Sapling Week 5/6 #7

[Jaclyn Schwartz 1I]

Hi! I know this is an old question from a couple of weeks ago but i was wondering if anyone could help me solve for #7? Im confused about the heat of vaportization and what equation to use. Any help would be appreciated!

Three liquid samples of known masses are heated to their boiling points with the use of a heater rated at 525.0 W . Once the boiling points of each sample are reached, the samples are heated for an additional 6.88 min , which results in the vaporization of some of each sample. After 6.88 min , the samples are cooled and the masses of the remaining liquids are determined. The process is performed at constant pressure. The results are recorded in the table.

Liquid Boiling point (°C) Initial mass (g) Final mass (g)
CH3OH 64.5 503.0 321.26
C4H10 −1.00 675.6 92.44
C3H6O 56.2 735.8 316.25
Calculate the molar enthalpy of vaporization, ΔHvap , and the molar entropy of vaporization, ΔSvap , for each sample. Assume that all of the heat from the heater goes into the sample.

ΔHvap of CH3OH:

ΔSvap of CH3OH:

ΔHvap of C4H10:

ΔSvap of C4H10:

ΔHvap of C3H6O:

ΔSvap of C3H6O:

[Hazelle Gunawan 3G]

Multiply 525 W by 6.88min * 60 sec/min to get total heat in J. Then convert each of your substances to moles and calculate how many moles of each substance were vaporized. Take this number and divide it by the number you got for heat in J and that's your molar enthalpy of vaporization. Then divide that number by temperature to get molar entropy of vaporization. Hope that helps!

[FrancescaHawkins2H]

Everything stated in the post above sounds correct for the most part, but I think you would also need to convert J to kJ when finding heat. Additionally, you would divide heat by moles, rather than the other way around.