$2{\mathrm{NH}}_3\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)$

If degree of dissociation of ammonia at equilibrium is 0.7 then observed molecular weight of reaction mixture at equilibrium :

The equilibrium reaction is,

$2{\mathrm{NH}}_3\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)$

$\alpha \mathit{=}\frac{{\mathit{M}}_{\mathit{t}}\mathit{-}{\mathit{M}}_{\mathit{0}}}{{\mathit{M}}_{\mathit{0}}}$

Where,

$\mathrm{\alpha }$= degree of dissociation
${\mathrm{M}}_{\mathrm{t}}$ = theoretical molecular mass
${\mathrm{M}}_0$ = observed molecular mass.

$1+(2-1)(0.7)=\frac{17}{{\mathrm{M}}_{\mathrm{obs}}}$

${\mathrm{M}}_{\mathrm{obs}}=10$