Considering the definition of enthalpy of vaporization, the amount of heat that would be involved in condensing 15.8 g of CH₃OH is 18.8 kJ.
When a material changes phase from solid to liquid, or from liquid to gas, a certain amount of energy is involved in this phase change.
In case of phase change from liquid to gas, this amount of energy is known as enthalpy of vaporization (symbolized as ∆Hvap).
So, the enthalpy of vaporization is the amount of energy required to change a unit of mass (for example, moles or kg) of a substance from phase liquid to phase gas at constant temperature and pressure.
In this case, you know that enthalpy of vaporization is ∆Hvap = 38.0 kJ/mol.
On the other side, you want to vaporize 15.8 g of CH₃OH. Then, being the molar mass of the compound 32 [tex]\frac{g}{mole}[/tex] (that is, the amount of mass contained in one mole), the number of moles to be vaporized is:
[tex]15.8 gramsx\frac{1 mole}{32 grams} =0.49375 moles[/tex]
Then, the amount of heat that would be involved in condensing 15.8 g of CH₃OH is calculated taking into account the enthalpy of vaporization by:
0.49375 moles× 38.0 [tex]\frac{kJ}{mol}[/tex]= 18.76 kJ ≅ 18.8 kJ
Finally, the amount of heat that would be involved in condensing 15.8 g of CH₃OH is 18.8 kJ.
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