Sel Test 7.9A How do we calculate H(vapor)?

[Jennifer Aguayo 1I]

A sample of benzene, C6H6, was heated to 80 degrees Celsius, its normal boiling point. The heating was continued until 15.4 kJ had been supplied; as a result, 39.1g of boiling benzene was vaporized. What is the enthalpy of vaporization of benzene at its boiling point?
For this problem I understand that we use the equation:
Delta H= H(vapor) - H(liquid)
But I am not sure how to start it, can someone help me please.

Thank you!

[Niharika Reddy 1D]

I'm not entirely sure if this is how you do it, but this is how I approached the problem and got the same answer as the textbook.

ΔH_vap is the energy as heat required to vaporize a mole of benzene (units: kJ/mol).
We are given the amount of energy required to vaporize 39.1 g of benzene (15.4 kJ). Converting the mass of benzene to moles using the molar mass of benzene (78.11 g/mol) gives 0.500 mol benzene. Dividing 15.4 kJ by the amount (moles) of benzene that that much heat vaporized gives the heat of vaporization of benzene in kJ/mol:

ΔH_vap = 15.4 kJ / 0.500 mol = 30.8 kJ/mol

[Jennifer Aguayo 1I]

Okay that makes sense.
So for this type of question we do not have to do use Delta H=Hm(vapor)- Hm(Liquid)? How would we know when to use this equation?

[Justin Le 2I]

For this equation:
Delta H= H(vapor) - H(liquid)
Are you talking about the delta H of the entire heating process?

[Jennifer Aguayo 1I]

Hi Justin,
Sorry for the late reply. No I was asking how to solve Delta Hvap .