Chapter 11.13, Problem 70AAP

(a)

To determine

The density of strontium oxide $\left(\text{SrO}\right)$ in grams per cubic centimeter.

Answer to Problem 70AAP

The density of strontium oxide $\left(\text{SrO}\right)$ in grams per cubic centimeter is $\text{4}\text{.96}\text{g}/{\text{cm}}^3$.

Explanation of Solution

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

 $a=2\left(r_{{\text{Sr}}^{+}}+R_{{\text{O}}^{2-}}\right)$                                                                                                   (I)

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

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

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

Here, number of atoms and molar mass of ${\text{Sr}}^{+}$ ion are $n_{{\text{Sr}}^{+}}$ and $M_{{\text{Sr}}^{+}}$, 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 strontium oxide 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 $0.127\text{nm}$ for $r_{{\text{Sr}}^{+}}$ and $0.132\text{nm}$ for $R_{{\text{O}}^{2-}}$ in Equation (I).

 $\begin{array}{c}a=2\left(0.127\text{nm}+0.132\text{nm}\right)\\ =0.518\text{nm}\\ =5.18\times {10}^{-8}\text{cm}\end{array}$

Substitute $4$ for $n_{{\text{Sr}}^{+}}$, $87.62\text{g}/\text{mol}$ for $M_{{\text{Sr}}^{+}}$, $4$ 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).

 $\begin{array}{c}m=\frac{\left(4\right)\left(87.62\text{g}/\text{mol}\right)+\left(4\right)\left(16\text{g}/\text{mol}\right)}{6.023\times {10}^{23}\text{ions}/\text{mol}}\\ =6.89\times {10}^{-22}\text{g}\end{array}$

Substitute $6.89\times {10}^{-22}\text{g}$ for $m$ and $5.18\times {10}^{-8}\text{cm}$ for $a$ in Equation (III).

 $\begin{array}{c}\rho =\frac{6.89\times {10}^{-22}\text{g}}{{\left(5.18\times {10}^{-8}\text{cm}\right)}^3}\\ =\text{4}\text{.96}\text{g}/{\text{cm}}^3\end{array}$

Thus, the density of strontium oxide $\left(\text{SrO}\right)$ in grams per cubic centimeter is $\text{4}\text{.96}\text{g}/{\text{cm}}^3$.

(b)

To determine

The density of vanadium oxide $\left(\text{VO}\right)$ in grams per cubic centimeter.

Answer to Problem 70AAP

The density of vanadium oxide $\left(\text{VO}\right)$ in grams per cubic centimeter is $7.27\text{g}/{\text{cm}}^3$.

Explanation of Solution

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

 $a=2\left(r_{{\text{V}}^{2+}}+R_{{\text{O}}^{2-}}\right)$                                                                                                   (IV)

Here, ionic radius of ${\text{V}}^{2+}$ ion is $r_{{\text{V}}^{2+}}$.

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

 $m=\frac{n_{{\text{V}}^{2+}}{M}_{{\text{V}}^{2+}}+n_{{\text{O}}^{2-}}{M}_{{\text{O}}^{2-}}}{N_A}$                                                                                         (V)

Here, number of atoms and molar mass of ${\text{V}}^{2+}$ ion are $n_{{\text{V}}^{2+}}$ and $M_{{\text{V}}^{2+}}$ respectively.

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

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

Here, volume of unit cell is $V$.

Conclusion:

Substitute $0.065\text{nm}$ for $r_{{\text{V}}^{2+}}$ and $0.132\text{nm}$ for $R_{{\text{O}}^{2-}}$ in Equation (IV).

 $\begin{array}{c}a=2\left(0.065\text{nm}+0.132\text{nm}\right)\\ =0.394\text{nm}\\ =3.94\times {10}^{-8}\text{cm}\end{array}$

Substitute $4$ for $n_{{\text{V}}^{2+}}$, $50.94\text{g}/\text{mol}$ for $M_{{\text{V}}^{2+}}$, $4$ 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 (V).

 $\begin{array}{c}m=\frac{\left(4\right)\left(50.94\text{g}/\text{mol}\right)+\left(4\right)\left(16\text{g}/\text{mol}\right)}{6.023\times {10}^{23}\text{ions}/\text{mol}}\\ =4.45\times {10}^{-22}\text{g}\end{array}$

Substitute $4.45\times {10}^{-22}\text{g}$ for $m$ and $3.94\times {10}^{-8}\text{cm}$ for $a$ in Equation (VI).

 $\begin{array}{c}\rho =\frac{4.45\times {10}^{-22}\text{g}}{{\left(3.94\times {10}^{-8}\text{cm}\right)}^3}\\ =\text{7}\text{.27}\text{g}/{\text{cm}}^3\end{array}$

Thus, the density of vanadium oxide $\left(\text{VO}\right)$ in grams per cubic centimeter is $7.27\text{g}/{\text{cm}}^3$.