Heat of vaporization question

This is the real problem:

What is the heat of vaporization (kJ/mol) if it takes 3,452 J of heat to completely vaporize 2.68 moles of the liquid at its boiling point?

since the heat of vaporization is KJ/mol
all you need to do is convert 3452 to kj which 3.452KJ and divide by 2.68mol
3.452KJ/2.68mol = 1.29KJ/mol

What is the heat of vaporization (kJ/mol) if it takes 3,452 J of heat to completely vaporize 2.68 moles of the liquid at its boiling point?

The units of the heat of vaporization indicate the process!

Heat of vaporization = Energy on kilojoules ÷ moles
Heat of vaporization = 3425 ÷ 2.68