Question

the heat of vaporization of water is 40.66 kj/mol. how much heat is absorbed when 3.48 g of water boils at atmospheric pressure?

Answer 1

Answer:6.48 kJ

Explanation:The molar heat of vaporization,

Δ

H

vap

, sometimes called the molar enthalpy of vaporization, tells you how much energy is needed in order to boil

1

mole of a given substance at its boiling point.

In water's case, a molar heat of vaporization of

40.66 kJ mol

−

1

means that you need to supply

40.66 kJ

of heat in order to boil

1

mole of water at its normal boiling point, i.e. at

100

∘

C

.