# The heat of fusion of water is 336 J/g. What is the molar heat of fusion (j/mol) of water?

May 22, 2017

$6050 \frac{J}{m o l}$, or $6.05 \frac{k J}{m o l}$

#### Explanation:

Use the fact that one mole of water is equal to $18.02$ grams of water:

$\left(\frac{336 J}{1 \cancel{g {H}_{2} O}}\right) \left(\frac{18.02 \cancel{g {H}_{2} O}}{1 m o l {H}_{2} O}\right) = 6050 \frac{J}{m o l}$

A more conventional unit for enthalpy of fusion, enthalpy of vaporization, etc. is kilojoules per mole, and is found by dividing the units in joules per mole by ${10}^{3}$:

$\left(\frac{6050 \cancel{J}}{1 m o l {H}_{2} O}\right) \left(\frac{1 k J}{{10}^{3} \cancel{J}}\right) = 6.05 \frac{k J}{m o l}$