How much heat is required to vaporize 80.6 g of water at 100 C? The heat of vaporization of water at 100 C is 40.7 kJ/mole.?

1 Answer
Jul 17, 2016

The heat being added to a substance during a phase change does not raise the Temperature, instead, it's being used to break the bonds in the solution.

Explanation:

So, to answer the question, you must convert grams of water to moles.

80.6 g * (1 mol)/(18 g) = x " moles" of H_2O

Now, multiply moles by the heat of vaporization, 40.7 kJ/mole and you should get your answer.

This is the amount of heat applied to water to completely break the bonds between water molecules so it can completely vaporize.