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

The key thing needed is to know the molar mass of water: $18$ $g m o {l}^{-} 1$. If each gram of water takes $2260$ $J$ to vaporise it, and a mole is $18$ $g$, then each mole takes $18 \times 2260 = 40 , 680$ $J m o {l}^{-} 1$ or $40.68$ $k J m o {l}^{-} 1$.