Chapter 12, Problem 12.114QP

Interpretation Introduction

Interpretation:

Given the atomic radius of solid Argon, its density has to be calculated.

Concept Introduction:

In a crystalline solid, the components are neatly stacked and closely packed in a regular pattern. The components are imagined as spheres.  The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing.  Cubic close packing structure has face-centered cubic (FCC) unit cell.

In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.

Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells.  Thus the number of atoms per unit cell in FCC unit cell is,

$\begin{array}{l}\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{6}\text{×}\frac{\text{1}}{\text{2}}\text{atoms}\text{in}\text{faces}\text{=}\text{1}\text{+}\text{3}\text{=}\text{4}\text{atoms}\\ \\ \text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\\ \end{array}$

Answer to Problem 12.114QP

The density of solid Argon is $1.69g/c{m}^3.$

Explanation of Solution

Solid Argon has cubic close packing structure with face-centered cubic unit cells. The atomic radius of solid Argon is given.  The formula for the edge length of the cubic unit cell is known as $\text{a}\text{=}\text{2R}\sqrt{\text{2}}$ in which ‘a’ corresponds to edge length of the unit cell and ‘R’ corresponds to radius of the atom in the unit cell.  The value is substituted and ‘a³’value is calculated.  This value corresponds to the volume of the unit cell.

Accordingly calculate volume of FCC unit cell as shown below –

$\text{given}\text{data:}\text{R}\text{=}\text{191}\text{pm}\text{=}\text{191}\text{×}{\text{10}}^{\text{-12}}\text{m}$

$\begin{array}{l}\text{volume}\text{of}\text{unit}\text{cell}\text{=}{\text{a}}^{\text{3}}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ {\text{a}}^{\text{3}}\text{=}{\left(\text{2R}\sqrt{\text{2}}\right)}^{\text{3}}\\ \text{=}{\left(\text{2}\text{×}\text{191}\text{×}{\text{10}}^{\text{-12}}\text{m}\text{×}\text{1}\text{.414}\right)}^{\text{3}}\\ \text{=}\text{1}\text{.57}\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}\end{array}$

Each unit cell contains 4 Ar atoms.  Therefore four times the average mass of one Argon atom gives mass of a unit cell of solid Argon.  Hence the mass of a unit cell of solid Argon.

$\begin{array}{l}\text{Average mass of one Ar atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{Ar}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{39.95\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{6}\text{.63}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

$\text{Each}\text{unit}\text{cell}\text{has}\text{4}\text{Ar}\text{atoms}\text{.}\text{Therefore,}$

$\begin{array}{l}\text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{4}\text{×}\text{average}\text{mass}\text{of}\text{one}\text{Ar}\text{atom}\\ \text{=}\text{4}\text{×}\text{6}\text{.63}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}\text{2}\text{.65}\text{×}{\text{10}}^{\text{-22}}\text{g}\end{array}$

Mass and volume of the unit cell is calculated in the previous steps. By substituting the values in the formula,$\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$ the density of solid Argon is determined.

The density of solid Argon is,

$\begin{array}{l}\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{=}\frac{\text{2}\text{.65}\text{×}{\text{10}}^{\text{-22}}\text{g}}{\text{1}\text{.57}\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}}\\ \text{=}\text{1}\text{.69}{\text{g/cm}}^{\text{3}}\end{array}$

Conclusion

The density of solid Argon was determined using the relation between edge length of the FCC unit cell and atomic radius.