The following information is given for ethanol C2H5OH at 1 a

The following information is given for ethanol, C₂H₅OH, at 1 atm:

At a pressure of 1 atm, _____ kJ of heat are needed to vaporize a 39.8 g sample of liquid ethanol at its normal boiling point of 78.4 °C.

boiling point = 78.4 °C	Hvap(78.4 °C) = 38.6 kJ/mol
melting point = -115 °C	Hfus(-115 °C) = 5.02 kJ/mol
specific heat liquid = 2.46 J/g°C

Solution

Molar mass of C2H5OH,
MM = 2*MM(C) + 6*MM(H) + 1*MM(O)
= 2*12.01 + 6*1.008 + 1*16.0
= 46.068 g/mol


mass(C2H5OH)= 39.8 g

use:
number of mol of C2H5OH,
n = mass of C2H5OH/molar mass of C2H5OH
=(39.8 g)/(46.068 g/mol)
= 0.8639 mol

Hvap = 38.6 KJ/mol

use:
Q = number of mol*Hvap
= 0.8639 mol * 38.6 KJ/mol
= 33.3 KJ
Answer: 33.3 KJ