Answer:
[tex]\boxed{\text{2.17 g/cm}^{3}}[/tex]
Explanation:
1. Ions per unit cell
(a) Chloride
8 corners + 6 faces
[tex]\text{No. of Cl$^{-}$ ions}\\\\= \text{8 corners} \times \dfrac{\frac{1 }{ 8} \text{ ion}} {\text{1 corner}} + \text{6 faces}\times \dfrac{\frac{1}{2} \text{ ion}}{\text{1 face}} = \text{1 ion + 3 ions = 4 ions}}[/tex]
(b) Chloride
12 edges + 1 centre
[tex]\text{No. of Na$^{+}$ ions}\\\\= \text{12 edges} \times \dfrac{\frac{1 }{ 4} \text{ ion}} {\text{1 edge}} + \text{1 centre}\times\dfrac{\text{1 ion}}{\text{1 centre}} = \text{3 ions + 1 ion = 4 ions}[/tex]
There are four formula units of NaCl in a unit cell.
2. Mass of unit cell
m = 4 × NaCl = 4 × 58.44 u = 233.76 u
[tex]m = \text{233.76 g} \times \dfrac{\text{1 g} }{6.022 \times 10^{23} \text{ u} } = 3.882 \times 10^{-22}\text{ g}[/tex]
3. Volume of unit cell
(a) Edge length
[tex]a = 2d_{\text{Na-Cl}} = 2 \times \text{2.819 \AA} = 5.638 \times10^{-10} \text{ m} = 5.638 \times10^{-8} \text{ cm}[/tex]
(b) Volume
[tex]V = a^{3} = \left( 5.638 \times (10^{-8} \text{ cm}\right)^{3} = 1.792 \times10^{-22} \text{ cm}^{3}[/tex]
4. Density
[tex]\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{3.882 \times 10^{-22}\text{ g}}{1.792 \times10^{-22} \text{ cm}^{3}}} = \text{2.17 g/cm}^{3}\\\\\text{The density of NaCl is }\boxed{\textbf{2.17 g/cm}^{3}}[/tex]
