CALCULATE the enthalpy of vapourisation per mole for ethanol given delta s =109.8J/KM AND THE BOILING POINT OF ETHANOL 78.5 CELSIUS

We know that,

Entropy of vaporization, 

Δ_vapS = Δ_vapH/ T_b where, Δ_vapH = enthalpy of vaporization and T_b= boiling point (temperature) in Kelvin 

Here, Δ_vapS = 109.8J/Kmol, and boiling point T_b = 273 + 78.5 =351.5 K

So, enthalpy of  vaporization of ethanol per mole is ,  Δ_vapH = Δ_vapS x T_b = 109.8 x 351.5 = 38594.7  J/mol