Question

In solid KCl, the smallest distance between the centersof a potassium ion and a chloride ion is 314 pm. Calculate the length of the edge of the unit cell and the density of KCl, assuming it has the same structure as sodium chloride.

Short answer

The length of the edge of the unit cell and density of KCl is 628 pm and $2.023\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$, respectively.

Step-by-step solution

Given Information

The smallest distance between K⁺ ion and Cl^- ion is 314 pm.

The KCl has the same structure as NaCl. The structure is fcc.

The edge length of the compound (a) =

$$\begin{gathered} 2\times \mathrm{smallest}\mathrm{distance}\mathrm{between}\mathrm{ions}\left(\mathrm{a}\right) \\ 2\times 314\mathrm{pm} \\ 628\mathrm{pm} \end{gathered}$$

The number of atoms presents per unit cell z=4

Avogadro's constant,_{${\mathrm{N}}_{\mathrm{A}}=6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}$}

Concept Introduction

The density of the compound can be define as

$\mathbf{d}\mathbf{=}\frac{\mathbf{Z}\mathbf{\times }\mathbf{M}}{{\mathbf{N}}_{\mathbf{A}}\mathbf{\times }{\mathbf{a}}^{\mathbf{3}}}$

where Z, M , a are the number of formula units in the unit cell, molar mass, and lattice paramter.

Describing the density of the KCl compound from the given data

The density of the compound can be calculated as,

$$\begin{gathered} \mathrm{d}=\frac{4\times 74.5518\mathrm{gm}/\mathrm{mole}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}\times {\left(628\mathrm{pm}\right)}^3} \\ \mathrm{d}=2.027\times {10}^6\mathrm{gm}/{\mathrm{cm}}^3 \\ \mathrm{d}=2.027\times {10}^3\mathrm{kg}/{\mathrm{cm}}^3 \end{gathered}$$

Therefore the density of KCl is$\mathrm{d}=2.027\times {10}^3\mathrm{kg}/{\mathrm{cm}}^3$