${\mathrm{S}}^0$ of ${\mathrm{N}}_2,{\mathrm{H}}_2$ and ${\mathrm{NH}}_3$ respectively are $190,130\mathrm{\backslash }111\mathrm{ }\mathrm{J}/\mathrm{K}/\mathrm{mol}$, then ${\mathrm{\Delta S}}_{\text{sys }}$ for formation of $8.5\mathrm{ }\mathrm{g}{\mathrm{NH}}_3$ is $[\mathrm{in}\mathrm{J}/\mathrm{K}]$

For the formation of $1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}$ of ammonia,

$1/2{ \mathrm{N}}_2\left(\mathrm{g}\right)+3/2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{NH}}_3\left(\mathrm{g}\right).$

So,

$$\begin{gathered} {\mathrm{\Delta S}}_{\text{sys }}={\mathrm{\Delta S}}_{\text{product }}-{\mathrm{\Delta S}}_{\text{reactant }} \\ \mathrm{\Delta S}={\mathrm{\Delta S}}_{{\mathrm{NH}}_3}-\left(1/2{\mathrm{\Delta S}}_{{\mathrm{N}}_2}+3/2{\mathrm{\Delta S}}_{{\mathrm{H}}_2}\right) \\ \mathrm{\Delta S}=111-(1/2\times 190+3/2\times 130) \\ \mathrm{\Delta S}=-179 \mathrm{J}/\mathrm{K}/\mathrm{mol} \end{gathered}$$

Now, $8.5\mathrm{ }\mathrm{g}$ ammonia $=8.5/17=0.5$ moles of ammonia

Thus, for 1 mole of ammonia, $\mathrm{\Delta S}=-179 \mathrm{J}/\mathrm{K}/\mathrm{mol}$

Therefore, for $0.5$ moles of ammonia $\mathrm{\Delta S}=-(179\times 1/2)\mathrm{J}/\mathrm{K}/\mathrm{mol}$

$=89.5 \mathrm{J}/\mathrm{K}/\mathrm{mol}$

Hence, option (D) is correct.