Question

The boiling point elevation constant for toluene is 3.32. The normal temperature of toluene is 110.7 then enthalpy of vapourization toluene will be

Answer 1

Hey buddy,

◆ Answer-
34 kJ/mol

◆ Explanation-
# Given-
Kb = 3.32 KKg/mol
Tb = 110.7 °C = 383.7 K
R = 8.314 Nm/K

# Solution-
Boiling point elevation constant is given by-
Kb = RTb^2 / 1000Evap
Evap = RTb^2 / 1000Kb
Evap = (8.314×383.7^2) / (1000×3.32)
Evap = 34 kJ/mol

Enthalpy of vaporization of toulene is 34 kJ/mol.

Hope this helps...