Considering the definition of enthalpy of vaporization, the correct answer is option c. the heat required to vaporize 15.8 g of CH₃OH at its boiling point 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 ∆H vap), also known as (latent) heat of vaporization or heat of evaporation.
In other words, 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 g / mole (that is, the amount of mass contained in one mole), the number of moles to be vaporized is:
[tex]15.8 gramsx\frac{1 mol}{32 grams}= 0.49375 moles[/tex]
Then, the heat required to vaporize 0.49375 moles of CH₃OH at its boiling point is calculated taking into account the enthalpy of vaporization by:
0.49375 moles× 38.0 kJ/mol= 18.76 kJ ≅ 18.8 kJ
Finally, the correct answer is option c. the heat required to vaporize 15.8 g of CH₃OH at its boiling point is 18.8 kJ.
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