text{ pm}\end{array}$

Substitute $\text{2}\text{.16 g}\cdot {\text{cm}}^{-3}$ for density, $83.80\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (5) to determine value of $r$ for $\text{Kr}$.

  $\begin{array}{c}r\left(\text{pm}\right)={\left(\left(2.92\times {10}^{-25}\text{ mol}\right)\left(\frac{83.80\text{ g}\cdot {\text{mol}}^{-1}}{\text{2}\text{.16 g}\cdot {\text{cm}}^{-3}}\right){\left(\frac{{10}^{10}\text{ pm}}{\text{1 cm}}\right)}^3\right)}^{\text{1/3}}\\ =225\text{ pm}\end{array}$

Substitute $\text{2}\text{.83 g}\cdot {\text{cm}}^{-3}$ for density, $131.30\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (5) to determine value of $r$ for $\text{Xe}$.

  $\begin{array}{c}r\left(\text{pm}\right)={\left(\left(2.92\times {10}^{-25}\text{ mol}\right)\left(\frac{131.30\text{ g}\cdot {\text{mol}}^{-1}}{\text{2}\text{.83 g}\cdot {\text{cm}}^{-3}}\right){\left(\frac{{10}^{10}\text{ pm}}{\text{1 cm}}\right)}^3\right)}^{\text{1/3}}\\ =238\text{ pm}\end{array}$

Substitute $4.4\text{ g}\cdot {\text{cm}}^{-3}$ for density, $222\text{ g}\cdot {\text{mol}}^{-1}$ for molar mass in equation (5) to determine value of $r$ for $\text{Rn}$.

  $\begin{array}{c}r\left(\text{pm}\right)={\left(\left(2.92\times {10}^{-25}\text{ mol}\right)\left(\frac{222\text{ g}\cdot {\text{mol}}^{-1}}{4.4\text{ g}\cdot {\text{cm}}^{-3}}\right){\left(\frac{{10}^{10}\text{ pm}}{\text{1 cm}}\right)}^3\right)}^{\text{1/3}}\\ =245\text{ pm}\end{array}$