Answer
See the answer below.
Work Step by Step
a) Performing a regression, it follows:
$ln\ P(mmHg)=-5311.93\ K/T(K)+18.53$
b,c) Solving for the temperature, we find:
$250\ mmHg: 135.25°C$
$650\ mmHg: 167.60°C$
$760\ mmHg: 173.40°C$, which is the normal boiling point.
d) From the Clausius-Clapeyron equation, it follows:
$\Delta H_v=-5311.93\ K×-0.008314\ kJ/mol.K=44.2\ kJ/mol$
