Chapter 12, Problem 80E

Interpretation Introduction

Interpretation:

The density of CsCl should be calculated.

Concept introduction:

As the caesium has a body centred cubic (bcc) crystal structure, nine atoms are associated with a bcc unit cell. One atom is located at each of the eight corners of the cube and one at the centre. The three atoms along a diagonal through the cube are in contact. The length of the cube diagonal (the distance from the farthest upper-right corner to the nearest lower left corner) is four times the atomic radius. Also, shown below is the fact the diagonal of a cube is equal to $\sqrt{3}$ $\times l$. The length of an edge, l, is what is given.

The right triangle must conform to the Pythagorean formula $a^2=b^2+c^2$.

From the Pythagorean formula, following the relationship between r and l can obtain. $l\sqrt{3}=2(r_{{\text{Cs}}^{\text{+}}}+r_{{\text{Cl}}^{\text{-}}})$

From the above equation, volume ( v ) of the unit cell can be calculated as by the following equation as this is a cube.

$v=l^3$

$v={\left(\frac{2(r_{C{s}^{+}}+r_{C{l}^{-}})}{\sqrt{3}}\right)}^3$

Density ( d ) defines as the mass per unit volume.

$d=\frac{m}{v}$

The following relationship is used to find the mass of a unit cell.

$m=n\frac{{\text{M}}_{\text{CsCl}}}{{\text{N}}_{\text{A}}}$

Here,

$\begin{array}{l}n\text{is}\text{}\text{molecules}\text{per}\text{unit}\text{cell}\text{.}\\ {\text{N}}_{\text{A}}\text{is avagadro}\text{constant}\\ {\text{M}}_{\text{CsCl}}\text{is relative molar mass of CsCl}\end{array}$