Question

Suppose that 181.1 J of heat energy is required to completely vaporize 3.9 g of a pure liquid. If the molar mass of the liquid is 128.34 g/mol, what is the molar enthalpy of vaporization of the liquid?

Answer 1

Moles evaporated$=\frac{3.9}{128.34}$

Molar enthalpy of evaporation$=\frac{128.34}{3.9}×\frac{181.1}{1000}$

$=5.9596kJ/mol$