# KF has NaCl structure. What is the distance between K+ and F– in KF if density is 2.48 g cm-3?

KF has NaCl structure. What is the distance between K+ and F in KF if density is 2.48 gm cm-3 ?

Given Values

Density = 2.48 gm/cm3

Formula mass (M) = KF = 39+19=58

KF has NaCl structure : Means, KF has FCC unit cell. So, Z = 4

Avogadro Number (NA) = 6.02 × 1023

Formula used :  $$D=\frac{Z \times M}{\left ( a \right )^{3}\times N_{A}}$$

Calculations

Let us first find the edge length (a) using the density of the unit cell

$$\left ( a \right )^{3}=\frac{Z \times M}{D \times N_{A}}$$

$$\left ( a \right )^{3}=\frac{4 \times 58}{2.48 \times 6.02 \times 10^{23}}$$ = 155.4 × 10-24 cm3

a = 5.38 × 10-8 cm = 538 pm

Let us first find the distance between Potassium ion and Fluoride ion

$$d = \frac{a}{2} = \frac{358~pm}{2} = 269~pm$$