Question

The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 2.87 g of water boils at atmospheric pressure?

Answer 1

`"6.48 kJ"`

Explanation:

The molar heat of vaporization, `DeltaH_"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"`.

`DeltaH_"vap" = color(blue)("40.66 kJ") color(white)(.)color(red)("mol"^(-1))`

You need `color(blue)("40.66 kJ")` of heat to boil `color(red)("1 mole")` of water at its normal boiling point.

Now, the first thing to do here is to convert the mass of water to moles by using its molar mass

`2.87 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.1593 moles H"_2"O"`

You can now use the molar heat of vaporization as a conversion factor to determine how much heat would be needed to boil `0.1593` moles of water at its boiling point

`0.1593 color(red)(cancel(color(black)("moles H"_2"O"))) * "40.66 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("6.48 kJ")))`

The answer is rounded to three sig figs, the number of sig figs you have for the mass of the sample.