Question

Suppose that 132.0 J of heat energy is required to completely vaporize 3.1 g of a pure liquid. If the molar mass of the liquid is 40.17 g/mol, what is the molar enthalpy of vaporization of the liquid?

Answer 1

If there are 132.0 J of energy per 3.1 grams of matter, then one mole of a mass of 40.17 grams accounts for:

$$Q_{molar} = {Q \over n}$$

$$n = {m \over M} = {3.1 \over 40.17} = 0.077 mol$$

$$Q_{molar} = {132.0 \over 0.077} = 1714.3 J/mol$$

If $Q = - \Delta H$, then:

$$\Delta H_{vaporization} = -1714.3 J/mol$$