The enthalpy of vaporization of liquid isopropyl alcohol is 45 kJ/mol. Calculate the energy required to vaporize 39.5 g of this compound.

29.625 kJ.

Explanation:

Enthalpy of vaporization, ∆Hvap which is also known as the latent heat of vaporization, is the amount of energy required to be added to a liquid substance to change the phase of a quantity of that substance into a gas.

Molar mass of Isopropyl alcohol (C3H80) = (12*3) + (1*8) + 16

= 60 g/mol.

Number of moles = mass*molar mass

= 39.5*60

= 0.658 moles.

Heat required = n * molar enthalpy of vapourisation.

∆Hvap = 45 kJ/mol.

= 0.658*45

= 29.625 kJ.