The enthalpy of vaporization of liquid water is 40.65 kJ/mol. Calculate the energy required to vaporize 12.5 g of liquid water.

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 =

Thus 1 mole of water requires heat = 40.65 kJ

0.694 moles of water requires heat =

Thus the energy required to vaporize 12.5 g of liquid water is 28.2 kJ

there is none because earth is stationary and flat.