Question

How much heat is needed to completely vapourize 15.2 g of water (H₂O) at its boiling point?

Answer 1

Molar mass of water = 18.02

Moles of $H2O = \frac{15.2}{18.02}=0.844 moles$

For water at 100∘C, the enthalpy change of vaporization is equal to

$∆Hvap=40.66kJ/mol$

$Heat = 0.844×40.66=34.32kJ$