Answer
$67.5\%$
Work Step by Step
The stoichiometric and actual reactions can be written as
Stoichiometric: $\mathrm{Na} \Leftrightarrow \mathrm{Na}^{+}+\mathrm{e}^{-}$(thus $v_{\mathrm{Na}}=1, v_{\mathrm{Na}^{+}}=1$ and $v_{\mathrm{e}^{-}}=1$ )
Actual: $\quad \mathrm{Na} \longrightarrow \underbrace{x \mathrm{Na}}_{\text {react. }}+\underbrace{y \mathrm{Na}^{+}+y \mathrm{e}^{-}}_{\text {products }}$
Na balance: $\quad 1=x+y$ or $y=1-x$
Total number of moles: $\quad N_{\text {total }}=x+2 y=2-x$
The equilibrium constant relation becomes,
Substituting, $$
0.668=\frac{(1-x)^2}{x}\left(\frac{0.8}{2-x}\right)
$$ Solving for $x$, $$
x=0.3254
$$ Thus the fraction of $\mathrm{Na}$ which dissociates into $\mathrm{Na}^{+}$and $\mathrm{e}^*$ is
$$ 1-0.3254=0.6746 \text { or } \mathbf{6 7 . 5} \%
$$
