If the vapor pressure of ethanol at 34.7degree c is 100
mmhg, and its heat of vaporization is 38.6 kj/mol. at what
temperaturewould ethanol boil at sea level?

we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:

 ln (P2/P1) = -)

where

P1 and P2 are the vapour pressures at temperatures  T1 and T2

Δ vapH  = the enthalpy of vaporization of the ETHANOL

R  = the Universal Gas Constant

In this problem,

P 1 = 100 mmHg

;  T 1 = 34.7 °C = 307.07 K

P 2 = 760mmHg

T 2 =T⁻²=?

Δ vap H = 38.6 kJ/mol

R = 0.008314 kJ⋅K -1 mol -1

ln ( 760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1 /0.008314 )

0.0004368=(0.00325 - T⁻²)  

T⁻²=0.002813

T² = 355.47K