Question

In sodium chloride, the distance between the center of the sodium ion and the centerof an adjacent chloride ion is 2.819 angstroms. Calculate the density in g/cm3of an ideal NaCl crystal from this information and what you learned from this lab.Hints: To calculate mass, determine how many equivalent ions are in a unit cell. To determinevolume of the unit cell, start by determining the length of on side of the unit cell.

Answer 1

Answer:

$\boxed{\text{2.17 g/cm}^{3}}$

Explanation:

1. Ions per unit cell

(a) Chloride

8 corners + 6 faces

$\text{No. of Cl ^{-} ions}\\\\= \text{8 corners} \times \dfrac{\frac{1 }{ 8} \text{ ion}} {\text{1 corner}} + \text{6 faces}\times \dfrac{\frac{1}{2} \text{ ion}}{\text{1 face}} = \text{1 ion + 3 ions = 4 ions}}$

(b) Chloride

12 edges + 1 centre

$\text{No. of Na ^{+} ions}\\\\= \text{12 edges} \times \dfrac{\frac{1 }{ 4} \text{ ion}} {\text{1 edge}} + \text{1 centre}\times\dfrac{\text{1 ion}}{\text{1 centre}} = \text{3 ions + 1 ion = 4 ions}$

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

$m = \text{233.76 g} \times \dfrac{\text{1 g} }{6.022 \times 10^{23} \text{ u} } = 3.882 \times 10^{-22}\text{ g}$

3. Volume of unit cell

(a) Edge length

$a = 2d_{\text{Na-Cl}} = 2 \times \text{2.819 \AA} = 5.638 \times10^{-10} \text{ m} = 5.638 \times10^{-8} \text{ cm}$

(b) Volume

$V = a^{3} = \left( 5.638 \times (10^{-8} \text{ cm}\right)^{3} = 1.792 \times10^{-22} \text{ cm}^{3}$

4. Density

$\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{3.882 \times 10^{-22}\text{ g}}{1.792 \times10^{-22} \text{ cm}^{3}}} = \text{2.17 g/cm}^{3}\\\\\text{The density of NaCl is }\boxed{\textbf{2.17 g/cm}^{3}}$