Answer
See the answer below.
Work Step by Step
a) $80.1°C$, the temperature in which the vapor pressure is equal to atmosphere's pressure, which is 760 mmHg.
b) Performing a regression, we obtain:
$log_{10} P(mmHg)=-1747.4/T(K)+7.835$
Or reversing it, it follows:
$T(°C)=-1747.4/(log_{10} P(mmHg)-7.835)-273$
For 250mmHg: $48.4°C$
For 600mmHg: $72.5°C$
c) From the regression: $\Delta H_v=-1747.4\ K×0.008314\ kJ/mol.K/log_{10} e=33.45\ kJ/mol$
The $log_{10} e$ factor is necessary to convert the base 10 logarithmic to a natural log one.
