Question

How many kj of heat are needed to completely vaporize 1.30 moles of h2o? the heat of vaporization for water at the boiling point is 40.6 kj/mole?

Answer 1

Explanation:

Heat of vaporization is defined as the amount of heat required to change one mole of a liquid into vapor state without any change in the temperature.

It is known that for 1 mole of water, latent heat of vaporization is 40.6 kJ/mol.

Therefore, heat of vaporization for 1.30 moles will be calculated as follows.

              $1.30 moles \times 40.6 kJ/mol$

               = 52.78 kJ

Thus, we can conclude that the 52.78 kJ of heat are needed to completely vaporize 1.30 moles of $H_{2}O$.

Answer 2

The Kj of heat that  are needed to completely vaporize 1.30  moles of H2O  if  the heat of vaporization  for water is 40.6 Kj/mole  is calculated as  below

Q(heat) = moles x heat  of vaporization)

=1.30 mol  x40.6 kj/mol= 52.78 Kj  is needed