For cobalt, Co, the heat of vaporization at its normal boiling point of 3097 °C is 389.1 kJ/mol. The entropy change when 1.85 moles of liquid Co vaporizes at 3097 °C, 1 atm is J/K.

Explanation:

First, for 1.85 mol Co the enthalpy will be calculated as follows.

   Enthalpy of vaporization, =

                                 = 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.

                      = 213.6 J/K

Thus, we can conclude that for vaporization is 213.6 J/K.

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