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The heat of vaporization of water is #2260# #Jg^-1#. How do you calculate the molar heat of vaporization #(Jmol^-1)# of water?

Answer 1

The key thing needed is to know the molar mass of water: #18# #gmol^-1#. If each gram of water takes #2260# #J# to vaporise it, and a mole is #18# #g#, then each mole takes #18xx2260 = 40,680# #Jmol^-1# or #40.68# #kJmol^-1#.

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Answer 2

Cameron Andrews

To calculate the molar heat of vaporization (J/mol) of water, you need to know the heat of vaporization of water in joules per gram (J/g) and the molar mass of water (H2O), which is approximately 18.015 g/mol.

The formula to calculate the molar heat of vaporization is:

Molar heat of vaporization (J/mol) = Heat of vaporization (J/g) * Molar mass of water (g/mol)

Given that the heat of vaporization of water is 2260 J/g, the calculation would be:

Molar heat of vaporization = 2260 J/g * 18.015 g/mol = 40,717.9 J/mol

Therefore, the molar heat of vaporization of water is approximately 40,717.9 J/mol.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.