Chapter 11.13, Problem 67AAP

To determine

The density of cesium bromide $\left(\text{CsBr}\right)$ in grams per cubic centimeter.

Answer to Problem 67AAP

The density of cesium bromide $\left(\text{CsBr}\right)$ in grams per cubic centimeter is $\text{4}\text{.87}\text{g}/{\text{cm}}^3$.

Explanation of Solution

Write the expression to calculate lattice constant of cesium bromide structure $\left(a\right)$.

 $a=\frac{2}{\sqrt{3}}\left(r_{{\text{Cs}}^{\text{+}}}+R_{{\text{Br}}^{-}}\right)$                                                                                                  ...... (I)

Here, ionic radius of ${\text{Cs}}^{\text{+}}$ ion is $r_{{\text{Cs}}^{\text{+}}}$ and ionic radius of ${\text{Br}}^{-}$ ion is $R_{{\text{Br}}^{-}}$.

Write the expression to calculate mass of unit cell of cesium bromide $\left(m\right)$.

 $m=\frac{n_{{\text{Cs}}^{\text{+}}}{M}_{{\text{Cs}}^{\text{+}}}+n_{{\text{Br}}^{-}}{M}_{{\text{Br}}^{-}}}{N_A}$                                                                                        ...... (II)

Here, number of atoms and molar mass of ${\text{Cs}}^{\text{+}}$ ion are $n_{{\text{Cs}}^{\text{+}}}$ and $M_{{\text{Cs}}^{\text{+}}}$, number of atoms and molar mass of ${\text{Br}}^{-}$ ion are $n_{{\text{Br}}^{-}}$ and $M_{{\text{Br}}^{-}}$ and Avogadro's number is $N_A$.

Write the expression to calculate density of cesium bromide in grams per cubic centimeter $\left(\rho \right)$.

 $\begin{array}{c}\rho =\frac{m}{V}\\ =\frac{m}{a^3}\end{array}$                                                                                                                  ...... (III)

Here, volume of unit cell is $V$.

Conclusion:

Substitute $0.165\text{nm}$ for $r_{{\text{Cs}}^{\text{+}}}$ and $0.196\text{nm}$ for $R_{{\text{Br}}^{-}}$ in Equation (I).

 $\begin{array}{c}a=\frac{2}{\sqrt{3}}\left(0.165\text{nm}+0.196\text{nm}\right)\\ =0.417\text{nm}\\ =4.17\times {10}^{-8}\text{cm}\end{array}$

Substitute $1$ for $n_{{\text{Cs}}^{\text{+}}}$, $132.9\text{g}/\text{mol}$ for $M_{{\text{Cs}}^{\text{+}}}$, $1$ for $n_{{\text{Br}}^{-}}$, $79.90\text{g}/\text{mol}$ for $M_{{\text{Br}}^{-}}$ and $6.023\times {10}^{23}\text{ions}/\text{mol}$ for $N_A$ in Equation (II).

 $\begin{array}{c}m=\frac{\left(1\right)\left(132.9\text{g}/\text{mol}\right)+\left(1\right)\left(79.9\text{g}/\text{mol}\right)}{6.023\times {10}^{23}\text{ions}/\text{mol}}\\ =3.53\times {10}^{-22}\text{g}\end{array}$

Substitute $3.53\times {10}^{-22}\text{g}$ for $m$ and $4.17\times {10}^{-8}\text{cm}$ for $a$ in Equation (III).

 $\begin{array}{c}\rho =\frac{3.53\times {10}^{-22}\text{g}}{{\left(4.17\times {10}^{-8}\text{cm}\right)}^3}\\ =\text{4}\text{.87}\text{g}/{\text{cm}}^3\end{array}$

Thus, the density of cesium bromide $\left(\text{CsBr}\right)$ in grams per cubic centimeter is $\text{4}\text{.87}\text{g}/{\text{cm}}^3$.