Question

The molar heat of vaporization of water is 40.7kj/mol. How much heat must be absorbed to convert 50.0 grams of liquid water at 100°C to steam at 100°C?

Answer 1

You have to use the formula q=nΔH(vap)
q=the heat absorbed or released
n=the number of moles of the sample (in this case 2.78mol (50g/18g/mol))
ΔH(vap)=the heat of vaporization for the sample (in this case 40.7kJ/mol)

When you plug everything in you should find that 113.05kJ is needed to vaporize 50g of water. (the temperature does not change during a phase change since all of the heat absorbed is being used to break the intermolecular forces and none of it is being used to speed up the molecules).

I hope this helps. Let me know if anything is unclear.