Question

How much heat is required to vaporize 17.5 g of water at 100 ∘C? The heat of vaporization of water at 100 ∘C is 40.7 kJ/mole.

Answer 1

The heat of vaporization of water at 100^∘C is 40.7 kJ/mole

Amount of water = 17.5 g

Molar mass of water = 18 g/mol

Total number of moles of water = 17.5 g / 18 gmol^-1 = 0.972 mol.

So the amount of heat required to completely vaporize 17.5 g of water is = 0.972 mol x 40.7 kJ/mol = 39.56 kJ

So the amount of heat required to vaporize 17.5 g of water at 100^∘C is 39.56 kJ