Explanation:
The given data is as follows.
Enthalpy of vaporization, [tex]\Delta H_{vap}[/tex] = 38.56 kJ/mol
Temperature (T) = [tex]78.4^{o}C[/tex]
= (78.4 + 273) K
= 351.4 K
Now, we will calculate the change in entropy using the formula as follows.
[tex]\Delta H_{vap} = T \Delta S_{vap}[/tex]
[tex]\Delta S_{vap} = \frac{\Delta H_{vap}}{T}[/tex]
= [tex]\frac{38.56 kJ/mol}{351.4 K}[/tex]
= 0.109 kJ/mol K
Thus, w ecan conclude that the entropy change for vaporization is 0.109 kJ/mol K.
