Question

question 8 sodium chloride (nacl) has a density of 2.17 g/cm³. the atomic weights of sodium and chlorine are 22.99 g/mol and 35.45 g/mol, respectively. (a) determine the unit cell edge length using the theoretical density formula. 0.563 nm (b) determine the unit cell edge length assuming that the na⁺ and cl⁻ ions just touch. note that the answers of (a) and (b) may be slightly different. select 0.134 0.566 0.770 0.882 < 0.140 nm cation anion ionic radius (nm) ionic radius (nm) na⁺ 0.102 cl⁻ 0.181 mg²⁺ 0.072 o²⁻ 0.140 fe²⁺ 0.077

Explanation:

Step1: Recall density formula for ionic - solid

The density formula for an ionic solid is $
ho=\frac{nM}{N_{A}V}$, where $
ho$ is the density, $n$ is the number of formula - units per unit cell, $M$ is the molar mass, $N_{A}=6.022\times 10^{23}\text{ mol}^{-1}$ is Avogadro's number, and $V = a^{3}$ is the volume of the unit cell ($a$ is the unit - cell edge length). For NaCl, $n = 4$ (face - centered cubic structure), $M=(22.99 + 35.45)\text{ g/mol}=58.44\text{ g/mol}$, and $
ho = 2.17\text{ g/cm}^3$.

Step2: Rearrange the density formula to solve for $a$

Starting from $
ho=\frac{nM}{N_{A}a^{3}}$, we can rearrange it to get $a^{3}=\frac{nM}{N_{A}
ho}$. Substitute $n = 4$, $M = 58.44\text{ g/mol}$, $N_{A}=6.022\times 10^{23}\text{ mol}^{-1}$, and $
ho = 2.17\text{ g/cm}^3$ into the formula.
\[a^{3}=\frac{4\times58.44\text{ g/mol}}{6.022\times 10^{23}\text{ mol}^{-1}\times2.17\text{ g/cm}^3}\]
\[a^{3}=\frac{233.76}{1.2967\times 10^{24}}\text{ cm}^3\]
\[a^{3}=1.8027\times 10^{-22}\text{ cm}^3\]
\[a=\sqrt[3]{1.8027\times 10^{-22}\text{ cm}^3}\]
\[a = 5.65\times 10^{-8}\text{ cm}\]
Convert to nm: $1\text{ cm}=10^{7}\text{ nm}$, so $a = 0.565\text{ nm}$.

For part (b), in a face - centered cubic unit cell of NaCl, the relationship between the unit - cell edge length $a$ and the ionic radii of $Na^{+}$ and $Cl^{-}$ is $a = 2(r_{Na^{+}}+r_{Cl^{-}})$. Given $r_{Na^{+}} = 0.102\text{ nm}$ and $r_{Cl^{-}} = 0.181\text{ nm}$, then $a = 2(0.102 + 0.181)\text{ nm}=2\times0.283\text{ nm}=0.566\text{ nm}$.

Answer:

(a) $0.565$ nm
(b) $0.566$ nm