Answer:
180.56 Kilo joules of energy is removed in the form of heat when 1.00 kg of freon-11 is evaporated.
Explanation:
Molar mass of freon-11 = 137.35 g/mol
Enthalpy of vaporization of freon-11= [tex]\Delta H_{vap}=24.8 kJ/mol[/tex]
Mass of freon-11 evaporated = 1.00 kg = 1000 g
Moles of freon-11 evaporated = [tex]\frac{1000 g}{137.35 g/mol}=7.2807 mol[/tex]
Energy in the form of heat removed when 1.00 kg of freon-11 gets evaporated:
[tex]7.28067 mol\times \Delta H_{vap}=7.2807 mol\times 24.8 kJ/mol[/tex]
[tex]=180.56 kJ[/tex]