Question

Hi Im having some trouble calculating the heat of vaporization of avolatile liquid using my lab results. I am getting a very smallvalue as my answer which is why I believe it is wrong. I had tograph my data on excel involving lnPvap (y) and 1/T (K-) in orderto attain the straight line equation.

Data: lnPvap 1/T (K-)
2.88 3.14*10^-3
2.09 3.23*10^-3
1.77 3.30*10^-3
1.31 3.35*10^-3

The resulting straight line equation is : y=-0.503x + 3.27

So I calculate the heat of vaporization by multiplying the slope(-.503) by the R constant of the Clausius-Clapeyron equation(8.314). As A result I get 4.18 J/mol but then I convert it tokJ/mol and get 4.18*10^-3. This value seems too small to becorrect? Am I making a mistake anywhere?

If I do have the correct heat of vaporization, then my problemcomes in with the follow-up question, which asks me to calculatethe normal boiling point of the volatile liquid using the straightline equation. I substitute in values and end up with this:
y=-.503x+3.27
ln (1)=-0.503(1/T) + 3.27 (The value I use for Pvap is 1.00 atm dueto the fact that is asking for the normal boiling point)

The answer I get after calculating is .153 and I know that cannotbe correct.

Sorry for the long question but I am really stumped, I must bemaking a step somewhere but I cant seem to pinpoint at one point.Thank you in advance for any help.

(The unknown volatile liquid is one of the following compounds:ethanol, iso-propanol, cyclohexane)