# Question 44dab

Jan 25, 2017

$\text{3.57 kJ}$

#### 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 ${\text{40.66 kJ mol}}^{- 1}$. This means that in order to boil $1$ mole of water at atmospheric pressure, you need to provide it with $\text{40.66 kJ}$ 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

$0.0877 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) * "40.66 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("3.57 kJ}}}}$

The answer is rounded to three sig figs.