J.R. S. answered • 05/05/21

Ph.D. in Biochemistry--University Professor--Chemistry Tutor

You asked the same question 2 days ago, and I provided an answer. If you don't think it is correct, or have a problem with the explanation, please leave a comment so that I or someone else can correct it. It doesn't do anyone any good to simply re-post the question. Here is my answer again.

Clausius Clapeyron equation:

ln (P2/P1) = -∆Hvap/R (1/T2 - 1/T1)

P1 = 92.0 torr

P2 = 206 torr

R = 8.314 J/mol-K = 0.008314 kJ/Kmol

T1 = 23.0ºC + 273 = 296K

T2 = 45ºC + 273 = 318K

∆Hvap = ?

ln (206/92) = - ∆Hvap/0.008314 (1/318 - 1/296)

0.806 = ∆Hvap / 0.008314 (0.00314 - 0.00338) = ∆Hvap/0.008314 (-0.00024)

∆Hvap = 27.9 kJ/mol

(be sure to check my math)

To find the normal boiling point of the liquid, repeat the above calculations substituting the following:

P2 = 760 torr

Solve for T2