Chapter 3.15, Problem 80SEP

(a)

To determine

The lattice constant of the sodium chloride $\left(\text{NaCl}\right)$.

Answer to Problem 80SEP

The lattice constant of the sodium chloride $\left(\text{NaCl}\right)$ is $0.552\text{nm}$.

Explanation of Solution

Write the expression to calculate lattice constant of sodium chloride $\left(a\right)$.

 $a=2r_{{\text{Cl}}^{-}}+2r_{{\text{Na}}^{+}}$                                                                                                       (I)

Here, ionic radius of ${\text{Cl}}^{-}$ ion is $r_{{\text{Cl}}^{-}}$ and ionic radius of ${\text{Na}}^{+}$ ion is $r_{{\text{Na}}^{+}}$.

Conclusion:

Substitute $0.095\text{nm}$ for $r_{{\text{Cl}}^{-}}$ and $0.181\text{nm}$ for $r_{{\text{Na}}^{+}}$ in Equation (I).

 $\begin{array}{c}a=2\left(0.095\text{nm}\right)+2\left(0.181\text{nm}\right)\\ =0.552\text{nm}\\ =5.52\times {10}^{-8}\text{cm}\end{array}$

Thus, the lattice constant of the sodium chloride $\left(\text{NaCl}\right)$ is $0.552\text{nm}$.

(b)

To determine

The density of sodium chloride $\left(\text{NaCl}\right)$ in grams per cubic centimeter.

Answer to Problem 80SEP

The density of sodium chloride $\left(\text{NaCl}\right)$ in grams per cubic centimeter is $\text{2}\text{.31}\text{g}/{\text{cm}}^3$.

Explanation of Solution

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

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

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

Write the expression to calculate density of sodium chloride 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:

The sodium chloride unit cell contains four ${\text{Na}}^{\text{2+}}$ ions and four ${\text{Cl}}^{-}$ ions respectively.

Substitute $4$ for $n_{{\text{Na}}^{\text{2+}}}$, $22.99\text{g}/\text{mol}$ for $M_{{\text{Na}}^{\text{2+}}}$, $4$ for $n_{{\text{Cl}}^{-}}$, $35.459\text{g}/\text{mol}$ for $M_{{\text{Cl}}^{-}}$, and $6.023\times {10}^{23}\text{ions}/\text{mol}$ for $N_A$ in Equation (II).

 $\begin{array}{c}m=\frac{\left(4\right)\left(22.99\text{g}/\text{mol}\right)+\left(4\right)\left(35.459\text{g}/\text{mol}\right)}{6.023\times {10}^{23}\text{ions}/\text{mol}}\\ =3.88\times {10}^{-22}\text{g}\end{array}$

Substitute $3.88\times {10}^{-22}\text{g}$ for $m$ and $5.52\times {10}^{-8}\text{cm}$ for $a$ in Equation (III).

 $\begin{array}{c}\rho =\frac{3.88\times {10}^{-22}\text{g}}{{\left(5.52\times {10}^{-8}\text{cm}\right)}^3}\\ =2.31\text{g}/{\text{cm}}^3\end{array}$

Thus, the density of sodium chloride $\left(\text{NaCl}\right)$  in grams per cubic centimeter is $\text{2}\text{.31}\text{g}/{\text{cm}}^3$.