Chapter 11.13, Problem 74AAP

To determine

The density of zirconium dioxide $\left({\text{ZrO}}_2\right)$ in grams per cubic centimeter.

Answer to Problem 74AAP

The density of zirconium dioxide $\left({\text{ZrO}}_2\right)$ in grams per cubic centimeter is $\text{6}\text{.32}\text{g}/{\text{cm}}^3$.

Explanation of Solution

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

 $a=\frac{4}{\sqrt{3}}\left(r_{{\text{Zr}}^{\text{4+}}}+R_{{\text{O}}^{2-}}\right)$                                                                                                ...... (I)

Here, ionic radius of ${\text{Zr}}^{4+}$ ion is $r_{{\text{Zr}}^{\text{4+}}}$ and ionic radius of ${\text{O}}^{2-}$ ion is $R_{{\text{O}}^{2-}}$.

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

 $m=\frac{n_{{\text{Zr}}^{\text{4+}}}{M}_{{\text{Zr}}^{\text{4+}}}+n_{{\text{O}}^{2-}}{M}_{{\text{O}}^{2-}}}{N_A}$                                                                                    ...... (II)

Here, number of atoms and molar mass of ${\text{Zr}}^{4+}$ ion are $n_{{\text{Zr}}^{4+}}$ and $M_{{\text{Zr}}^{4+}}$, number of atoms and molar mass of ${\text{O}}^{2-}$ ion are $n_{{\text{O}}^{2-}}$ and $M_{{\text{O}}^{2-}}$ and Avogadro's number is $N_A$.

Write the expression to calculate density of zirconium dioxide 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 $0.087\text{nm}$ for $r_{{\text{Zr}}^{4+}}$ and $0.132\text{nm}$ for $R_{{\text{O}}^{2-}}$ in Equation (I).

 $\begin{array}{c}a=\frac{4}{\sqrt{3}}\left(0.087\text{nm}+0.132\text{nm}\right)\\ =0.506\text{nm}\\ =5.06\times {10}^{-8}\text{cm}\end{array}$

There will be 4 zirconium ions and 8 oxygen ions present in a single unit cell of zirconium dioxide.

Substitute $4$ for $n_{{\text{Zr}}^{4+}}$, $91.22\text{g}/\text{mol}$ for $M_{{\text{Zr}}^{4+}}$, $8$ for $n_{{\text{O}}^{2-}}$, $16\text{g}/\text{mol}$ for $M_{{\text{O}}^{2-}}$ and $6.023\times {10}^{23}\text{ions}/\text{mol}$ for $N_A$ in Equation (II).

 $\begin{array}{c}m=\frac{\left(4\right)\left(91.22\text{g}/\text{mol}\right)+\left(8\right)\left(16\text{g}/\text{mol}\right)}{6.023\times {10}^{23}\text{ions}/\text{mol}}\\ =8.19\times {10}^{-22}\text{g}\end{array}$

Substitute $8.19\times {10}^{-22}\text{g}$ for $m$ and $5.06\times {10}^{-8}\text{cm}$ for $a$ in Equation (III).

 $\begin{array}{c}\rho =\frac{8.19\times {10}^{-22}\text{g}}{{\left(5.06\times {10}^{-8}\text{cm}\right)}^3}\\ =\text{6}\text{.32}\text{g}/{\text{cm}}^3\end{array}$

Thus, the density of zirconium dioxide $\left({\text{ZrO}}_2\right)$ in grams per cubic centimeter is $\text{6}\text{.32}\text{g}/{\text{cm}}^3$.