Question

Question #44dab

Answer 1

`"3.57 kJ"`

Explanation:

A substance's heat of vaporization, sometimes called molar enthalpy of vaporization, essentially tells you how much heat is needed in order to boil `1` mole of that substance at its boiling point.

In your case, water is said to have a heat of vaporization equal to `"40.66 kJ mol"^(-1)`. This means that in order to boil `1` mole of water at atmospheric pressure, you need to provide it with `"40.66 kJ"` of heat.

Notice that the problem gives you grams of water, so start by converting this to moles. To do that, use the molar mass of water

`1.58 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.0877 moles H"_2"O"`

Now you can use the heat of vaporization as a conversion factor to get the amount of heat needed to boil your sample of water

`0.0877 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)("3.57 kJ")))`

The answer is rounded to three sig figs.