Explanation:
First, for 1.85 mol Co the enthalpy will be calculated as follows.
Enthalpy of vaporization, [tex]\Delta H_{vap}[/tex] = [tex]1.85 mol \times \frac{389.1 kJ}{mol}[/tex]
= 719.8 kJ
Now, we will convert the temperature into Kelvin as follows.
(3097 + 273) K
= 3370 K
Therefore, entropy of vaporization will be calculated as follows.
[tex]\Delta H_{vap} = T \times \Delta S_{vap}[/tex]
[tex]\Delta S_{vap} = \frac{719.8 \times 10^{3} J}{3370 K}[/tex]
= 213.6 J/K
Thus, we can conclude that [tex]\Delta S[/tex] for vaporization is 213.6 J/K.
