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Melting and Boiling point of NaCl respectively are 1080K & 1600K. $\Delta S$ for stage -I & II in

$\begin{gathered} NaC{l_{(s)}}\xrightarrow{I}NaC{l_{(l)}}\xrightarrow{{II}}NaC{l_{(g)}} \\ \Delta {H_{fusion}} = 30KJ\,,\,\,\Delta {H_{vapourization}} = 160KJ \\ \end{gathered}$

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I = \frac{1}{{36}}(KJ/mol/K);II = \frac{1}{{10}}(KJ/mol/K)

I = \frac{1}{{36}}(KJ/mol/K);\,II = 10(KJ/mol/K)

I = 36(KJ/mol/K);\,II = \frac{1}{{10}}(KJ/mol/K)

answer is A.

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Detailed Solution

\begin{gathered} \Delta {S_I} = \Delta {S_{fusion}} = \frac{{\Delta {H_{fusion}}}}{{M.P}} = \frac{{30}}{{1080}} = \frac{1}{{36}}KJ/mole/K \\\\ \Delta {S_{II}} = \Delta {S_{vapourization}} = \frac{{\Delta {H_{vapourization}}}}{{B.P}} = \frac{{160}}{{1600}} = \frac{1}{{10}}KJ/mole/K \\ \end{gathered}

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