How much heat is required to vaporize 166g of water at it's boiling point, if water's heat of vaporization is equal to 40.7kJ/mol?
kJ

375.3KJ

Explanation:

The following data were obtained from the question:

Mass of water = 166g

Heat of Vaporisation (ΔHv) = 40.7kJ/mol

Heat (Q) =..?

Next, we shall determine the number of mole in 166g of water. This is illustrated below:

Mass of H2O = 166g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Number of mole = Mass/Molar Mass

Number of mole of H2O = 166/18

Number of mole of H2O = 9.22 moles.

Now, we can obtain the heat required to vaporise the water as shown below:

Q = n·ΔHv

Q = 9.22 mol x 40.7kJ/mol

Q = 375.3KJ

Therefore, the heat required to vaporise the water is 375.3KJ.

pv=nrt

explanation: