Answer
For methane,
$\Delta S_{fus}=10.34\,JK^{-1}mol^{-1}$
$\Delta S_{vap}=79.78\,JK^{-1}mol^{-1}$
For ethane,
$\Delta S_{fus}=31.64\,JK^{-1}mol^{-1}$
$\Delta S_{vap}=84.80\,JK^{-1}mol^{-1}$
For propane,
$\Delta S_{fus}=41.20\,JK^{-1}mol^{-1}$
$\Delta S_{vap}=87.12\,JK^{-1}mol^{-1}$
Work Step by Step
Recall that $\Delta S_{fus}=\frac{\Delta H_{fus}}{T_{m}}$ and
$\Delta S_{vap}=\frac{\Delta H_{vap}}{T_{b}}$
From the given data,
For methane,
$\Delta S_{fus}=\frac{0.9370\times1000\,J/mol}{(-182.5+273.15)K}=10.34\,JK^{-1}mol^{-1}$
$\Delta S_{vap}=\frac{8.907\times1000\,J/mol}{(-161.5+273.15)K}=79.78\,JK^{-1}mol^{-1}$
For ethane,
$\Delta S_{fus}=\frac{2.859\times1000\,J/mol}{(-182.8+273.15)K}=31.64\,JK^{-1}mol^{-1}$
$\Delta S_{vap}=\frac{15.65\times1000\,J/mol}{(-88.6+273.15)K}=84.80\,JK^{-1}mol^{-1}$
For propane,
$\Delta S_{fus}=\frac{3.525\times1000\,J/mol}{(-187.6+273.15)K}=41.20\,JK^{-1}mol^{-1}$
$\Delta S_{vap}=\frac{20.13\times1000\,J/mol}{(-42.1+273.15)K}=87.12\,JK^{-1}mol^{-1}$
