Melting and boiling point of $\mathrm{NaCl}$ respectively are $1080 \mathrm{K} \mathrm{and}1600 \mathrm{K}. \mathrm{\Delta S}$ for stage -I and II

$$\begin{gathered} {\mathrm{NaCl}}_{\left(\mathrm{s}\right)}\stackrel{\mathrm{I}}{⟶}{\mathrm{NaCl}}_{\left(\mathrm{l}\right)}\stackrel{\mathrm{II}}{⟶}{\mathrm{NaCl}}_{\left(\mathrm{g}\right)} \\ {\mathrm{\Delta H}}_{\mathrm{fus}}=30\mathrm{KJ} {\mathrm{\Delta H}}_{\mathrm{vap}}=160\mathrm{KJ} \end{gathered}$$

$∆\mathrm{S}\left(\mathrm{I}\right)$                                  $∆\mathrm{S}\left(\mathrm{II}\right)$

when solid changes to liquid or liquid changes to gas in equilibrium condition, the Gibbs free energy $∆\mathrm{G}=0$.

A quantity that measures the maximum amount of work done when the temperature and pressure are constant is known as Gibbs free energy. This happens in a thermodynamic system.

The formula is $\text{ΔG=ΔH-TΔS}$

$$\begin{gathered} =>0=\mathrm{\Delta H}-\mathrm{T\Delta S} \\ =>\mathrm{\Delta H}=\mathrm{T\Delta S} \\ =>\mathrm{\Delta S}=\mathrm{\Delta H}/\mathrm{T} \end{gathered}$$

The entropy in stage I and II can be calculated using this formula.

For stage-1

$$\begin{gathered} {\mathrm{\Delta S}}_{\mathrm{l}}={\mathrm{\Delta H}}_{\mathrm{fus}}/\mathrm{T} \\ =>{\mathrm{\Delta S}}_{\mathrm{l}}=\frac{30}{1080}=1/36(\mathrm{KJ}/\mathrm{mol}/\mathrm{K}) \end{gathered}$$

For stage II,

$$\begin{gathered} {\mathrm{\Delta S}}_{\mathrm{II}}={\mathrm{\Delta H}}_{\mathrm{vap}}/\mathrm{T} \\ =>{\mathrm{\Delta S}}_{\mathrm{II}}=\frac{160}{1600}=1/10(\mathrm{KJ}/\mathrm{mol}/\mathrm{K}) \end{gathered}$$

Therefore, option $\left(\mathrm{A}\right)$ is correct.