The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol. How much heat would be absorbed or released upon freezing 25.6 grams of methanol?

5.04 kJ

Explanation:

We are given the enthalpy of fusion of methanol as 3.16 kJ per mol of methanol. Thus, what we need is to determine is the number of mol in 25.4 g of methanol and multiply by this enthalpy.

mol methanol = mass / MW CH₃OH = 25.6 mol / 16.04 g/mol = 1.60 mol

ΔH = 1.60 mol x 3.16 kJ/mol = 5.04 kJ