Answer
See the answer below.
Work Step by Step
From the Clausius-Clapeyron equation we obtain:
$ln\dfrac{P_2}{P_1}=-\dfrac{\Delta H_v}{R}×\left(\dfrac 1{T_2}-\dfrac 1{T_1}\right)$
$ln(29.70/14.70)=-44.0\ kJ/mol/0.008314\ kJ/mol.K×(1/T_2-1/373)$
Solving for the temperature gives:
$T=392\ K=119.5°C$
