Answer: The energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid into its vapor state without change in its temperature.
Given : The enthalpy of vaporization of water is 40.65 kJ/mol.
n = number of moles = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{12.5g}{18g/mole}=0.694mole[/tex]
Thus 1 mole of water requires heat = 40.65 kJ
0.694 moles of water requires heat = [tex]\frac{40.65}{1}\times 0.694=28.2kJ[/tex]
Thus the energy required to vaporize 12.5 g of liquid water is 28.2 kJ
