Equilibrium constant can also be expressed in terms of ${\mathrm{K}}_{\mathrm{x}}$, when concentration of the species are taken in mole fraction

${\mathrm{F}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{F}\left(\mathrm{g}\right); {\mathrm{K}}_{\mathrm{x}} = \frac{{\mathrm{X}}_{\mathrm{F}}^2}{{\mathrm{X}}_{{\mathrm{F}}_2}}$

For the above equilibrium mixture, average molar mass at 1000 K was 36.74 g c Thus, ${\mathrm{K}}_{\mathrm{x}}$ is

Let, ${\mathrm{F}}_2\left(\mathrm{g}\right)=\mathrm{x}$ (mole fraction),

molar mass = 38.09 ${\mathrm{mol}}^{-1}$

F(g) = (1 - x), molar mass = 19.0 g  ${\mathrm{mol}}^{-1}$

$\therefore$   Average molar mass = $\frac{{\mathrm{M}}_1{\mathrm{x}}_1 + {\mathrm{M}}_2{\mathrm{x}}_2}{{\mathrm{x}}_1 + {\mathrm{x}}_2}$

36.74 = 38 x + 19 (1 - x)

$\therefore$    x = 0.9337 (mole faction of ${\mathrm{F}}_2$)

(1 - x) = 0.0633 (mole fraction of F)

$\therefore$   ${\mathrm{K}}_{\mathrm{x}} = \frac{{\mathrm{X}}_{\mathrm{F}}^2}{{\mathrm{X}}_{{\mathrm{F}}_2}}=\frac{{\left(0.0663\right)}^2}{0.9337} = 4.708\times {10}^{-3}$