Question

KF has the same type of crystal structure as NaCl. The unit cell of KF has an edge length of 5.39 $\stackrel{\mathbf{°}}{\mathbf{A}}$. Find the density of KF.

Short answer

The density of KF is$0.24\times {10}^7\mathrm{g}/{\mathrm{m}}^3$.

Step-by-step solution

Definition of crystal structure

In a crystal, the constituent particles form a structure by repeating a symmetric pattern along three dimensions. This symmetric structure of the particles in a crystal is named as crystal structure.

Definition of a unit cell

The smallest symmetric pattern from which the whole lattice can be formed by repeating in three dimensions is called the unit cell.

Finding the density of KF

Here, $\text{Density}=\frac{\mathrm{ZM}}{{\mathrm{NOa}}_3}$

NaCl's cubic crystal structure is face-centered.

For a face-centered cubic structure, Z, i.e., ions in a unit cell, is 4

Therefore, no. of ions in a unit cell for KF is 4.

$\begin{array}{rcl}\mathrm{M},\mathrm{molecular}\mathrm{weight}\mathrm{of}\mathrm{KF}& =& \left(39+19\right)\mathrm{g}/\mathrm{mol}=58\mathrm{g}/\mathrm{mol}\\ {\mathrm{N}}_0\mathrm{or}\mathrm{Avogadro}\mathrm{no}& =& 6.023\times {10}^{23}{\mathrm{mol}}^{-1}\\ \mathrm{a}\mathrm{or}\mathrm{edge}\mathrm{length}& =& 5.39\times {10}^{-10}\mathrm{m}\\ \mathrm{Density}\mathrm{of}\mathrm{KF}& =& 4\times \frac{58}{6.023}\times {10}^{23}\times {\left(5.39\times {10}^{-10}\right)}^3\mathrm{g}/{\mathrm{m}}^3\\ & =& 0.24\times {10}^7\mathrm{g}/{\mathrm{m}}^3\end{array}$