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.

it is important to conserve water because there is very limited amount of freshwater here on earth and if we are not smart about how we use it we will only be runnning out of our supply faster than neccessarry.